## Fun with (Kalman) Filters Part I

Suppose we wish to estimate the mean of a sample drawn from a normal distribution. In the Bayesian approach, we know the prior distribution for the mean (it could be a non-informative prior) and then we update this with our observations to create the posterior, the latter giving us improved information about the distribution of the mean. In symbols

$\displaystyle p(\theta \,\vert\, x) \propto p(x \,\vert\, \theta)p(\theta)$

Typically, the samples are chosen to be independent, and all of the data is used to perform the update but, given independence, there is no particular reason to do that, updates can performed one at a time and the result is the same; nor is the order of update important. Being a bit imprecise, we have

$\displaystyle p(z \,\vert\, x, y) = p(z, x, y)p(x, y) = p(z, x, y)p(x)p(y) = p((z \,\vert\, x) \,\vert\, y) = p((z \,\vert\, y) \,\vert\, x)$

The standard notation in Bayesian statistics is to denote the parameters of interest as $\theta \in \mathbb{R}^p$ and the observations as $x \in \mathbb{R}^n$. For reasons that will become apparent in later blog posts, let us change notation and label the parameters as $x$ and the observations as $y$.

Let us take a very simple example of a prior $X \sim {\cal{N}}(0, \sigma^2)$ where $\sigma^2$ is known and then sample from a normal distribution with mean $x$ and variance for the $i$-th sample $c_i^2$ where $c_i$ is known (normally we would not know the variance but adding this generality would only clutter the exposition unnecessarily).

$\displaystyle p(y_i \,\vert\, x) = \frac{1}{\sqrt{2\pi c_i^2}}\exp\bigg(\frac{(y_i - x)^2}{2c_i^2}\bigg)$

The likelihood is then

$\displaystyle p(\boldsymbol{y} \,\vert\, x) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi c_i^2}}\exp\bigg(\frac{(y_i - x)^2}{2c_i^2}\bigg)$

As we have already noted, instead of using this with the prior to calculate the posterior, we can update the prior with each observation separately. Suppose that we have obtained the posterior given $i - 1$ samples (we do not know this is normally distributed yet but we soon will):

$\displaystyle p(x \,\vert\, y_1,\ldots,y_{i-1}) = {\cal{N}}(\hat{x}_{i-1}, \hat{\sigma}^2_{i-1})$

Then we have

\displaystyle \begin{aligned} p(x \,\vert\, y_1,\ldots,y_{i}) &\propto p(y_i \,\vert\, x)p(x \,\vert\, y_1,\ldots,y_{i-1}) \\ &\propto \exp-\bigg(\frac{(y_i - x)^2}{2c_i^2}\bigg) \exp-\bigg(\frac{(x - \hat{x}_{i-1})^2}{2\hat{\sigma}_{i-1}^2}\bigg) \\ &\propto \exp-\Bigg(\frac{x^2}{c_i^2} - \frac{2xy_i}{c_i^2} + \frac{x^2}{\hat{\sigma}_{i-1}^2} - \frac{2x\hat{x}_{i-1}}{\hat{\sigma}_{i-1}^2}\Bigg) \\ &\propto \exp-\Bigg( x^2\Bigg(\frac{1}{c_i^2} + \frac{1}{\hat{\sigma}_{i-1}^2}\Bigg) - 2x\Bigg(\frac{y_i}{c_i^2} + \frac{\hat{x}_{i-1}}{\hat{\sigma}_{i-1}^2}\Bigg)\Bigg) \end{aligned}

Writing

$\displaystyle \frac{1}{\hat{\sigma}_{i}^2} \triangleq \frac{1}{c_i^2} + \frac{1}{\hat{\sigma}_{i-1}^2}$

and then completing the square we also obtain

$\displaystyle \frac{\hat{x}_{i}}{\hat{\sigma}_{i}^2} \triangleq \frac{y_i}{c_i^2} + \frac{\hat{x}_{i-1}}{\hat{\sigma}_{i-1}^2}$

# More Formally

Now let’s be a bit more formal about conditional probability and use the notation of $\sigma$-algebras to define ${\cal{F}}_i = \sigma\{Y_1,\ldots, Y_i\}$ and $M_i \triangleq \mathbb{E}(X \,\vert\, {\cal{F}}_i)$ where $Y_i = X + \epsilon_i$, $X$ is as before and $\epsilon_i \sim {\cal{N}}(0, c_k^2)$. We have previously calculated that $M_i = \hat{x}_i$ and that ${\cal{E}}((X - M_i)^2 \,\vert\, Y_1, \ldots Y_i) = \hat{\sigma}_{i}^2$ and the tower law for conditional probabilities then allows us to conclude ${\cal{E}}((X - M_i)^2) = \hat{\sigma}_{i}^2$. By Jensen’s inequality, we have

$\displaystyle {\cal{E}}(M_i^2) = {\cal{E}}({\cal{E}}(X \,\vert\, {\cal{F}}_i)^2)) \leq {\cal{E}}({\cal{E}}(X^2 \,\vert\, {\cal{F}}_i))) = {\cal{E}}(X^2) = \sigma^2$

Hence $M$ is bounded in $L^2$ and therefore converges in $L^2$ and almost surely to $M_\infty \triangleq {\cal{E}}(X \,\vert\, {\cal{F}}_\infty)$. The noteworthy point is that if $M_\infty = X$ if and only if $\hat{\sigma}_i$ converges to 0. Explicitly we have

$\displaystyle \frac{1}{\hat{\sigma}_i^2} = \frac{1}{\sigma^2} + \sum_{k=1}^i\frac{1}{c_k^2}$

which explains why we took the observations to have varying and known variances. You can read more in Williams’ book (Williams 1991).

# A Quick Check

We have reformulated our estimation problem as a very simple version of the celebrated Kalman filter. Of course, there are much more interesting applications of this but for now let us try “tracking” the sample from the random variable.

> {-# OPTIONS_GHC -Wall                     #-}
> {-# OPTIONS_GHC -fno-warn-type-defaults   #-}
> {-# OPTIONS_GHC -fno-warn-unused-do-bind  #-}
> {-# OPTIONS_GHC -fno-warn-missing-methods #-}
> {-# OPTIONS_GHC -fno-warn-orphans         #-}

> module FunWithKalmanPart1 (
>     obs
>   , nObs
>   , estimates
>   , uppers
>   , lowers
>   ) where
>
> import Data.Random.Source.PureMT
> import Data.Random

> var, cSquared :: Double
> var       = 1.0
> cSquared  = 1.0
>
> nObs :: Int
> nObs = 100

