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(Edit:) The short version: Calculate $$E[e^{cY}]$$ if $c < 0$ and $Y$ is lognormally distributed, i.e. $\log(Y) \sim N(\tilde\mu, \tilde\sigma^2)$.

The long version: I want to calculate $$E[e^{cX_T} \mid X_t = x]$$ for $c < 0$, where $X$ is a geometric Brownian motion, i.e. $$dX_t = rX_tdt + \sigma X_t dW_t$$ for $r \in \mathbb{R}, \sigma > 0$ and a Brownian motion $W$.

My ideas so far:

(i) If $X_t = x$, then $X_T = x e^{(r-\sigma^2/2)(T-t) + \sigma (W_T-W_t)}$ is log-normally distributed, so the expecation is $$ \frac{1}{2\pi\sigma\sqrt{T-t}} \int_0^\infty e^{cy} \frac{1}{y} e^{-\frac{(\log(y/x) - (r-\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}}dy.$$ But I do not really know what to do with this expression.

(ii) Itô: $$ \begin{align} de^{cX_t} &= ce^{cX_t} dX_t + \frac{c^2}{2} e^{cX_t} d[X]_t\\ &= ce^{cX_t} dX_t + \frac{c^2}{2} e^{cX_t} \sigma^2 X_t^2 dt\\ &= (crX_t + \frac{1}{2} c^2 \sigma^2 X_t^2) e^{cX_t}dt + c \sigma X_t dW_t.\end{align}.$$ Again, I am stuck.

Does anyone have an idea how to proceed?

Elias Schoof
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  • Have you come across moment generating functions? – Chinny84 Jul 25 '14 at 13:01
  • See the solution here – SBF Jul 25 '14 at 13:19
  • If $c > 0$, the expectation above is indeed the moment-generating function of the lognormal distribution. Unfortunately, that function does not exist (i.e. the expectation is infinity). For $c < 0$, I could not find any calculations on Google. – Elias Schoof Jul 25 '14 at 13:48
  • Ilya, thank you for the link! I am not really sure how to apply this to my problem, as I have to work with $X_t = x \exp(... + \sigma W_t)$ and not $W_t$ directly. – Elias Schoof Jul 25 '14 at 13:51
  • @EliasStrehle Do you want to calculate the expectation or the conditional expectation? – saz Jul 25 '14 at 15:21
  • As long as $T$ can be chosen arbitrary, that does not really matter, if I am not mistaken. X is a time-homogenous Markov process, so if $X_t = x > 0$, then $$E[f(X_T) \mid \mathcal{F_t}] = E[f(X_T) \mid X_t = x] = E[f(X_{T-t}) \mid X_0 = x] = v(x)$$ for a function $v$, correct? – Elias Schoof Jul 25 '14 at 16:59
  • The left-hand side is a random variable whereas the right-hand side is a deterministic value. This equality does not hold. – saz Jul 25 '14 at 18:28
  • Sorry, you are right of course. My notation is very sloppy. $Y_t = E[f(X_T) \mid \mathcal{F}t]$ _is a random variable. But the value of $Y_t$ only depends on the value of $X_t$ (this follows from the Markov property; we can disregard all information that is contained in $\mathcal{F}_t$, except for the information what $X_t$ is). This is why we can write $Y_t = v(X_t)$ here. In the equations above I merely assumed that $X_t = x$. – Elias Schoof Jul 25 '14 at 20:14

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