How to compute $E[W(T)\exp(W(T)]$

Question

I have got this interview question twice. Does anyone know from which interview question book or another source this question comes from? It may be some well known source as two different interviewers asked the same question, but unfortunately I don't know the source.

Here is what I did during the interviews:

Apply Itô to $f(W)=W\exp(W)$

$\frac{\partial f}{\partial W}=\exp(W)+W \exp(W)$

$\frac{\partial^2 f}{\partial W^2}=2\exp(W)+W \exp(W)$

$df=(\exp(W)+W \exp(W))dW + (\exp(W)+1/2 W \exp(W)) dt$

Given that we will compute the expected value, we ignore the stochastic integral term:

$df= (\exp(W)+1/2f) dt$

Answer 1

Hereunder is how I would solve that. I would say this is some sort recurring exercice in probability classes at university.

Solution based on the derivation of the characteristic function $e^{\lambda W_T}$, as $W_T$ is a gaussian random variable of mean 0 and standard deviation $\sqrt{T}$. Then, $E[e^{\lambda W_T}] = e^{\lambda^2 \times T/2}$

Deriving the rhs expression once in $\lambda$ gives $2\lambda\frac{T}{2} \times e^{\lambda^2 \frac{T}{2}} $.

Finally, observing that $E[W_T e^{W_T}]$ corresponds to the derivation of $E[e^{\lambda W_T}]$ evaluated in $\lambda=1$, we can conclude that $E[W_T e^{W_T}] = Te^{T/2}$

Answer 2

Notice that $$e^{W^{Q}(T)}$$ looks almost like Doleans exponential $$e^{W^{Q}(T)-\frac{1}{2}T}$$ Therefore $$E^Q[W^{Q}(T)e^{W^{Q}(T)}]=E^Q[W^{Q}(T)e^{W^{Q}(T)}]e^{-\frac{1}{2}T}e^{+\frac{1}{2}T}$$ $$=E^Q[W^{Q}(T)e^{W^{Q}(T)-\frac{1}{2}T}]e^{\frac{1}{2}T}$$ We now define new probability measure $\bar{Q}$ via the Radon Nikodym derivative: $$\frac{{d\bar{Q}}}{dQ}=e^{W(T)-\frac{1}{2}T}$$ Under $\bar{Q}$ measure $$W^{\bar{Q}}(t)=W^{Q}(t)-t$$ is brownian motion. Therefore: $$E^Q[W^{Q}(T)e^{W^{Q}(T)-\frac{1}{2}T}]e^{\frac{1}{2}T}=E^{\bar{Q}}[W^{\bar{Q}}(T)+T]e^{\frac{1}{2}T}=Te^{\frac{1}{2}T}$$

Answer 3

To continue your thought:

$$f(x)=x{\rm e}^x $$

$$df(W_t) = \left({\rm e}^{W_t} + f(W_t)\right) dW_t + \left({\rm e}^{W_t} + 1/2f(W_t) \right)dt $$

We now integrate from $0$ to $T$ ($W_0=0$):

$$f(W_T) = \int_0^T \left({\rm e}^{W_t} + f(W_t) \right) dW_t + \int_0^T {\rm e}^{W_t} dt + 1/2 \int_0^T f(W_t) dt. $$

Then take expectations on both sides and obtain (commuting integration and expectation for the time integrals):

$$E\left[f(W_T)\right] = \int_0^T {\rm e}^{t/2} dt + 1/2 \int_0^T E\left[ f(W_t)\right] dt $$

Introducing deterministic function:

$$ y(u) := E\left[f(W_u)\right], $$

we get

$$ y(T) = \int_0^T {\rm e}^{t/2} dt + 1/2 \int_0^T y(t) dt. $$

Taking the derivative wrt $T$ on both sides, gives ODE:

$$ y'(T) = {\rm e}^{T/2} + 1/2 y(T), $$

$$y(0) = 0,$$

with solution:

$$y(T) = T{\rm e}^{T/2}.$$

Answer 4

So let‘s add the brute force solution as well:

$W_T\sim N(0,T)$ so

$$ \begin{align} E\left(W_Te^{W_T}\right)&=\int_{-\infty}^{\infty}xe^x\frac{e^{-\frac{x^2}{2T}}}{\sqrt{2\pi T}}dx\\ &= \int_{-\infty}^{\infty}x\frac{e^{-\frac{x^2-2Tx}{2T}}}{\sqrt{2\pi T}}dx \\ &= \int_{-\infty}^{\infty}x\frac{e^{-\frac{x^2-2Tx+T^2-T^2}{2T}}}{\sqrt{2\pi T}}dx\\ &= e^{1/2T}\int_{-\infty}^{\infty}x\frac{e^{-\frac{1}{2}\left(\frac{x-T}{\sqrt{T}}\right)^2}}{\sqrt{2\pi T}}dx \end{align} $$

Now let $z=(x-T)/\sqrt{T}$ and $x=T+z\sqrt{T}$ and $dx=dz\sqrt{T}$ then

$$ \begin{align} E\left(W_Te^{W_T}\right)&=e^{1/2T}\int_{-\infty}^{\infty}x\frac{e^{-\frac{1}{2}\left(\frac{x-T}{\sqrt{T}}\right)^2}}{\sqrt{2\pi T}}dx\\&= e^{1/2T}\int_{-\infty}^{\infty}(T+\sqrt{T}z)\frac{e^{-\frac{1}{2}\left(z\right)^2}}{\sqrt{2\pi}}dz\\ &=Te^{\frac{1}{2}T} \end{align} $$

Answer 5

There are many roads to Rome. Here is the road of overkill, but which nevertheless gives a glimpse of the uses of Malliavin calculus:

Note that $$ E_0 \left[W(T) e^{W(T)} \right] = E_0 \left[e^{W(T)} \int_0^T dW(t) \right] $$ The integration by parts formula of Malliavin calculus reads $$ E_0 \left[ F \int_0^T h(t)dW(t) \right] = E_0 \left[ \int_0^T (D^W_t F) h(t) dt \right] $$ where $D_t^W$ denotes the Malliavin derivative with respect to $W$.

In this problem $F = e^{W(T)}$ and $h(t) = 1$. Furthermore, $$ D^W_t e^{W(T)} = 1_{[0,T]} (t) e^{W(T)} $$ with $1_{[0,T]}(t) = 1$ if $t \in [0,T]$ and $0$ otherwise.

Hence, \begin{align} E_0 \left[ \int_0^T (D^W_t F) h(t) dt \right] &= E_0 \left[ \int_0^T e^{W(T)} dt \right] \\ &= T E_0 \left[e^{W(T)} \right] \\ &= T e^{T/2} \end{align}