> createObs :: RVar (Double, [Double])
> createObs = do
>   x <- rvar (Normal 0.0 var)
>   ys <- replicateM nObs $rvar (Normal x cSquared) > return (x, ys) > > obs :: (Double, [Double]) > obs = evalState (sample createObs) (pureMT 2) > > updateEstimate :: (Double, Double) -> (Double, Double) -> (Double, Double) > updateEstimate (xHatPrev, varPrev) (y, cSquared) = (xHatNew, varNew) > where > varNew = recip (recip varPrev + recip cSquared) > xHatNew = varNew * (y / cSquared + xHatPrev / varPrev) > > estimates :: [(Double, Double)] > estimates = scanl updateEstimate (y, cSquared) (zip ys (repeat cSquared)) > where > y = head$ snd obs
>     ys = tail $snd obs > > uppers :: [Double] > uppers = map (\(x, y) -> x + 3 * (sqrt y)) estimates > > lowers :: [Double] > lowers = map (\(x, y) -> x - 3 * (sqrt y)) estimates  # Bibliography Williams, David. 1991. Probability with Martingales. Cambridge University Press. Posted in Uncategorized | 2 Comments ## Gibbs Sampling in Haskell This is really intended as a draft chapter for our book. Given the diverse natures of the intended intended audiences, it is probably a bit light on explanation of the Haskell (use of monad transformers) for those with a background in numerical methods. It is hoped that the explanation of the mathematics is adequate for those with a background in Haskell but not necessarily in numerical methods. As always, any feedback is gratefully accepted. # Introduction Imagine an insect, a grasshopper, trapped on the face of a clock which wants to visit each hour an equal number of times. However, there is a snag: it can only see the value of the hour it is on and the value of the hours immediately anti-clockwise and immediately clockwise. For example, if it is standing on 5 then it can see the 5, the 4, and the 6 but no others. It can adopt the following strategy: toss a fair coin and move anti-clockwise for a head and move clockwise for a tail. Intuition tells us that over a large set of moves the grasshopper will visit each hour (approximately) the same number of times. Can we confirm our intuition somehow? Suppose that the strategy has worked and the grasshopper is now to be found with equal probability on any hour. Then at the last jump, the grasshopper must either have been at the hour before the one it is now on or it must have been at the hour after the one it is now on. Let us denote the probability that the grasshopper is on hour $n$ by $\pi(n)$ and the (conditional) probability that the grasshopper jumps to state $n$ given it was in state $m$ by $p(n \, |\, m)$. Then we have $\displaystyle \pi'(n) = p(n \, |\, n - 1)\pi(n - 1) + p(n \, |\, n + 1)\pi(n + 1)$ Substituting in where $N$ is a normalising constant (12 in this case) we obtain $\displaystyle \pi'(n) = \frac{1}{2}\frac{1}{N} + \frac{1}{2}\frac{1}{N} = \frac{1}{N}$ This tells us that the required distribution is a fixed point of the grasshopper’s strategy. But does the strategy actually converge to the fixed point? Let us perform an experiment. First we import some modules from hmatrix. > {-# LANGUAGE FlexibleContexts #-}  > module Chapter1 where  > import Data.Packed.Matrix > import Numeric.LinearAlgebra.Algorithms > import Numeric.Container  > import Data.Random > import Control.Monad.State > import qualified Control.Monad.Writer as W > import qualified Control.Monad.Loops as ML > import Data.Random.Source.PureMT  Let us use a clock with 5 hours to make the matrices sufficiently small to fit on one page. Here is the strategy encoded as a matrix. For example the first row says jump to position 1 with probablity 0.5 or jump to position 5 with probability 0.5. > eqProbsMat :: Matrix Double > eqProbsMat = (5 >< 5) > [ 0.0, 0.5, 0.0, 0.0, 0.5 > , 0.5, 0.0, 0.5, 0.0, 0.0 > , 0.0, 0.5, 0.0, 0.5, 0.0 > , 0.0, 0.0, 0.5, 0.0, 0.5 > , 0.5, 0.0, 0.0, 0.5, 0.0 > ]  We suppose the grasshopper starts at 1 o’clock. > startOnOne :: Matrix Double > startOnOne = ((1 >< 5) [1.0, 0.0, 0.0, 0.0, 0.0])  If we allow the grasshopper to hop 1000 times then we see that it is equally likely to be found on any hour hand with a 20% probability. ghci> eqProbsMat (5><5) [ 0.0, 0.5, 0.0, 0.0, 0.5 , 0.5, 0.0, 0.5, 0.0, 0.0 , 0.0, 0.5, 0.0, 0.5, 0.0 , 0.0, 0.0, 0.5, 0.0, 0.5 , 0.5, 0.0, 0.0, 0.5, 0.0 ] ghci> take 1$ drop 1000  iterate (<> eqProbsMat) startOnOne [(1><5) [ 0.20000000000000007, 0.2, 0.20000000000000004, 0.20000000000000004, 0.2 ]]  In this particular case, the strategy does indeed converge. Now suppose the grasshopper wants to visit each hour in proportion the value of the number on the hour. Lacking pen and paper (and indeed opposable thumbs), it decides to adopt the following strategy: toss a fair coin as in the previous strategy but only move if the number is larger than the one it is standing on; if, on the other hand, the number is smaller then choose a number at random from between 0 and 1 and move if this value is smaller than the ratio of the proposed hour and the hour on which it is standing otherwise stay put. For example, if the grasshopper is standing on 5 and gets a tail then it will move to 6 but if it gets a head then four fifths of the time it will move to 4 but one fifth of the time it will stay where it is. Suppose that the strategy has worked (it is not clear that is has) and the grasshopper is now to be found at 12 o’clock 12 times as often as at 1 o’clock, at 11 o’clock 11 times as often as at 1 o’clock, etc. Then at the last jump, the grasshopper must either have been at the hour before the one it is now on, the hour after the one it is now on or the same hour it is now on. Let us denote the probability that the grasshopper is on hour $n$ by $\pi(n)$. $\displaystyle \pi'(n) = p(n \, |\, n - 1)\pi(n - 1) + p(n \, |\, n)\pi(n) + p(n \, |\, n + 1)\pi(n + 1)$ Substituting in at 4 say \displaystyle \begin{aligned} \pi'(4) &= \frac{1}{2}\pi(3) + \frac{1}{2}\frac{1}{4}\pi(4) + \frac{1}{2}\frac{4}{5}\pi(5) \\ &= \frac{1}{2}\bigg(\frac{3}{N} + \frac{1}{4}\frac{4}{N} + \frac{4}{5}\frac{5}{N}\bigg) \\ &= \frac{1}{N}\frac{8}{2} \\ &= \frac{4}{N} \\ &= \pi(4) \end{aligned} The reader can check that this relationship holds for all other hours. This tells us that the required distribution is a fixed point of the grasshopper’s strategy. But does this strategy actually converge to the fixed point? Again, let us use a clock with 5 hours to make the matrices sufficiently small to fit on one page. Here is the strategy encoded as a matrix. For example the first row says jump to position 1 with probablity 0.5 or jump to position 5 with probability 0.5. > incProbsMat :: Matrix Double > incProbsMat = scale 0.5
>   (5 >< 5)
>     [ 0.0,         1.0,     0.0,        0.0, 1.0
>     , 1.0/2.0, 1.0/2.0,     1.0,        0.0, 0.0
>     , 0.0,     2.0/3.0, 1.0/3.0,        1.0, 0.0
>     , 0.0,         0.0, 3.0/4.0,    1.0/4.0, 1.0
>     , 1.0/5.0,     0.0,     0.0,    4.0/5.0, 1.0/5.0 + 4.0/5.0
>     ]


We suppose the grasshopper starts at 1 o’clock.

If we allow the grasshopper to hop 1000 times then we see that it is equally likely to be found on any hour hand $n$ with a probability of $n$ times the probability of being found on 1.

ghci> incProbsMat
(5><5)
[  0.0,                0.5,                 0.0,   0.0, 0.5
, 0.25,               0.25,                 0.5,   0.0, 0.0
,  0.0, 0.3333333333333333, 0.16666666666666666,   0.5, 0.0
,  0.0,                0.0,               0.375, 0.125, 0.5
,  0.1,                0.0,                 0.0,   0.4, 0.5 ]

ghci> take 1 $drop 1000$ iterate (<> incProbsMat) startOnOne
[(1><5)
[ 6.666666666666665e-2, 0.1333333333333333, 0.19999999999999996, 0.2666666666666666, 0.33333333333333326 ]]


In this particular case, the strategy does indeed converge.

Surprisingly, this strategy produces the desired result and is known as the Metropolis Algorithm. What the grasshopper has done is to construct a (discrete) Markov Process which has a limiting distribution (the stationary distribution) with the desired feature: sampling from this process will result in each hour being sampled in proportion to its value.

# Markov Chain Theory

Let us examine what is happening in a bit more detail.

The grasshopper has started with a very simple Markov Chain: one which jumps clockwise or anti-clockwise with equal probability and then modified it. But what is a Markov Chain?

A time homogeneous Markov chain is a countable sequence of random variables
$X_0, X_1, \ldots$ such that

$\displaystyle \mathbb{P} (X_{n+1} = j \,|\, X_0 = i_0, X_1 = i_1, \dots X_n = i) = \mathbb{P} (X_{n+1} = j \,|\, X_n = i)$

We sometimes say that a Markov Chain is discrete time stochastic process with the above property.

So the very simple Markov Chain can be described by

$\displaystyle q(i, j) = \begin{cases} \mathbb{P} (X_{n+1} = j \,|\, X_n = i) = \frac{1}{2} & \text{if } j = i + 1 \mod N \\ \mathbb{P} (X_{n+1} = j \,|\, X_n = i) = \frac{1}{2} & \text{if } j = i - 1 \mod N \\ \mathbb{P} (X_{n+1} = j \,|\, X_n = i) = 0 & \text{otherwise } \end{cases}$

The grasshopper knows that $\pi(i) = i/N$ so it can calculate $\pi(j)/\pi(i) = j/i$ without knowing $N$. This is important because now, without knowing $N$, the grasshopper can evaluate

$\displaystyle p(i, j) = \begin{cases} q(i,j)\bigg[\frac{\pi(j) q(j,i)}{\pi(i) q(i,j)} \land 1 \bigg] & \text{if } j \ne i \\ 1 - \sum_{k : k \ne i} q(i,k) \bigg[\frac{\pi(k) q(k,i)}{\pi(i) q(i,k)} \land 1 \bigg] & \text{if } j = i \end{cases}$

where $\land$ takes the maximum of its arguments. Simplifying the above by substituing in the grasshopper’s probabilities and noting that $j = i \pm 1 \mod N$ is somewhat obscure way of saying jump clockwise or anti-clockwise we obtain

$\displaystyle q(i, j) = \begin{cases} \frac{1}{2} (\frac{j}{i} \land 1) & \text{if } j \text{ is 1 step clockwise} \\ \frac{1}{2} (\frac{j}{i} \land 1) & \text{if } j \text{ is 1 step anti-clockwise} \\ 1 - \frac{1}{2}(\frac{j^c}{i} \land 1) - \frac{1}{2}(\frac{j^a}{i} \land 1) & \text{if } j = i \text{ and } j^c \text{ is one step clockwise and } j^a \text{ is one step anti-clockwise} \\ 0 & \text{otherwise} \end{cases}$

## The Ergodic Theorem

In most studies of Markov chains, one is interested in whether a chain has a stationary distribution. What we wish to do is take a distribution and create a chain with this distribution as its stationary distribution. We will still need to show that our chain does indeed have the correct stationary distribution and we state the relevant theorem somewhat informally and with no proof.

