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Let $B_t$ be Brownian motion on $\mathbb R$. Let $u$ be a polynomial in $x$ and $t$ which satisfies the heat equation $\partial_t u = -(1/2)\Delta_x u$. Theorem 8.5.8 on page 323 of Durrett's "Probability: theory and examples" (http://stat-www.berkeley.edu/~aldous/205A/PTE4_Jan2010.pdf) asserts that under these hypotheses, $u(t,B_t)$ is a martingale. The proof seems to consist of showing that $\partial_t E_x u(t,B_t) = 0$. However, I am not sure why this implies that $E_x(u(t,B_t) | \mathcal F_s) = u(s,B_s)$ for each $t > s$ (which I believe is the definiton of a martingale). Here $\{\mathcal F_t\}$ is the filtration os $\sigma$-algebras corresponding to Brownian motion. Can someone explain how we get this implication?

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You have typos, it should be $\partial_t E_x u(t,B_t)$ and $E_x(u(t,B_t)|\mathcal F_s)$, IIUC. I think you're right it only shows $\partial_t E_x u(t,B_t)=0$. To evaluate $E_x(u(t,B_t)|\mathcal F_s)$ you would need to change $p_t(x,y)$ to $p'_t(x,y)=p_{t-s}(x+B_s,y)$, but that transition probability still satisfies the heat equation, so $\partial_t E_x(u(t,B_t)|\mathcal F_s) = 0$. –  Kirill Jan 6 '13 at 3:49
you have another typo the equation that $u$ satisfies is $\partial_tu=-(1/2)\Delta_xu$ not $\partial_tu=(1/2)\Delta_xu$ –  Paul Jan 6 '13 at 3:54
Typos have been fixed. –  user15464 Jan 6 '13 at 3:58
I tried to deduce this from the Markov property: $E_x(Y\circ \theta_s |\mathcal F_s) = E_{B_s}(Y)$ (where $\theta_s$ is the shift operator) but I seem to get the wrong thing. Namely, let $Y = u(t-s, B_{t-s})$ in the Markov property to get $E_x(u(t , B_t) | \mathcal F_s) = E_x(Y\circ \theta_s |\mathcal F_s) = E_{B_s}(u(t-s , B_{t-s})$. Since this expectation is independent of $t$ (as shown in the proof in Durrett), this should equal $E_{B_s}(u(0 , B_0)) = u(0 ,B_s)$, rather than $u(s,B_s)$, which is what we need. Where is the flaw in this reasoning? –  user15464 Jan 6 '13 at 4:22

1 Answer 1

up vote 3 down vote accepted

Perhaps, a better way to convince yourself is to note that, by Itô's lemma we have

$$ u(t,B_t) = u(0,B_0) +\int_0^t \left(\partial_t +\frac{1}{2} \Delta \right)u(s,B_s)d~s +\int_0^t \partial_x u(s,B_s)d~B_s $$

then if $u(t,x)$ is solution of $ \left(\partial_t +\frac{1}{2} \Delta \right)u(t,x)=0$, we have

$$ u(t,B_t) = u(0,B_0) +\int_0^t \partial_x u(s,B_s)d~B_s $$

which is a $\mathcal F_t$- martingale since it's a stochastic integral.

In the proof you refered, which is a "with hands" proof, ie without uses stochastic calculus results as Itô's lemma (it's natural because the autor seems to want to motivate this stochastic results giving an intuition with simple probabilistic-analytical proofs) he has shown that

$$\partial_t \mathbb E_x \left\{ u(t,B_t)\right\}=\int_{\mathbb R}p_t(x,y)\left( \partial_t +\frac{1}{2}\Delta_x\right)u(t,y)~dy =0$$

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