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1 vote

Determine $E[W_p W_q W_r]$

Note that $\{W_t \mid t \geq 0\}$ is a martingale. Then, for $0<p<q<r$, \begin{align*} E(W_pW_qW_r) &= E\Big( E(W_pW_qW_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q E(W_r \mid \mathcal{F}_q)\Big)\\ &=E …
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Show that $E[B_t|\mathscr{F}_s] = B_s$ for $B_t = W_t^3 - 3 t W_t$

\begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qqu …
Gordon's user avatar
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Independence of increments of the stochastic process $\frac{1}{t}\int_0^t u dW_u $

Note that, for $t>s>0$, \begin{align*} X_t-X_s &= \frac{1}{t}\int_0^t udW_u - \frac{1}{s}\int_0^s udW_u\\ &=\frac{1}{t}\bigg(\int_s^t u dW_u + \int_0^s udW_u \bigg)- \frac{1}{s}\int_0^s udW_u\\ &=\fra …
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Solving a backwards heat equation using stochastic calculus

Based on the form of your equation, we can consider the SDE \begin{align*} dX_t = \sigma dW_t, \end{align*} where $W$ is a standard Brownian motion. Since, for $0 \leq t \leq T$, \begin{align*} X_T …
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