### Theorem

An irreducible, aperiodic and positive recurrent Markov chain has a unique stationary distribution.

Roughly speaking

• Irreducible means it is possible to get from any state to any other state.

• Aperiodic means that returning to a state having started at that state occurs at irregular times.

• Positive recurrent means that the first time to hit a state is finite (for every state and more pedantically except on sets of null measure).

Note that the last condition is required when the state space is infinite – see Skrikant‘s lecture notes for an example and also for a more formal definition of the theorem and its proof.

### Algorithm

Let $\pi$ be a probability distribution on the state space $\Omega$ with $\pi(i) > 0$ for all $i$ and let $(Q, \pi_0)$ be an ergodic Markov chain on $\Omega$ with transition probabilities $q(i,j) > 0$ (the latter condition is slightly stronger than it need be but we will not need fully general conditions).

Create a new (ergodic) Markov chain with transition probabilities

$\displaystyle p_{ij} = \begin{cases} q(i,j)\bigg[\frac{\pi(j) q(j,i)}{\pi(i) q(i,j)} \land 1 \bigg] & \text{if } j \ne i \\ 1 - \sum_{k : k \ne i} q(i,k) \bigg[\frac{\pi(j) q(j,i)}{\pi(i) q(i,j)} \land 1 \bigg] & \text{if } j = i \end{cases}$

where $\land$ takes the maximum of its arguments.

Calculate the value of interest on the state space e.g. the total magnetization for each step produced by this new chain.

Repeat a sufficiently large number of times and take the average. This gives the estimate of the value of interest.

### Convergence

Let us first note that the Markov chain produced by this algorithm almost trivially satisfies the detailed balance condition, for example,

\displaystyle \begin{aligned} \pi(i) q(i,j)\bigg[\frac{\pi(j) q(j, i)}{\pi(i)q(i,j)} \land 1\bigg] &= \pi(i)q(i,j) \land \pi(j)q(j,i) \\ &= \pi(j)q(j,i)\bigg[\frac{\pi(i) q(i, j)}{\pi(j)q(j,i)} \land 1\bigg] \end{aligned}

Secondly since we have specified that $(Q, \pi_0)$ is ergodic then clearly $(P, \pi_0)$ is also ergodic (all the transition probabilities are $> 0$).

So we know the algorithm will converge to the unique distribution we specified to provide estimates of values of interest.

# Gibbs Sampling

## Random Scan

For simplicity let us consider a model with two parameters and that we sample from either parameter with equal probability. In this sampler, We update the parameters in a single step.

$\displaystyle \begin{cases} \text{Sample } \theta_1^{(i+1)} \sim \pi(\theta_1 \,\big|\, \theta_2^{(i)}) & \text{with probability } \frac{1}{2} \\ \text{Sample } \theta_2^{(i+1)} \sim \pi(\theta_2 \,\big|\, \theta_1^{(i)}) & \text{with probability } \frac{1}{2} \end{cases}$

The transition density kernel is then given by

$\displaystyle q\big(\boldsymbol{\theta}^{(i+1)}, \boldsymbol{\theta}^{(i)}\big) = \frac{1}{2}\pi(\theta_1^{(i+1)} \,\big|\, \theta_2^{(i)})\delta({\theta_2^{(i)},\theta_2^{(i+1)}}) + \frac{1}{2}\pi(\theta_2^{(i+1)} \,\big|\, \theta_1^{(i)})\delta({\theta_1^{(i)},\theta_1^{(i+1)}})$

where $\delta$ is the Dirac delta function.

### Detailed balance

This sampling scheme satisifies the detailed balance condition. We have

\displaystyle \begin{aligned} \pi(\theta_1, \theta_2) \bigg[ \frac{1}{2}\pi(\theta_1' \,\big|\, \theta_2)\delta({\theta_2,\theta_2'}) + \frac{1}{2}\pi(\theta_2' \,\big|\, \theta_1)\delta({\theta_1,\theta_1'})\bigg] &= \\ \frac{1}{2}\bigg[\pi(\theta_1, \theta_2) \pi(\theta_1' \,\big|\, \theta_2)\delta({\theta_2,\theta_2'}) + \pi(\theta_1, \theta_2) \pi(\theta_2' \,\big|\, \theta_1)\delta({\theta_1,\theta_1'})\bigg] &= \\ \frac{1}{2}\bigg[\pi(\theta_1, \theta_2') \pi(\theta_1' \,\big|\, \theta_2)\delta({\theta_2,\theta_2'}) + \pi(\theta_1', \theta_2) \pi(\theta_2' \,\big|\, \theta_1)\delta({\theta_1,\theta_1'})\bigg] &= \\ \frac{1}{2}\bigg[ \pi(\theta_2')\pi(\theta_1 \,\big|\, \theta_2') \frac{1}{2}\pi(\theta_1' \,\big|\, \theta_2)\delta({\theta_2,\theta_2'}) + \pi(\theta_1')\pi(\theta_2 \,\big|\, \theta_1') \pi(\theta_2' \,\big|\, \theta_1)\delta({\theta_1,\theta_1'}) \bigg] &= \\ \frac{1}{2}\bigg[ \pi(\theta_1', \theta_2')\pi(\theta_1 \,\big|\, \theta_2') \delta({\theta_2',\theta_2}) + \pi(\theta_1', \theta_2')\pi(\theta_2 \,\big|\, \theta_1') \delta({\theta_1',\theta_1}) \bigg] &= \\ \pi(\theta_1', \theta_2')\bigg[ \frac{1}{2}\pi(\theta_1 \,\big|\, \theta_2') \delta({\theta_2',\theta_2}) + \frac{1}{2}\pi(\theta_2 \,\big|\, \theta_1') \delta({\theta_1',\theta_1}) \bigg] & \end{aligned}

In other words

$\displaystyle \pi\big({\boldsymbol{\theta}}\big)q\big(\boldsymbol{\theta}', \boldsymbol{\theta}\big) = \pi\big({\boldsymbol{\theta'}}\big)q\big(\boldsymbol{\theta}, \boldsymbol{\theta}'\big)$

Hand waving slightly, we can see that this scheme satisfies the premises of the ergodic theorem and so we can conclude that there is a unique stationary distribution and $\pi$ must be that distribution.

# Systematic Scan

Most references on Gibbs sampling do not describe the random scan but instead something called a systematic scan.

Again for simplicity let us consider a model with two parameters. In this sampler, we update the parameters in two steps.

\displaystyle \begin{aligned} \text{Sample } \theta_1^{(i+1)} & \sim & \pi(\theta_1 \,\big|\, \theta_2^{(i)}) \\ \text{Sample } \theta_2^{(i+1)} & \sim & \pi(\theta_2 \,\big|\, \theta_1^{(i+1)}) \end{aligned}

We observe that this is not time-homegeneous; at each step the transition matrix flips between the two transition matrices given by the individual steps. Thus although, as we show below, each individual transtion satisifies the detailed balance condition, we cannot apply the ergodic theorem as it only applies to time-homogeneous processes.

The transition density kernel is then given by

$\displaystyle q\big(\boldsymbol{\theta}^{(i)}, \boldsymbol{\theta}^{(i+1)}\big) = q_1\big(\boldsymbol{\theta}^{(i)}, \tilde{\boldsymbol{\theta}}\big) q_2\big(\tilde{\boldsymbol{\theta}}, \boldsymbol{\theta}^{(i+1)}\big)$

where $\tilde{\boldsymbol{\theta}} = (\theta_1^{(i+1)}, \theta_2^{(i)})^\top$.

