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I need a check on the following exercise, which is the same situation that the user @saz proved here with all the details. The fact is that the notation is a bit a problem for me and I'd like to be sure that what I've worked out is fine.


Let $f$ be a $C^2$ function on $\mathbb{R}$ such that $f,f',f''$ are bounded and consider the random variable $W_t$ of density $p_t(x)=\frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}}$.

We know from the lectures that that $$E[f(W_t) - f(W_0)] = \frac{1}{2} E[\int_0^t f''(W_u)du]$$ (Use the fact that $p_t(x)$ solves the heat equation and Fubini thm)

Prove now that the process defined by $$M_t=f(W_t)-f(W_0) - \frac{1}{2} \int_0^t f''(W_u)du$$ is a $(F_t)_t$ martingale.


To this aim, I fix $0<s<t$ and compute as usual: $$E[M_t - M_s|F_s] = E[f(W_t) - f(W_s)- \frac{1}{2} \int_s^t f''(W_u)du|F_s]$$

First, I note that by the Freezing lemma (F.L.), the first two summands are respectively $$E[f(W_t)]$$ $$E[f(W_s)]$$

Now I focus on $E[\int_s^t f''(W_u)du|F_s]$, which is equal, by Fubini, to: $$\int_s^t E[f''(W_u)|F_s] du = \\ \int_s^t E[f''( (W_u-W_s) + W_s) |F_s ]=_{F.L.} \\ \int_s^t E[f''(W_u)]du =_{FUB} \\ E[\int_s^t f''(B_u) du]$$

Collecting everything together:$$E[f(W_t) - f(W_0) - \frac{1}{2} \int_s^t f''(W_u) du]$$

Now, repeating exactly the argument I used to show that $\star$, I conclude that $$E[f(W_t) - f(W_0) - \frac{1}{2} \int_s^t f''(W_u) du]=0$$

and hence the thesis follows.

Is everything okay? Especially the application of the freezing lemma on the integral term

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  • $\begingroup$ Sorry it's a typo. I have fixed it now :) $\endgroup$
    – andereBen
    Jan 13, 2020 at 20:29

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