This is rather an unfamiliar territory for me, but I managed to do the first part of the question, so there is no reason why I cannot do the second, especially with some help ))) Besides, I really want to understand how I can use MGF here.

Question

If $\displaystyle v_t=e^{-w_t}$ and $\displaystyle w_t={\gamma}z_t+\frac{1}{2}\gamma^2t$, ie $\displaystyle dw_t=\frac{1}{2}\gamma^2dt+{\gamma}dz_t$,

where $\displaystyle z_t$ is standard Brownian motion, show that

$\displaystyle dv_t=-{\gamma}v_tdt$.

Show also that $\displaystyle E(e^{-w_t})=1$. (Hint: consider the moment generating function of the standard normal variable $\displaystyle \frac{z_t}{t}$.

Answer.

I use Ito's lemma to prove the first:

first and second derivatives of $\displaystyle v_t: f'=(e^{-w_t})'=-e^{-w_t}; f''=e^{-w_t}$

$\displaystyle dv_t=(\frac{1}{2}\gamma^2(-e^{-w_t})+\frac{1}{2}e^{-w_t}\gamma^2)dt+\gamma(-e^{-w_t})dz_t=-{\gamma}v_tdz_t$ (as the first bracket is zero.)

Now, for the second part, I could try substitution the formula for w_t :

$\displaystyle E(e^{-w_t})=E(e^{-{\gamma}z_t-\frac{1}{2}\gamma^2t})=e^{-1/2\gamma^2t}E(e^{-{\gamma}Z_t})=e^{-1/2{\gamma}^2t}M_{Z_t}(-\gamma)=e^{-1/2\gamma^2t}e^{1/2{\gamma}^2}$ - somehow I feel I should get $\displaystyle e^0=1$ but I don't see it yet. Perhaps I am neglecting properties of Brownian motion?

(in my MGF for Zt I assumed variance of Zt as 1. If, say, std deviation of $\displaystyle z_t=\sqrt{t}$ then I get my result 1.

...UPDATED.

Got it! Checked again the definiton of Standard Brownian motion and the variance is t!!

So, $\displaystyle e^{-1/2{\gamma}^2t}M_{Z_t}(-\gamma)=e^{-1/2\gamma^2t}e^{1/2{\gamma}^2t}=e^0=1$

Now tell me if I was wrong!

PS Who would have thought studying maths could be such an emotional roller-coaster