Thus

$\displaystyle q\big(\boldsymbol{\theta}, \boldsymbol{\theta}'\big) = \pi(\theta_1' \,\big|\, \theta_2) \pi(\theta_2' \,\big|\, \theta_1')$

### Detailed balance

Suppose that we have two states $\boldsymbol{\theta} = (\theta_1, \theta_2)^\top$ and $\boldsymbol{\theta}' = (\theta_1', \theta_2')^\top$ and that $\theta_2 \neq \theta_2'$. Then $q_1\big(\boldsymbol{\theta}, \boldsymbol{\theta}'\big) = 0$. Trivially we have

$\displaystyle \pi\big({\boldsymbol{\theta}}\big)q_1\big(\boldsymbol{\theta}, \boldsymbol{\theta}'\big) = \pi\big({\boldsymbol{\theta'}}\big)q_1\big(\boldsymbol{\theta}', \boldsymbol{\theta}\big)$

Now suppose that $\theta_2 = \theta_2'$

\displaystyle \begin{aligned} \pi(\theta_1, \theta_2)q_1((\theta_1, \theta_2), (\theta_1', \theta_2)) & = \pi(\theta_1, \theta_2)\pi(\theta_1' \,\big|\, \theta_2) \\ & = \pi(\theta_1 \,\big|\, \theta_2)\pi(\theta_1', \theta_2) \\ & = \pi(\theta_1 \,\big|\, \theta_2')\pi(\theta_1', \theta_2') \\ & = \pi(\theta_1', \theta_2')q_1((\theta_1', \theta_2), (\theta_1, \theta_2)) \end{aligned}

So again we have

$\displaystyle \pi\big({\boldsymbol{\theta}}\big)q_1\big(\boldsymbol{\theta}, \boldsymbol{\theta}'\big) = \pi\big({\boldsymbol{\theta'}}\big)q_1\big(\boldsymbol{\theta}', \boldsymbol{\theta}\big)$

Similarly we can show

$\displaystyle \pi\big({\boldsymbol{\theta}}\big)q_2\big(\boldsymbol{\theta}, \boldsymbol{\theta}'\big) = \pi\big({\boldsymbol{\theta'}}\big)q_2\big(\boldsymbol{\theta}', \boldsymbol{\theta}\big)$

But note that

\displaystyle \begin{aligned} \pi(\theta_1, \theta_2) q_1((\theta_1, \theta_2), (\theta_1', \theta_2)) q_2((\theta_1', \theta_2), (\theta_1', \theta_2')) & = \pi(\theta_1, \theta_2) \pi(\theta_1' \,\big|\, \theta_2) \pi(\theta_2' \,\big|\, \theta_1') \\ & = \pi(\theta_1', \theta_2) \pi(\theta_1 \,\big|\, \theta_2) \pi(\theta_2' \,\big|\, \theta_1') \\ & = \pi(\theta_1' \,\big|\, \theta_2) \pi(\theta_1 \,\big|\, \theta_2) \pi(\theta_2', \theta_1') \end{aligned}

whereas

\displaystyle \begin{aligned} \pi(\theta_1', \theta_2') q_1((\theta_1', \theta_2'), (\theta_1, \theta_2')) q_2((\theta_1, \theta_2'), (\theta_1, \theta_2)) & = \pi(\theta_1', \theta_2') \pi(\theta_1 \,\big|\, \theta_2') \pi(\theta_2 \,\big|\, \theta_1) \\ & = \pi(\theta_1, \theta_2') \pi(\theta_1' \,\big|\, \theta_2') \pi(\theta_2 \,\big|\, \theta_1) \\ & = \pi(\theta_2' \,\big|\, \theta_1) \pi(\theta_1' \,\big|\, \theta_2') \pi(\theta_2, \theta_1) \end{aligned}

and these are not necessarily equal.

So the detailed balance equation is not satisfied, another sign that we cannot appeal to the ergodic theorem.

# An Example: The Bivariate Normal

Let us demonstrate the Gibbs sampler with a distribution which we actually know: the bivariate normal.

$\displaystyle \begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix} \bigg| y \sim N \begin{bmatrix} \begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix} & \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix} \end{bmatrix}$

The conditional distributions are easily calculated to be

\displaystyle \begin{aligned} \theta_1 \,\vert\, \theta_2, y &\sim {\cal{N}}(y_1 + \rho(\theta_2 - y_2), 1 - \rho^2) \\ \theta_2 \,\vert\, \theta_1, y &\sim {\cal{N}}(y_2 + \rho(\theta_1 - y_1), 1 - \rho^2) \end{aligned}

Let’s take a correlation of 0.8, a data point of (0.0, 0.0) and start the chain at (2.5, 2.5).

> rho :: Double
> rho = 0.8
>
> y :: (Double, Double)
> y = (0.0, 0.0)
>
> y1, y2 :: Double
> y1 = fst y
> y2 = snd y
>
> initTheta :: (Double, Double)
> initTheta = (2.5, 2.5)


We pre-calculate the variance needed for the sampler.

> var :: Double
> var = 1.0 - rho^2


In Haskell and in the random-fu package, sampling from probability distributions is implemented as a monad. We sample from the relevant normal distributions and keep the trajectory using a writer monad.

> gibbsSampler :: Double -> RVarT (W.Writer [(Double,Double)]) Double
> gibbsSampler oldTheta2 = do
>   newTheta1 <- rvarT (Normal (y1 + rho * (oldTheta2 - y2)) var)
>   lift $W.tell [(newTheta1, oldTheta2)] > newTheta2 <- rvarT (Normal (y2 + rho * (newTheta1 - y1)) var) > lift$ W.tell [(newTheta1, newTheta2)]
>   return $newTheta2  It is common to allow the chain to “burn in” so as to “forget” its starting position. We arbitrarily burn in for 10,000 steps. > burnIn :: Int > burnIn = 10000  We sample repeatedly from the sampler using the monadic form of iterate. Running the monadic stack is slightly noisy but nonetheless straightforward. We use mersenne-random-pure64 (albeit indirectly via random-source) as our source of entropy. > runMCMC :: Int -> [(Double, Double)] > runMCMC n = > take n$
>   drop burnIn $> snd$
>   W.runWriter (evalStateT (sample (ML.iterateM_ gibbsSampler (snd initTheta))) (pureMT 2))


We can look at the trajectory of our sampler for various run lengths.

For bigger sample sizes, plotting the distribution sampled re-assures us that we are indeed sampling from a bivariate normal distribution as the theory predicted.

# Applications to Bayesian Statistics

Some of what is here and here excluding JAGS and STAN (after all this is a book about Haskell).

Applications to Physics

# Applications to Physics

Most of what is here.

## A Monoidal Category Example

I have never felt entirely comfortable with Haskell’s arrows and skimming the literature for their categorical basis didn’t reveal anything as straightforward as monads or applicatives. It did however lead me to start thinking about monoidal categories and since I always want an example, I thought I would write up Hilbert spaces.

Let $H_1$ and $H_2$ be Hilbert spaces then as vector spaces we can form the tensor product $H_1 \otimes H_2$. The tensor product can be defined as the free vector space on $H_1$ and $H_2$ as sets (that is purely formal sums of $(u,v)$) modulo a relation $\sim$ defined by

\displaystyle \begin{aligned} (u_1,v) + (u_2,v) \sim (u_1 + u_2,v) \\ (u,v_1) + (u,v_2) \sim (u,v_1 + v_2) \\ c(u,v) \sim (cu,v) \\ c(u,v) \sim (u,cv) \end{aligned}

Slightly overloading notation, we can define an inner product on the tensored space by

$\displaystyle \langle u_1 \otimes v_1, u_2 \otimes v_2\rangle = \langle u_1, v_1 \rangle \langle u_2, v_2\rangle$

Of course this might not be complete so we define the tensor product on Hilbert spaces to be the completion of this inner product.

For Hilbert spaces to form a monoidal category, we take the arrows (in the categorical sense) to be linear continuous maps and the bifunctor to be the tensor product. We also need an identity object $I$ which we take to be $\mathbb{R}$ considered as a Hilbert space. We should check the coherence conditions but the associativity of the tensor product and the fact that our Hilbert spaces are over the $\mathbb{R}$ make this straightforward.

Now for some slightly interesting properties of this category.

• The tensor product is not the product in the categorical sense. If $\{u_i\}$ and $\{v_i\}$ are (orthonormal) bases for $H_1$ and $H_2$ then $\{u_i \otimes v_j\}$ is a (orthonormal) basis for $H_1 \otimes H_2$. Thus a linear combination of basis vectors in the tensor product cannot be expressed as the tensor of basis vectors in the component spaces.

• There is no diagonal arrow $\Delta : X \rightarrow X \otimes X$. Suppose there were such a diagonal then for arbitrary $\lambda$ we would have $\Delta(\lambda u) = (\lambda u) \otimes (\lambda u) = \lambda^2 (u \otimes u)$ and since $\Delta$ must be linear this is not possible.

Presumably the latter is equivalent to the statement in quantum mechanics of “no cloning”.

# Introduction

I have seen Hölder’s inequality and Minkowski’s inequality proved in several ways but this seems the most perspicuous (to me at any rate).

# Young’s Inequality

If $a, b \ge 0$ and $p,q \ge 1$ such that

$\displaystyle \frac{1}{p} + \frac{1}{q} = 1$

then

$\displaystyle ab \le \frac{a^p}{p} + \frac{b^q}{q}$

A $p$ and $q$ satisfying the premise are known as conjugate indices.

Proof

Since $\log$ is convex we have

$\displaystyle t\log{x} + (1 - t)\log{y} \le \log{(tx + (1 - t)y)}$

Substituting in appropriate values gives

$\displaystyle \frac{1}{p}\log{a^p} + \frac{1}{q}\log{b^q} \le \log{\bigg(\frac{a^p}{p} + \frac{b^q}{q}\bigg)}$

or

$\displaystyle \log{a} + \log{b} \le \log{\bigg(\frac{a^p}{p} + \frac{b^q}{q}\bigg)}$

Now take exponents.

