I was looking at another thread, and the following two distinct solutions to $E[e^{W_s}e^{W_t}]$ (assume that $W_0 = 0$ and $t>s$) were given, with both giving identical answers (I have slightly re-written the notation in the second solution for the reader's sake):
1.
by direct computation I find:
$$
\begin{array}
\mathbb{E}[e^{W_t}e^{W_s}] &= \mathbb{E}[e^{W_t – W_s} e^{2W_s}] \\
&= \mathbb{E}[e^{W_t – W_s}]\mathbb{E}[e^{2W_s}] \\
&= e^{\frac12(t-s)} e^{2s} \\
&= e^{\frac12 t + \frac32 s}.
\end{array}
$$
Given the facts that for $X_t = e^{W_t}$,
$(a)$ $e^{-\frac t2}X_t$ is a martingale and
$(b)$ For any constant $\lambda \in\mathbb R$, the process $Y_t= e^{\lambda W_t – \frac12 \lambda^2t}$ is a martingale,
we have
\begin{align}
E[e^{W_s}e^{W_t}] &= e^{\frac{t}{2}}E[e^{W_s}e^{W_t – \frac{t}{2}}] \\
&= e^{\frac{t}{2}}E[e^{W_s}E[e^{W_t – \frac{t}{2}}|\mathcal{F}_s]]
\quad \text{$W_s$ is $\mathcal{F}_s$-mesurable and thanks to a)}\\
&= e^{\frac{t}{2}}E[e^{2W_s – \frac{s}{2}}] \\
&= e^{\frac{t+3s}{2}}E[e^{2W_s – 2s}] \quad \text{thanks to b) with $\lambda=2$}\\
&= e^{\frac{t+3s}{2}}\\
\end{align}
On the other hand, I took quite a different approach, getting a different answer (note that I write $\exp(x)$ instead of $e^x$):
Let $Z \sim N(0,1)$. Then its moment generating function is,
\begin{align*}
E[\exp{(\lambda Z)}] &= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp\left({\lambda z})\exp({-z^2/2}\right)dz \\
&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\exp\left({(-z^2 + 2\lambda z)/{2}}\right)dz \\
&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\exp\left({{-(z + \lambda)^2}/{2} + {\lambda^2}/{2}}\right)dz \\
&= \exp\left({\frac{\lambda^2}{2}}\right)\int_{-\infty}^{\infty}\frac{\exp\left({{-(z + \lambda)^2}/2}\right)}{\sqrt{2 \pi}}dz .
\end{align*}Since the function inside the last integral is precisely the pdf of $N(\lambda , 1)$, its integral equals 1. Therefore we have,
$$E[\exp{(\lambda Z)}] = \exp\left({\frac{\lambda^2}{2}}\right).$$Now,
\begin{align*}
E[\exp\left({W_s}\right)\exp\left({W_t}\right)] &= E[\exp\left({W_s + W_t}\right)] \\
&= E[\exp\left({(\sqrt{s} + \sqrt{t})Z}\right)] \text{ since $W_t \sim N(0,t)$}\\
&= \exp\left({\frac{(\sqrt{s} + \sqrt{t})^2}{2}}\right) \\
&= \exp\left( \frac{s + t + 2\sqrt{st}}{2} \right). \\
\end{align*}
Since $t > s$, we can write $t = a s$ for some $a > 1$. Hence,
\begin{align*}
E[\exp\left({W_s}\right)\exp\left({W_t}\right)] &= \exp\left( \frac{s + t + 2\sqrt{s(as)}}{2} \right) = \exp\left( \frac{s + t + 2s\sqrt{a}}{2} \right) \\
&> \exp\left( \frac{s + t + 2s \cdot \sqrt{1}}{2} \right) = \exp\left( \frac{t + 3s}{2} \right),
\end{align*}
and hence the discrepancy is clear (with the two answers being equal only if $s = t$).
Naturally, my question is where my mistake is. Thanks!
Best Answer
The issue with your second computation is that $W_s + W_t$ does not agree in distribution with $(\sqrt{s} + \sqrt{t}) Z$ for a standard Gaussian $Z$. We can see this by a computation. First: $$(\sqrt{s} + \sqrt{t}) Z \sim N(0, s + t + 2\sqrt{st})$$ Now observing that $(W_s, W_t)$ is a bivariate normal vector, we have that $W_s + W_t$ is normally distributed with: $$\begin{align*} E(W_s + W_t) &= 0 \\ \mathrm{Var}(W_s + W_t) &= \mathrm{Var}(W_s) + \mathrm{Var}(W_t) + 2 \mathrm{Cov}(W_s, W_t) = s + t + 2 s \land t \end{align*}$$ In particular the variances (hence, the distributions) of these two random variables do not agree, and thus: $$E(\exp (W_s + W_t)) \neq E( \exp ((\sqrt{s} + \sqrt{t}) Z))$$