Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
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 …
3
votes
Accepted
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 …
3
votes
Accepted
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 …
1
vote
Accepted
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 …