$\blacksquare$

# Hölders’s Inequality

Let $p$ and $q$ be conjugate indices with $1 < p < \infty$ and let $f \in L^p(\Omega)$ and $g \in L^q(\Omega)$ then $fg \in L^1(\Omega)$ and

$\displaystyle \|fg\|_{L^1} \le \|f\|_{L^p}\|g\|_{L^q}$

Proof

By Young’s inequality

$\displaystyle \int_\Omega \frac{|f(x)|}{\|f\|_{L^p}} \frac{|g(x)|}{\|g\|_{L^q}} \le \int_\Omega \frac{1}{p}\frac{|f(x)|^p}{\|f\|_{L^p}^p} + \frac{1}{q}\frac{|g(x)|^q}{\|g\|_{L^q}^q} = \frac{1}{p} + \frac{1}{q} = 1$

$\blacksquare$

By applying a counting measure to $\Omega$ we also obtain

$\displaystyle \sum |x_i y_i| \le \big(\sum |x_i|^p\big)^{1/p} \big(\sum |y_i|^q\big)^{1/q}$

# Minkowski’s Inequality

$\displaystyle \|f + g\|_{L^p} \le \|f\|_{L^p} + \|g\|_{L^p}$

Proof

By Hölder’s inequality

$\displaystyle \int_\Omega |f + g|^p \le \int_\Omega |f||f + g|^{p-1} + \int_\Omega |g||f + g|^{p-1} \le \|f\|_{L^p}A + \|g\|_{L^p}A$

where

$\displaystyle A = \||f + g|^{p-1}\|_{L^q} = \big(\int_\Omega |f(x) + g(x)|^p\big)^{1/q}$

and $A$ is finite since $L^p$ is a vector space.

$\blacksquare$

# Introduction

It’s possible to Gibbs sampling in most languages and since I am doing some work in R and some work in Haskell, I thought I’d present a simple example in both languages: estimating the mean from a normal distribution with unknown mean and variance. Although one can do Gibbs sampling directly in R, it is more common to use a specialised language such as JAGS or STAN to do the actual sampling and do pre-processing and post-processing in R. This blog post presents implementations in native R, JAGS and STAN as well as Haskell.

## Preamble

> {-# OPTIONS_GHC -Wall                      #-}
> {-# OPTIONS_GHC -fno-warn-type-defaults    #-}
> {-# OPTIONS_GHC -fno-warn-unused-do-bind   #-}
> {-# OPTIONS_GHC -fno-warn-missing-methods  #-}
> {-# OPTIONS_GHC -fno-warn-orphans          #-}

> {-# LANGUAGE NoMonomorphismRestriction     #-}

> module Gibbs (
>     main
>   , m
>   , Moments(..)
>   ) where
>
> import qualified Data.Vector.Unboxed as V
> import qualified Control.Monad.Loops as ML
> import Data.Random.Source.PureMT
> import Data.Random
> import Data.Histogram ( asList )
> import Data.Histogram.Fill
> import Data.Histogram.Generic ( Histogram )
> import Data.List
> import qualified Control.Foldl as L
>
> import Diagrams.Backend.Cairo.CmdLine
>
> import LinRegAux
>
> import Diagrams.Backend.CmdLine
> import Diagrams.Prelude hiding ( sample, render )


The length of our chain and the burn-in.

> nrep, nb :: Int
> nb   = 5000
> nrep = 105000


Data generated from ${\cal{N}}(10.0, 5.0)$.

> xs :: [Double]
> xs = [
>     11.0765808082301
>   , 10.918739177542
>   , 15.4302462747137
>   , 10.1435649220266
>   , 15.2112705014697
>   , 10.441327659703
>   , 2.95784054883142
>   , 10.2761068139607
>   , 9.64347295100318
>   , 11.8043359297675
>   , 10.9419989262713
>   , 7.21905367667346
>   , 10.4339807638017
>   , 6.79485294803006
>   , 11.817248658832
>   , 6.6126710570584
>   , 12.6640920214508
>   , 8.36604701073303
>   , 12.6048485320333
>   , 8.43143879537592
>   ]


# A Bit of Theory

## Gibbs Sampling

For a multi-parameter situation, Gibbs sampling is a special case of Metropolis-Hastings in which the proposal distributions are the posterior conditional distributions.

Referring back to the explanation of the metropolis algorithm, let us describe the state by its parameters $i \triangleq \boldsymbol{\theta}^{(i)} \triangleq (\theta^{(i)}_1,\ldots, \theta^{(i)}_n)$ and the conditional posteriors by $\pi\big({\theta}_{k}^{(j)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(i)}\big)$ where ${\boldsymbol{\theta}}^{(i)}_{-k} = \big(\theta_1^{(i)},\ldots,\theta_{k-1}^{(i)},\theta_{k+1}^{(i)}\ldots\theta_n^{(i)}\big)$ then

\displaystyle \begin{aligned} \frac{\pi\big(\boldsymbol{\theta}^{(j)}\big)q\big(\boldsymbol{\theta}^{(j)}, \boldsymbol{\theta}^{(i)}\big)} {\pi(\boldsymbol{\theta}^{(i)})q(\boldsymbol{\theta}^{(i)}, \boldsymbol{\theta}^{(j)})} &= \frac{ \pi\big({\theta}_{k}^{(j)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(j)}\big)\pi\big({\boldsymbol{\theta}}_{-k}^{(j)}\big)\pi\big({\theta}_{k}^{(i)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(j)}\big) } { \pi\big({\theta}_{k}^{(i)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(i)}\big)\pi\big({\boldsymbol{\theta}}_{-k}^{(i)}\big)\pi\big({\theta}_{k}^{(j)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(i)}\big) } \\ &= \frac{ \pi\big({\theta}_{k}^{(j)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(j)}\big)\pi\big({\boldsymbol{\theta}}_{-k}^{(j)}\big)\pi\big({\theta}_{k}^{(i)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(j)}\big) } { \pi\big({\theta}_{k}^{(i)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(j)}\big)\pi\big({\boldsymbol{\theta}}_{-k}^{(j)}\big)\pi\big({\theta}_{k}^{(j)} \,\big|\, {\boldsymbol{\theta}}_{-k}^{(j)}\big) } \\ &= 1 \end{aligned}

where we have used the rules of conditional probability and the fact that $\boldsymbol{\theta}_i^{(-k)} = \boldsymbol{\theta}_j^{(-k)}$

Thus we always accept the proposed jump. Note that the chain is not in general reversible as the order in which the updates are done matters.

## Normal Distribution with Unknown Mean and Variance

It is fairly standard to use an improper prior

\displaystyle \begin{aligned} \pi(\mu, \tau) \propto \frac{1}{\tau} & & -\infty < \mu < \infty\, \textrm{and}\, 0 < \tau < \infty \end{aligned}

The likelihood is

$\displaystyle p(\boldsymbol{x}\,|\,\mu, \sigma) = \prod_{i=1}^n \bigg(\frac{1}{\sigma\sqrt{2\pi}}\bigg)\exp{\bigg( -\frac{(x_i - \mu)^2}{2\sigma^2}\bigg)}$

re-writing in terms of precision

$\displaystyle p(\boldsymbol{x}\,|\,\mu, \tau) \propto \prod_{i=1}^n \sqrt{\tau}\exp{\bigg( -\frac{\tau}{2}{(x_i - \mu)^2}\bigg)} = \tau^{n/2}\exp{\bigg( -\frac{\tau}{2}\sum_{i=1}^n{(x_i - \mu)^2}\bigg)}$

Thus the posterior is

$\displaystyle p(\mu, \tau \,|\, \boldsymbol{x}) \propto \tau^{n/2 - 1}\exp{\bigg( -\frac{\tau}{2}\sum_{i=1}^n{(x_i - \mu)^2}\bigg)}$

We can re-write the sum in terms of the sample mean $\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$ and variance $s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2$ using

\displaystyle \begin{aligned} \sum_{i=1}^n (x_i - \mu)^2 &= \sum_{i=1}^n (x_i - \bar{x} + \bar{x} - \mu)^2 \\ &= \sum_{i=1}^n (x_i - \bar{x})^2 - 2\sum_{i=1}^n (x_i - \bar{x})(\bar{x} - \mu) + \sum_{i=1}^n (\bar{x} - \mu)^2 \\ &= \sum_{i=1}^n (x_i - \bar{x})^2 - 2(\bar{x} - \mu)\sum_{i=1}^n (x_i - \bar{x}) + \sum_{i=1}^n (\bar{x} - \mu)^2 \\ &= (n - 1)s^2 + n(\bar{x} - \mu)^2 \end{aligned}

Thus the conditional posterior for $\mu$ is

\displaystyle \begin{aligned} p(\mu \,|\, \tau, \boldsymbol{x}) &\propto \exp{\bigg( -\frac{\tau}{2}\bigg(\nu s^2 + \sum_{i=1}^n{(\mu - \bar{x})^2}\bigg)\bigg)} \\ &\propto \exp{\bigg( -\frac{n\tau}{2}{(\mu - \bar{x})^2}\bigg)} \\ \end{aligned}

which we recognise as a normal distribution with mean of $\bar{x}$ and a variance of $(n\tau)^{-1}$.

