# Tag Info

0

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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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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First, your statement that $dW_t=\sigma\,dt^{1/2}$ is incorrect. In fact, it's not even meaningful (you can see this by noticing that the expression on the left-hand-side is an "increment" of Brownian motion, and hence random, while the expression on the right-hand side is deterministic). What you mean to say is that $W$ is a Brownian motion, and hence ...

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I don't know what you mean by "any scaling" rule. For the square-root of time I can say that it only needs uncorrelated returns. Assume that the return from time point $1$ to $T$ is called $r_{1,T}$ and that it is given as $r_{1,T} = r_1 + r_2 + \cdots + r_T = \sum_{t=1}^T r_t$ where $r_t, t=1,\ldots,T$ are the one-period (e.g. one day) returns. The ...

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