Let $W_t$ denote univariate Brownian motion and define $X_t = \exp(\alpha W_t)$, where $\alpha \in \mathbb{C}$ is a constant. Find the stochastic differential equation (SDE) satisfied by $dX_t$ and hence or otherwise prove that

$\mathbb{E}e^{\gamma V} = e^{\gamma \mu + \gamma^2 \sigma^2/2}$

when $V \sim N(\mu, \sigma^2)$, for any complex number $\gamma \in \mathbb{C}$.

I think the SDE satisfied by $dX_t$ is $dX_t = \mu e^{\alpha W_t}dt + \sigma e^{\alpha W_t}dW_t$.

I'm not sure where to go from here.

  • 1
    $\begingroup$ If you are going to minus 1 this can you please explain to me why so I can avoid making the same mistake in the future? Thanks. $\endgroup$ – math_apprentice Jan 8 '17 at 16:03

The way you are presenting your question is a bit confusing. In particular, you're giving two different definitions for $X_t$. Here I'm going to use \begin{align*} Y_t &= \exp(\alpha W_t)\\ X_t &= X_0\exp((\mu-\frac{\sigma^2}{2})t + \sigma W_t) \end{align*} Note that in your question, I think there's a typo with the brackets in the definition of $X_t$. Also, everything I do here is under the assumption that $\alpha$, $\mu$ and $\sigma$ are in $\mathbb{R}$. I'm not sure how it applies when they are in $\mathbb{C}$, but from what I see, I don't think you need that now.

Now using Ito's lemma (https://en.wikipedia.org/wiki/It%C3%B4's_lemma) on $f(t,x)=e^{\alpha x}$ and $g(t,x) = e^{(\mu-\frac{\sigma^2}{2})t + \sigma x}$, you have that \begin{align} dY_t = \alpha Y_t dW_t + \frac{\alpha^2}{2} Y_t dt\\ dX_t = \mu X_t dt + \sigma X_t dW_t. \end{align} I'm unsure where you get the SDE you give for $dX_t$. Maybe there's something missing in your question.

Finally, for the expectation, what you have is the moment generating function of $V \sim N(\mu,\sigma^2)$ (see https://en.wikipedia.org/wiki/Normal_distribution#Moments). You can use this result to obtain $E[X_t]=X_0 e^{\mu t}$ as follows: \begin{align*} E[X_t] &= E[X_0 \exp((\mu-\frac{\sigma^2}{2})t + \sigma W_t)]\\ &= X_0 \exp((\mu-\frac{\sigma^2}{2})t) E[\exp(\sigma W_t)]\\ &= X_0 \exp((\mu-\frac{\sigma^2}{2})t) \exp(\frac{\sigma^2 t}{2})\\ &= X_0 e^{\mu t}. \end{align*} The third equality follows from $W_t \sim N(0,t)$. Analogously, $E[Y_t] = e^{\frac{\alpha^2 t}{2}}$.

  • $\begingroup$ I have edited my question for clarity. $X_t = \exp (\alpha W_t)$. I agree I don't think in matters it is in $\mathbb{C}$. The equation you have given for $dX_t$ is correct so thanks for that. The problem I am having is getting from $dX_t$ to $\mathbb{E} e^{\gamma V}$. I've tried computing $\mathbb{E}(dX_t)$ but I get $\frac{\alpha^3}{2} \exp (2 \alpha W_t)$ and I don't see how this helps. $\endgroup$ – math_apprentice Jan 8 '17 at 19:52
  • $\begingroup$ The correct SDE for $X_t$ as defined in your question is given by $dY_t$ in my answer. For $E[e^{\gamma V}]$, why don't you directly use the density or the moment generating function of $V$? You can also use the fact that $V$ has the same distribution as $\mu+\sigma Z$, where $Z \sim N(0,1)$. $\endgroup$ – user392529 Jan 8 '17 at 22:21

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