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Long time lurker, first time posting! Have a problem, that looks familiar but I can't put my finger on it.

Need to calculate $\mathbb{E} [\exp(aW_T^2)|F_t]$ where $W_t$ is an $F_t$ adapted standard Brownian motion and $t \leq T$. Any help on exponentials of squared Brownian motion is very appreciated!

SBF
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POSSY
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    Starting point: write $W_T$ as $W_t + (W_T - W_t)$. – Nate Eldredge Jul 23 '14 at 21:14
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    Middle point: recall that $\exp(x+y) = \exp(x)\exp(y)$ and what to do with conditional expectation of $\mathscr F_t$-measurable and $\mathscr F_t$-independent random variables. The decomposition @NateEldredge proposed together with the latter two properties works in many examples of conditional expectations with Brownian motion. You may want to try $\Bbb E[W^3_T|\mathscr F_t]$ after that. – SBF Jul 23 '14 at 21:30
  • What I have is: $\mathbb{E}(exp aW_T^2) = \mathhbb{E} exp(a(W_T - W_t + W_t)^2) = exp(aW_t)\mathbb{E}exp(2a(W_t(W_T-W_t)))\mathbb{E}exp(a(W_T-W_t)^2) = exp(aW_t)exp(.5a^2(W_T-W_t)^2)$ – POSSY Jul 23 '14 at 23:07
  • What I have is: $\mathbb{E}(exp aW_T^2) = \mathbb{E} exp(a(W_T - W_t + W_t)^2) = exp(aW_t)\mathbb{E}exp(2a(W_t(W_T-W_t)))\mathbb{E}exp(a(W_T-W_t)^2) = exp(aW_t)exp(.5a^2(T-t)^2)$ I'm sure of the last and first expectation but it is $\mathbb{E}exp(2a(W_t(W_T-W_t)))$ I'm concerned with, is this equal to one? (Sorry for such silly questions... years since my stoch calculus was used last!) – POSSY Jul 23 '14 at 23:13

2 Answers2

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If $2aT\geqslant1$, then $\exp(aW_T^2)$ is not integrable hence $E(\exp(aW_T^2)\mid F_t)$ does not exist. From now on, assume that $2aT\lt1$. Define $Z$ by $$W_T-W_t=\sqrt{T-t}\cdot Z,$$ then $Z$ is standard normal and independent of $F_t$ and $$W_T^2=W_t^2+2\sqrt{T-t}W_tZ+(T-t)Z^2,$$ hence $$ E(\exp(aW_T^2)\mid F_t)=\exp(aW_t^2)\cdot E(\exp(2aW_t\sqrt{T-t}\cdot Z)\mid F_t)\cdot E(\exp(a(T-t)Z^2)). $$ For every $b$, $E(\exp(bZ))=\exp(\frac12b^2)$ hence $$E(\exp(2aW_t\sqrt{T-t}\cdot Z)\mid F_t)=\exp(2a^2(T-t)W_t^2).$$ Since $2a(T-t)\lt1$, $$E(\exp(a(T-t)Z^2))=\frac1{\sqrt{1-2a(T-t)}}. $$ Finally, if $2aT\lt1$ then $$ E(\exp(aW_T^2)\mid F_t)=\frac{\exp(a(1+2a(T-t))W_t^2)}{\sqrt{1-2a(T-t)}}. $$

Did
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I think the previous answer is false, as the expectation of the product is the product of the expectations if and only if both random variables are independent, which is obviously not the case.

However the first part is true, so I will only try to correct the second part:

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = \mathbb{E} \left[ e^{u(W_T - W_t)^2} e^{2 u W_t (W_T - W_t)} \mid \mathcal{F}_t \right] e^{u(W_t)^2}$

As $(W_T - W_t) \perp \!\!\! \perp \mathcal{F}_t$ and $(W_T - W_t) \sim \mathcal{N}(0, T-t)$ we have :

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = e^{u(W_t)^2} \int_\mathbb{R} e^{u(T - t) x^2 + 2 u W_t \sqrt{(T - t)} x} \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx$

You then get this (I leave you the intermediate calculations) :

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = \frac{\exp \left( u(W_t)^2 + \frac{2 u^2 (W_t)^2 (T - t)}{1 - 2 u (T - t)} \right)}{\sqrt{2 \pi}} \int_\mathbb{R} e^{-\frac{1}{2} \left( \sqrt{1 - 2 u (T - t)}x - \frac{2 u W_t \sqrt{(T - t)} }{\sqrt{1 - 2 u (T - t)}} \right)^2}dx$

And then after the right (obvious here) change of variables and a simplification in the exponential on the left, you end up with :

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = \frac{ \exp \left( \frac{u(W_t)^2}{1 - 2 u (T - t)} \right)}{\sqrt{1 - 2 u (T - t)}}$

If you require further explanations, I will be glad to give it.

Louis-Amand
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