I have a question concerning the variance of the Itô integral on general time intervals, i.e. I want to calculate \begin{align*}\operatorname{Var}\left(\int_s^T f(t)dW_t\right),\end{align*} where $f: [0, T]\to \mathbb{R}$ is deterministic, $W_t$ is the standard Wiener-process and $0<s<T$. For $s=0$ it is a standard fact that this equals $\int_0^Tf(t)^2dt$. I suspect that this should also hold for $s>0$ as it is true for step processes. Is this true or am I missing something?
1 Answer
$\begingroup$
$\endgroup$
3
For any $s\in[0,T)$, the variance in question is $$\int_s^T f(t)^2\,dt.$$ This is shown just as in the case $s=0$.
Alternatively, the case $s\in(0,T)$ can be reduced to the case $s=0$ by noting that
the random function $[0,T-s]\ni u\mapsto W^s_u:=W_{s+u}-W_s$ is a standard Wiener process on the interval $[0,T-s]$;
$\int_s^T f(t)\,dW_t=\int_0^{T-s} f(s+u)\,dW^s_u$;
the variance of $\int_0^{T-s} f(s+u)\,dW^s_u$ is $\int_0^{T-s} f(s+u)^2\,du=\int_s^T f(t)^2\,dt$.
-
$\begingroup$ There are three steps to your alternative argument, and (though it is clear in retrospect) I had trouble parsing where one ended and the next began. Would you be willing to consider a comma in place of the first "and" (or perhaps even a list) to make it easier to parse? $\endgroup$– LSpiceCommented Oct 23, 2023 at 17:14
-
1$\begingroup$ @LSpice : Thank you for your comment. Does it look better now? $\endgroup$ Commented Oct 23, 2023 at 17:27
-