The conditional posterior for $\tau$ is

\displaystyle \begin{aligned} p(\tau \,|\, , \mu, \boldsymbol{x}) &\propto \tau^{n/2 -1}\exp\bigg(-\tau\frac{1}{2}\sum_{i=1}^n{(x_i - \mu)^2}\bigg) \end{aligned}

which we recognise as a gamma distribution with a shape of $n/2$ and a scale of $\frac{1}{2}\sum_{i=1}^n{(x_i - \mu)^2}$

In this particular case, we can calculate the marginal posterior of $\mu$ analytically. Writing $z = \frac{\tau}{2}\sum_{i=1}^n{(x_i - \mu)^2}$ we have

\displaystyle \begin{aligned} p(\mu \,|\, \boldsymbol{x}) &= \int_0^\infty p(\mu, \tau \,|\, \boldsymbol{x}) \textrm{d}\tau \\ &\propto \int_0^\infty \tau^{n/2 - 1}\exp{\bigg( -\frac{\tau}{2}\sum_{i=1}^n{(x_i - \mu)^2}\bigg)} \textrm{d}\tau \\ &\propto \bigg( \sum_{i=1}^n{(x_i - \mu)^2} \bigg)^{-n/2} \int_0^\infty z^{n/2 - 1}\exp{-z}\textrm{d}\tau \\ &\propto \bigg( \sum_{i=1}^n{(x_i - \mu)^2} \bigg)^{-n/2} \\ \end{aligned}

Finally we can calculate

\displaystyle \begin{aligned} p(\mu \,|\, \boldsymbol{x}) &\propto \bigg( (n - 1)s^2 + n(\bar{x} - \mu)^2 \bigg)^{-n/2} \\ &\propto \bigg( 1 + \frac{n(\mu - \bar{x})^2}{(n - 1)s^2} \bigg)^{-n/2} \\ \end{aligned}

This is the non-standardized Student’s t-distribution $t_{n-1}(\bar{x}, s^2/n)$.

Alternatively the marginal posterior of $\mu$ is

$\displaystyle \frac{\mu - \bar{x}}{s/\sqrt{n}}\bigg|\, x \sim t_{n-1}$

where $t_{n-1}$ is the standard t distribution with $n - 1$ degrees of freedom.

Following up on a comment from a previous blog post, let us try using the foldl package to calculate the length, the sum and the sum of squares traversing the list only once. An improvement on creating your own strict record and using foldl’ but maybe it is not suitable for some methods e.g. calculating the skewness and kurtosis incrementally, see below.

> x2Sum, xSum, n :: Double
> (x2Sum, xSum, n) = L.fold stats xs
>   where
>     stats = (,,) <> > (L.premap (\x -> x * x) L.sum) <*> > L.sum <*> > L.genericLength  And re-writing the sample variance $s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2$ using \displaystyle \begin{aligned} \sum_{i=1}^n (x_i - \bar{x})^2 &= \sum_{i=1}^n (x_i^2 - 2x_i\bar{x} + \bar{x}^2) \\ &= \sum_{i=1}^n x_i^2 - 2\bar{x}\sum_{i=1}^n x_i + \sum_{i=1}^n \bar{x}^2 \\ &= \sum_{i=1}^n x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 \\ &= \sum_{i=1}^n x_i^2 - n\bar{x}^2 \\ \end{aligned} we can then calculate the sample mean and variance using the sums we have just calculated. > xBar, varX :: Double > xBar = xSum / n > varX = n * (m2Xs - xBar * xBar) / (n - 1) > where m2Xs = x2Sum / n  In random-fu, the Gamma distribution is specified by the rate paratmeter, $\beta$. > beta, initTau :: Double > beta = 0.5 * n * varX > initTau = evalState (sample (Gamma (n / 2) beta)) (pureMT 1)  Our sampler takes an old value of $\tau$ and creates new values of $\mu$ and $\tau$. > gibbsSampler :: MonadRandom m => Double -> m (Maybe ((Double, Double), Double)) > gibbsSampler oldTau = do > newMu <- sample (Normal xBar (recip (sqrt (n * oldTau)))) > let shape = 0.5 * n > scale = 0.5 * (x2Sum + n * newMu^2 - 2 * n * newMu * xBar) > newTau <- sample (Gamma shape (recip scale)) > return Just ((newMu, newTau), newTau)


From which we can create an infinite stream of samples.

> gibbsSamples :: [(Double, Double)]
> gibbsSamples = evalState (ML.unfoldrM gibbsSampler initTau) (pureMT 1)


As our chains might be very long, we calculate the mean, variance, skewness and kurtosis using an incremental method.

> data Moments = Moments { mN :: !Double
>                        , m1 :: !Double
>                        , m2 :: !Double
>                        , m3 :: !Double
>                        , m4 :: !Double
>                        }
>   deriving Show

> moments :: [Double] -> Moments
> moments xs = foldl' f (Moments 0.0 0.0 0.0 0.0 0.0) xs
>   where
>     f :: Moments -> Double -> Moments
>     f m x = Moments n' m1' m2' m3' m4'
>       where
>         n = mN m
>         n'  = n + 1
>         delta = x - (m1 m)
>         delta_n = delta / n'
>         delta_n2 = delta_n * delta_n
>         term1 = delta * delta_n * n
>         m1' = m1 m + delta_n
>         m4' = m4 m +
>               term1 * delta_n2 * (n'*n' - 3*n' + 3) +
>               6 * delta_n2 * m2 m - 4 * delta_n * m3 m
>         m3' = m3 m + term1 * delta_n * (n' - 2) - 3 * delta_n * m2 m
>         m2' = m2 m + term1


In order to examine the posterior, we create a histogram.

> numBins :: Int
> numBins = 400

> hb :: HBuilder Double (Data.Histogram.Generic.Histogram V.Vector BinD Double)
> hb = forceDouble -<< mkSimple (binD lower numBins upper)
>   where
>     lower = xBar - 2.0 * sqrt varX
>     upper = xBar + 2.0 * sqrt varX


And fill it with the specified number of samples preceeded by a burn-in.

> hist :: Histogram V.Vector BinD Double
> hist = fillBuilder hb (take (nrep - nb) $drop nb$ map fst gibbsSamples)


Now we can plot this.

And calculate the skewness and kurtosis.

> m :: Moments
> m = moments (take (nrep - nb) $drop nb$ map fst gibbsSamples)

ghci> import Gibbs
ghci> putStrLn $show$ (sqrt (mN m)) * (m3 m) / (m2 m)**1.5
8.733959917065126e-4

ghci> putStrLn $show$ (mN m) * (m4 m) / (m2 m)**2
3.451374739494607


We expect a skewness of 0 and a kurtosis of $3 + 6 / \nu - 4 = 3.4$ for $\nu = 19$. Not too bad.

# The Model in JAGS

JAGS is a mature, declarative, domain specific language for building Bayesian statistical models using Gibbs sampling.

Here is our model as expressed in JAGS. Somewhat terse.

model {
for (i in 1:N) {
x[i] ~ dnorm(mu, tau)
}
mu ~ dnorm(0, 1.0E-6)
tau <- pow(sigma, -2)
sigma ~ dunif(0, 1000)
}

To run it and examine its results, we wrap it up in some R

## Import the library that allows R to inter-work with jags.
library(rjags)

## Read the simulated data into a data frame.

jags <- jags.model('example1.bug',
data = list('x' = fn[,1], 'N' = 20),
n.chains = 4,

## Burnin for 10000 samples
update(jags, 10000);

mcmc_samples <- coda.samples(jags, variable.names=c("mu", "sigma"), n.iter=20000)

png(file="diagrams/jags.png",width=400,height=350)
plot(mcmc_samples)
dev.off()



And now we can look at the posterior for $\mu$.

# The Model in STAN

STAN is a domain specific language for building Bayesian statistical models similar to JAGS but newer and which allows variables to be re-assigned and so cannot really be described as declarative.

Here is our model as expressed in STAN. Again, somewhat terse.

data {
int<lower=0> N;
real x[N];
}

parameters {
real mu;
real<lower=0,upper=1000> sigma;
}
model{
x     ~ normal(mu, sigma);
mu    ~ normal(0, 1000);
}

Just as with JAGS, to run it and examine its results, we wrap it up in some R.

library(rstan)

## Read the simulated data into a data frame.

## Running the model
fit1 <- stan(file = 'Stan.stan',
data = list('x' = fn[,1], 'N' = 20),
pars=c("mu", "sigma"),
chains=3,
iter=30000,
warmup=10000)

png(file="diagrams/stan.png",width=400,height=350)
plot(fit1)
dev.off()

Again we can look at the posterior although we only seem to get medians and 80% intervals.

# PostAmble

Write the histogram produced by the Haskell code to a file.

