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The solution to the SDE

$dx= -kx\ dt + cx \ dW$

is

$x(t) = x_0 e^{(c - k^2/2)t}e^{-k W}$

with mean

$\langle x(t) \rangle = x_0 e^{(c - k^2/2)t}$

where $W(t)$ is the Wiener process.

Im looking to find the expression

$\langle [W(s) + W(t) - 2W(0)]^2 \rangle$

but am unsure of how to proceed.

Help would be appreciated.

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2 Answers

I would calculate it this way,

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

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Your solution $x_t$ is wrong.

Your mean is wrong too. Note that $\mathbb{E}\left[ e^{W_t}\right] = \frac{t}{2}$.

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