Doob’s Inequality II

Doob’s {{\cal L}^p} Inequality for Right (Left) Continuous Martingales

In all the proofs I have seen, some of the details are elided; here is my proof with all the details.

Proof: Let {D(m)} be an increasing set of finite sets whose union is dense in {[0,t]} and note that by right (left) continuity, we have:

\displaystyle  \sup_{s \in [0,t]}Z_s(\omega) = \sup_m \sup_{s \in D(m)}Z_s(\omega)


\displaystyle  \mathop{\mathbb E} (\sup_{s \in [0,t]}Z_s(\omega)) =

\displaystyle  \mathop{\mathbb E} (\sup_m \sup_{s \in D(m)} Z_s^p(\omega)) =

By monotone convergence

\displaystyle  \sup_m \mathop{\mathbb E} \sup_{s \in D(m)} Z_s^p(\omega) \leq

By Doob’s Inequality for discrete martingales

\displaystyle  \sup_m q^p \sup_{s \in D(m)}Z_s^p(\omega) =

\displaystyle  q^p\sup_{s \in [0,t]}Z_s^p(\omega)


Note that if {Z_s(\omega)} is not continuous then all bets are off. For example, consider the finite sets on the interval {[0,1]} of dyadic rationals:

\displaystyle  \begin{array}{rcl}  D_0 & = & \{0,1\}\\ D_1 & = & \{0, \frac{1}{2}, 1\} \\ D_2 & = & \{0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1\} \\ & \ldots & \end{array}

Then {D = \cup_nD_n} is dense in {[0,1]} and consider the function

\displaystyle  \begin{array}{rcl}  Z_s(\omega) & = & 0\ {\rm if} s \in D \\ Z_s(\omega) & = & 1\ {\rm otherwise} \end{array}


\displaystyle  1 = sup_{s \in [0,1]}Z_s(\omega) \neq \sup_m\sup_{s \in D(m)}Z_s(\omega) = 0


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