> displayHeader :: FilePath -> Diagram B R2 -> IO ()
>   mainRender ( DiagramOpts (Just 900) (Just 700) fn
>              , DiagramLoopOpts False Nothing 0
>              )

> main :: IO ()
> main = do
>     (barDiag
>      (zip (map fst $asList hist) (map snd$ asList hist)))


# Introduction

The other speaker at the Machine Learning Meetup at which I gave my talk on automatic differentiation gave a very interesting talk on A/B testing. Apparently this is big business these days as attested by the fact I got 3 ads above the wikipedia entry when I googled for it.

It seems that people tend to test with small sample sizes and to do so very often, resulting in spurious results. Of course readers of XKCD will be well aware of some of the pitfalls.

I thought a Bayesian approach might circumvent some of the problems and set out to write a blog article only to discover that there was no Haskell library for sampling from Student’s t. Actually there was one but is currently an unreleased part of random-fu. So I set about fixing this shortfall.

I thought I had better run a few tests so I calculated the sampled mean, variance, skewness and kurtosis.

I wasn’t really giving this my full attention and as a result ran into a few problems with space. I thought these were worth sharing and that is what this blog post is about. Hopefully, I will have time soon to actually blog about the Bayesian equivalent of A/B testing.

## Preamble

> {-# OPTIONS_GHC -Wall                      #-}
> {-# OPTIONS_GHC -fno-warn-type-defaults    #-}
> {-# OPTIONS_GHC -fno-warn-unused-do-bind   #-}
> {-# OPTIONS_GHC -fno-warn-missing-methods  #-}
> {-# OPTIONS_GHC -fno-warn-orphans          #-}
>
> {-# LANGUAGE NoMonomorphismRestriction     #-}
>
> module StudentTest (
>     main
>   ) where
>
> import qualified Data.Vector.Unboxed as V
> import Data.Random.Source.PureMT
> import Data.Random
> import Data.Random.Distribution.T
> import Data.Histogram.Fill
> import Data.Histogram.Generic ( Histogram )
> import Data.List


# Space Analysis

Let’s create a reasonable number of samples as the higher moments converge quite slowly.

> nSamples :: Int
> nSamples = 1000000


An arbitrary seed for creating the samples.

> arbSeed :: Int
> arbSeed = 8


Student’s t only has one parameter, the number of degrees of freedom.

> nu :: Integer
> nu = 6


Now we can do our tests by calculating the sampled values.

> ts :: [Double]
> ts =
>   evalState (replicateM nSamples (sample (T nu)))
>             (pureMT $fromIntegral arbSeed)  > mean, variance, skewness, kurtosis :: Double > mean = (sum ts) / fromIntegral nSamples > variance = (sum (map (**2) ts)) / fromIntegral nSamples > skewness = (sum (map (**3) ts) / fromIntegral nSamples) / variance**1.5 > kurtosis = (sum (map (**4) ts) / fromIntegral nSamples) / variance**2  This works fine for small sample sizes but not for the number we have chosen. ./StudentTest +RTS -hc Stack space overflow: current size 8388608 bytes. Use +RTS -Ksize -RTS' to increase it. It seems a shame that the function in the Prelude has this behaviour but never mind let us ensure that we consume values strictly (they are being produced lazily). > mean' = (foldl' (+) 0 ts) / fromIntegral nSamples > variance' = (foldl' (+) 0 (map (**2) ts)) / fromIntegral nSamples > skewness' = (foldl' (+) 0 (map (**3) ts) / fromIntegral nSamples) / variance'**1.5 > kurtosis' = (foldl' (+) 0 (map (**4) ts) / fromIntegral nSamples) / variance'**2  We now have a space leak on the heap as using the ghc profiler below shows. What went wrong? If we only calculate the mean using foldl then all is well. Instead of 35M we only use 45K. Well that gives us a clue. The garbage collector cannot reclaim the samples as they are needed for other calculations. What we need to do is calculate the moments strictly altogether. Let’s create a strict record to do this. > data Moments = Moments { m1 :: !Double > , m2 :: !Double > , m3 :: !Double > , m4 :: !Double > } > deriving Show  And calculate the results strictly. > > m = foldl' (\m x -> Moments { m1 = m1 m + x > , m2 = m2 m + x**2 > , m3 = m3 m + x**3 > , m4 = m4 m + x**4 > }) (Moments 0.0 0.0 0.0 0.0) ts > > mean'' = m1 m / fromIntegral nSamples > variance'' = m2 m / fromIntegral nSamples > skewness'' = (m3 m / fromIntegral nSamples) / variance''**1.5 > kurtosis'' = (m4 m / fromIntegral nSamples) / variance''**2  Now we have what we want; the program runs in small constant space. > main :: IO () > main = do > putStrLn$ show mean''
>   putStrLn $show variance'' > putStrLn$ show skewness''
>   putStrLn show kurtosis''  Oh and the moments give the expected answers. ghci> mean'' 3.9298418844289093e-4 ghci> variance'' 1.4962681916693004 ghci> skewness'' 1.0113188204317015e-2 ghci> kurtosis'' 5.661776268997382  # Running the Code To run this you will need my version of random-fu. The code for this article is here. You will need to compile everything with profiling, something like ghc -O2 -main-is StudentTest StudentTest.lhs -prof -package-db=.cabal-sandbox/x86_64-osx-ghc-7.6.2-packages.conf.d Since you need all the packages to be built with profiling, you will probably want to build using a sandbox as above. The only slightly tricky aspect is building random-fu so it is in your sandbox. runghc Setup.lhs configure --enable-library-profiling --package-db=/HasBayes/.cabal-sandbox/x86_64-osx-ghc-7.6.2-packages.conf.d --libdir=/HasBayes/.cabal-sandbox/lib Posted in Haskell, Statistics | 5 Comments ## Bayesian Analysis: A Conjugate Prior and Markov Chain Monte Carlo # Introduction This is meant to be shorter blog post than normal with the expectation that the material will be developed further in future blog posts. A Bayesian will have a prior view of the distribution of some data and then based on data, update that view. Mostly the updated distribution, the posterior, will not be expressible as an analytic function and sampling via Markov Chain Monte Carlo (MCMC) is the only way to determine it. In some special cases, when the posterior is of the same family of distributions as the prior, then the posterior is available analytically and we call the posterior and prior conjugate. It turns out that the normal or Gaussian distribution is conjugate with respect to a normal likelihood distribution. This gives us the opportunity to compare MCMC against the analytic solution and give ourselves more confidence that MCMC really does deliver the goods. Some points of note: • Since we want to display the posterior (and the prior for that matter), for histograms we use the histogram-fill package. • Since we are using Monte Carlo we can use all the cores on our computer via one of Haskell’s parallelization mechanisms. ## Preamble > {-# OPTIONS_GHC -Wall #-} > {-# OPTIONS_GHC -fno-warn-name-shadowing #-} > {-# OPTIONS_GHC -fno-warn-type-defaults #-} > {-# OPTIONS_GHC -fno-warn-unused-do-bind #-} > {-# OPTIONS_GHC -fno-warn-missing-methods #-} > {-# OPTIONS_GHC -fno-warn-orphans #-}  > {-# LANGUAGE NoMonomorphismRestriction #-}  > module ConjMCMCSimple where > > import qualified Data.Vector.Unboxed as V > import Data.Random.Source.PureMT > import Data.Random > import Control.Monad.State > import Data.Histogram ( asList ) > import qualified Data.Histogram as H > import Data.Histogram.Fill > import Data.Histogram.Generic ( Histogram ) > import Data.List > import Control.Parallel.Strategies > > import Diagrams.Backend.Cairo.CmdLine > > import Diagrams.Backend.CmdLine > import Diagrams.Prelude hiding ( sample, render ) > > import LinRegAux  # A Simple Example ## Analytically Suppose the prior is $\mu \sim \cal{N}(\mu_0, \sigma_0)$, that is $\displaystyle \pi(\mu) \propto \exp{\bigg( -\frac{(\mu - \mu_0)^2}{2\sigma_0^2}\bigg)}$ Our data is IID normal, $x_i \sim \cal{N}(\mu, \sigma)$, where $\sigma$ is known, so the likelihood is $\displaystyle p(x\,|\,\mu, \sigma) \propto \prod_{i=1}^n \exp{\bigg( -\frac{(x_i - \mu)^2}{2\sigma^2}\bigg)}$ The assumption that $\sigma$ is known is unlikely but the point of this post is to demonstrate MCMC matching an analytic formula. This gives a posterior of \displaystyle \begin{aligned} p(\mu\,|\, \boldsymbol{x}) &\propto \exp{\bigg( -\frac{(\mu - \mu_0)^2}{2\sigma_0^2} - \frac{\sum_{i=1}^n(x_i - \mu)^2}{2\sigma^2}\bigg)} \\ &\propto \exp{\bigg[-\frac{1}{2}\bigg(\frac{\mu^2 \sigma^2 -2\sigma^2\mu\mu_0 - 2\sigma_0^2n\bar{x}\mu + \sigma_0^2 n\mu^2}{\sigma^2\sigma_0^2}\bigg)\bigg]} \\ &= \exp{\bigg[-\frac{1}{2}\bigg(\frac{ (n\sigma_0^2 + \sigma^2)\mu^2 - 2(\sigma^2\mu_0 - \sigma_0^2n\bar{x})\mu}{\sigma^2\sigma_0^2}\bigg)\bigg]} \\ &= \exp{\Bigg[-\frac{1}{2}\Bigg(\frac{ \mu^2 - 2\mu\frac{(\sigma^2\mu_0 - \sigma_0^2n\bar{x})}{(n\sigma_0^2 + \sigma^2)}}{\frac{\sigma^2\sigma_0^2}{(n\sigma_0^2 + \sigma^2)}}\Bigg)\Bigg]} \\ &\propto \exp{\Bigg[-\frac{1}{2}\Bigg(\frac{\big(\mu - \frac{(\sigma^2\mu_0 - \sigma_0^2n\bar{x})}{(n\sigma_0^2 + \sigma^2)}\big)^2}{\frac{\sigma^2\sigma_0^2}{(n\sigma_0^2 + \sigma^2)}}\Bigg)\Bigg]} \end{aligned} In other words $\displaystyle \mu\,|\, \boldsymbol{x} \sim {\cal{N}}\bigg(\frac{\sigma^2\mu_0 + n\sigma_0^2\bar{x}}{n\sigma_0^2 + \sigma^2}, \frac{\sigma^2\sigma_0^2}{n\sigma_0^2 + \sigma^2} \bigg)$ Writing $\displaystyle \sigma_n^2 = \frac{\sigma^2\sigma_0^2}{n\sigma_n^2 + \sigma^2}$ we get $\displaystyle \frac{1}{\sigma_n^2} = \frac{n}{\sigma^2} + \frac{1}{\sigma_0^2}$ Thus the precision (the inverse of the variance) of the posterior is the precision of the prior plus the precision of the data scaled by the number of observations. This gives a nice illustration of how Bayesian statistics improves our beliefs. Writing $\displaystyle \mu_n = \frac{\sigma^2\mu_0 + n\sigma_0^2\bar{x}}{n\sigma_0^2 + \sigma^2}$ and $\displaystyle \lambda = 1 / \sigma^2, \, \lambda_0 = 1 / \sigma_0^2, \, \lambda_n = 1 / \sigma_n^2$ we see that $\displaystyle \mu_n = \frac{n\bar{x}\lambda + \mu_0\lambda_0}{\lambda_n}$ Thus the mean of the posterior is a weight sum of the mean of the prior and the sample mean scaled by preciscion of the prior and the precision of the data itself scaled by the number of observations. Rather arbitrarily let us pick a prior mean of > mu0 :: Double > mu0 = 11.0  and express our uncertainty about it with a largish prior variance > sigma_0 :: Double > sigma_0 = 2.0  And also arbitrarily let us pick the know variance for the samples as > sigma :: Double > sigma = 1.0  We can sample from this in way that looks very similar to STAN and JAGS: > hierarchicalSample :: MonadRandom m => m Double > hierarchicalSample = do > mu <- sample (Normal mu0 sigma_0) > x <- sample (Normal mu sigma) > return x  and we didn’t need to write a new language for this. Again arbitrarily let us take > nSamples :: Int > nSamples = 10  and use > arbSeed :: Int > arbSeed = 2  And then actually generate the samples. > simpleXs :: [Double] > simpleXs = > evalState (replicateM nSamples hierarchicalSample) > (pureMT fromIntegral arbSeed)


