# Ito process generator and Dynkin's Formula

I have attempted to answer the following multi-part question but am especially having trouble with part iv) of the question. Any feedback would be greatly appreciated!.

Consider the Ito process $X(t) = \int_{0}^{t} sdB(s), t\leq1$

i) Compute the mean of X

If the Ito process is a martingale and thus has mean 0 if $\mathbb{E}\left( \int_{0}^{t} sdB(s) \right) < \infty$. This is satisfied so the mean is 0.

ii) What is the generator of X?

$$=\frac{1}{2}t^2 \frac{\partial^2}{\partial x^2}$$

iii) Let $L$ be the generator of $X$ and set $f(x,t)=e^{ux}$. Determine $Lf$

$$= \frac{1}{2}t^2 u^2e^{ux}$$

iv) Let $h(t) = \mathbb{E}[e^{uX(t)}]$. Use Dynkin's Formula to show:

$$h'(t) = \frac{1}{2}u^2t^2h(t), h(0)=1$$

v) Solve the ODE in iv) to obtain the mgf $h(t)$

$$h = exp\left\{\frac{1}{6}u^2t^3\right\}$$

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Since $h(t)=E(f(X_t,t))$, Dynkin tells you $h'(t)=E(Lf(X_t,t))$, iii) tells you $Lf(x,t)=\frac12u^2t^2f(x,t)$, hence $h'(t)=\frac12u^2t^2E(f(X_t,t))=\frac12u^2t^2h(t)$. What is unclear? –  Did Jun 23 '12 at 21:33
I was asked to prove it using $E[f(X_{t},t)] = f(X_{0},0) + \mathbb{E} \int_{0}^{t} (Lf + f_{t})(X_{u},u)du$ with a hint to differentiate the integral in the formula. I kept obtaining $E[e^{uX}] = e^{uX(0)} + \frac{1}{2}t^2u^2 \int_{0}^t E[e^{uX}]ds$ and can't rearrange it as such. Is my use of the formula incorrect? –  Lindah Jun 23 '12 at 21:45
The error is in the factor $t^2$ outside the integral term of Itô's formula. In fact you should use $Lf(X_s,s)=\frac12u^2s^2f(X_s,s)$ inside the integral hence one obtains $h(t)=1+\frac12u^2\int\limits_0^ts^2h(s)ds$ instead of your $h(t)=1+\frac12u^2t^2\int\limits_0^th(s)ds$. After that, the ODE is direct. –  Did Jun 24 '12 at 8:02
Hi did, thank you very much for the clarification! The operations make much more sense now –  Lindah Jun 24 '12 at 21:42