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The solution to the SDE

$$dx_t= -kx_t dt + cx_t dW_t$$


$$x_t = x_0 e^{\left(c - \frac{k^2}{2} \right)t}e^{-k W_t}$$

with mean

$$\mathbb{E} \left[ x_t \right] = x_0 e^{\left(c - \frac{k^2}{2}\right)t}$$

where $W_)$ is the Wiener process.

I'm looking to compute

$\mathbb{E} \left[ (W_s + W_t - 2W_0)^2 \right]$

but am unsure of how to proceed.

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I don't see how the computation of the expectation is related to the solution of the GBM SDE... – SRKX Jul 20 '14 at 21:11

I would calculate it this way,

$\mathbb{E}[(W_s+W_t−2W_0)^2] = \mathbb{E}\left[\left((W_s-W_0)+(W_t-W_0)\right)^2\right]\\ \hspace{4cm}=\mathbb{E}[(W_s-W_0)^2]+\mathbb{E}[(W_t-W_0)^2]+2\mathbb{E}[(W_s-W_0)(W_t-W_0)] \\ \hspace{4cm}=s+t+2\mathbb{E}[W_sW_t]\\ \hspace{4cm}=s+t+2\min(s,t)$

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How do you come to $E(W_s,W_t)=\min(s,t)$? – emcor Jul 19 '14 at 21:56
Assume WLOG that $s<t$ then $E[W_sW_t]=E[W_s (W_t - W_s )+W_s^2]$ independent increments and the distributional properties and you are done. – Henrik Jul 20 '14 at 14:44
@Henrik I see that formula, but it does not depend on $s<t$, you can rewrite it both ways getting $E(W_sW_t)=s=t$ which is obviously wrong if $s<t$? – emcor Jul 21 '14 at 9:59
@emcor: Henrik answer is right! The formula depends on $s<t$. Indeed, $(W_t-W_s)$ is independent of $W_s$, it's Markov property that says that $(W_t-W_s)$ behave as a new brownien process independent of all what happens in the past $F_s$. Due to this, $\mathbb{E}[W_s(W_t-W_s)]=\mathbb{E}[W_s]\mathbb{E}[(W_t-W_s)]=0$. – aajajim Jul 22 '14 at 12:04
@aajajim I still dont agree. Let $s>t$. Then $(W_t-W_s)=-(W_s-W_t)$ which also has expectation zero. – emcor Jul 22 '14 at 12:21

Your solution $x_t$ is wrong.

Your mean is wrong too. Note that $\mathbb{E}\left[ e^{W_t}\right] = e^\frac{t}{2}$. I corrected the typo that was pointed out by Richard.

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$W_t$ is Gaussian - right? Then $E[e^{W_t}]$ is just the moment generating function of a Gaussian random variable with variance $t$ evaluated at $1$. So your formula is wrong. – Richard Jul 16 '14 at 6:51
$E[e^{W_t}] = \exp(t^2/2)$. – Richard Jul 16 '14 at 7:10
Richard: thanks for pointing out the typo. If $X \sim \mathcal{N}(\mu, \sigma^2)$, $\mathbb{E}\left[e^X\right] = e^{\mu + \frac{1}{2} \sigma^2}$. – wsw Jul 19 '14 at 16:00
Richard: Note that $W_t \sim N\Big(0, \big(\sqrt{t}\big)^2\Big)$. Then $E\big(e^{W_t} \big) = e^{\frac{t}{2}}$. – Gordon Aug 12 '14 at 13:33

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