Using the formulae we did above we can calculate the posterior

> mu_1, sigma1, simpleNumerator :: Double
> simpleNumerator = fromIntegral nSamples * sigma_0**2 + sigma**2
> mu_1 = (sigma**2 * mu0 + sigma_0**2 * sum simpleXs) / simpleNumerator
> sigma1 = sigma**2 * sigma_0**2 / simpleNumerator


and then compare it against the prior

The red posterior shows we are a lot more certain now we have some evidence.

## Via Markov Chain Monte Carlo

The theory behinde MCMC is described in a previous post. We need to generate some proposed steps for the chain. We sample from the normal distribution but we could have used e.g. the gamma.

> normalisedProposals :: Int -> Double -> Int -> [Double]
> normalisedProposals seed sigma nIters =
>   evalState (replicateM nIters (sample (Normal 0.0 sigma)))
>   (pureMT $fromIntegral seed)  We also need samples from the uniform distribution > acceptOrRejects :: Int -> Int -> [Double] > acceptOrRejects seed nIters = > evalState (replicateM nIters (sample stdUniform)) > (pureMT$ fromIntegral seed)


And now we can calculate the (un-normalised) prior, likelihood and posterior

> prior :: Double -> Double
> prior mu = exp (-(mu - mu0)**2 / (2 * sigma_0**2))
>
> likelihood :: Double -> [Double] -> Double
> likelihood mu xs = exp (-sum (map (\x -> (x - mu)**2 / (2 * sigma**2)) xs))
>
> posterior :: Double -> [Double] -> Double
> posterior mu xs = likelihood mu xs * prior mu


The Metropolis algorithm tells us that we always jump to a better place but only sometimes jump to a worse place. We count the number of acceptances as we go.

> acceptanceProb :: Double -> Double -> [Double] -> Double
> acceptanceProb mu mu' xs = min 1.0 ((posterior mu' xs) / (posterior mu xs))

> oneStep :: (Double, Int) -> (Double, Double) -> (Double, Int)
> oneStep (mu, nAccs) (proposedJump, acceptOrReject) =
>   if acceptOrReject < acceptanceProb mu (mu + proposedJump) simpleXs
>   then (mu + proposedJump, nAccs + 1)
>   else (mu, nAccs)


Now we can actually run our simulation. We set the number of jumps and a burn in but do not do any thinning.

> nIters, burnIn :: Int
> nIters = 300000
> burnIn = nIters div 10


Let us start our chain at

> startMu :: Double
> startMu = 10.0


and set the variance of the jumps to

> jumpVar :: Double
> jumpVar = 0.4

> test :: Int -> [(Double, Int)]
> test seed =
>   drop burnIn $> scanl oneStep (startMu, 0)$
>   zip (normalisedProposals seed jumpVar nIters)
>       (acceptOrRejects seed nIters)


We put the data into a histogram

> numBins :: Int
> numBins = 400

> hb :: HBuilder Double (Data.Histogram.Generic.Histogram V.Vector BinD Double)
> hb = forceDouble -<< mkSimple (binD lower numBins upper)
>   where
>     lower = startMu - 1.5*sigma_0
>     upper = startMu + 1.5*sigma_0
>
> hist :: Int -> Histogram V.Vector BinD Double
> hist seed = fillBuilder hb (map fst $test seed)  Not bad but a bit lumpy. Let’s try a few runs and see if we can smooth things out. > hists :: [Histogram V.Vector BinD Double] > hists = parMap rpar hist [3,4..102]  > emptyHist :: Histogram V.Vector BinD Double > emptyHist = fillBuilder hb (replicate numBins 0) > > smoothHist :: Histogram V.Vector BinD Double > smoothHist = foldl' (H.zip (+)) emptyHist hists  Quite nice and had my machine running at 750% with +RTS -N8. ## Comparison Let’s create the same histogram but from the posterior created analytically. > analPosterior :: [Double] > analPosterior = > evalState (replicateM 100000 (sample (Normal mu_1 (sqrt sigma1)))) > (pureMT$ fromIntegral 5)
>
> histAnal :: Histogram V.Vector BinD Double
> histAnal = fillBuilder hb analPosterior


And then compare them. Because they overlap so well, we show the MCMC, both and the analytic on separate charts.

# PostAmble

Normally with BlogLiteratelyD, we can generate diagrams on the fly. However, here we want to run the simulations in parallel so we need to actually compile something.

ghc -O2 ConjMCMCSimple.lhs -main-is ConjMCMCSimple -threaded -fforce-recomp
> displayHeader :: FilePath -> Diagram B R2 -> IO ()
>   mainRender ( DiagramOpts (Just 900) (Just 700) fn
>              , DiagramLoopOpts False Nothing 0
>              )

> main :: IO ()
> main = do
>     (barDiag MCMC
>      (zip (map fst $asList (hist 2)) (map snd$ asList (hist 2)))
>      (zip (map fst $asList histAnal) (map snd$ asList histAnal)))
>
>     (barDiag MCMCAnal
>      (zip (map fst $asList (hist 2)) (map snd$ asList (hist 2)))
>      (zip (map fst $asList histAnal) (map snd$ asList histAnal)))
>
>      (zip (map fst $asList (hist 2)) (map snd$ asList (hist 2)))
>      (zip (map fst $asList histAnal) (map snd$ asList histAnal)))
>      (zip (map fst $asList smoothHist) (map snd$ asList smoothHist))
>      (zip (map fst $asList histAnal) (map snd$ asList histAnal)))
`