I'm trying to find the variance of the following:
$$ \int_{0}^{t} N\Bigl(0,\sigma^2e^{-C(t-\tau)}\sin^2\bigl(B(t-\tau)\bigr)\Bigr)d\tau $$
where $N$ is a Gaussian distribution with zero mean and variance:
$$ \sigma^2(t,\tau) = \sigma^2_0e^{-C(t-\tau)}\sin^2\bigl(B(t-\tau)\bigr) .$$
It reminds me a bit of a wiener process so I've tried to follow the same logic:
$$ E\Bigl[\bigl(\int_{0}^{t}N_\tau d\tau \bigr)^2\Bigr] = E\Bigl[\int_{0}^{t} \int_{0}^{t} N_\tau N_{\tau '} d\tau d\tau '\Bigr] $$ $$ = \int_{0}^{t} \int_{0}^{t} E\Bigl[N_\tau N_{\tau '}\Bigr] d\tau d\tau ' $$
$$ = \int_{0}^{t} \Biggl( \int_{0}^{\tau '} \sigma^2(t,\tau) d\tau + \int_{\tau '}^{t} \sigma^2(t,\tau ') d\tau \Biggr)d\tau '.$$
I'm a bit concerned about that last step though. It's been a while since I've looked a wiener functions and web trawling hasn't managed to remind me the logic behind that step and therefore if it is valid in this case.
So two questions really:
a) Is that last step valid?
b) If no to a) then how would you calculate the variance of:
$$ \int_{0}^{t} N\Bigl(0,\sigma^2e^{-C(t-\tau)}\sin^2\bigl(B(t-\tau)\bigr)\Bigr)d\tau . $$
Thanks in advance for your help.
I've worked out the answer to my problem. In short, I was close but not quite right. In long:
Because I my integration is over a function of tau not t, the auto covariance should be
$$ \sigma^2(t-\tau) $$
when,
$$ t-\tau < t-\tau' $$ $$ \tau' < \tau $$
And,
$$ \sigma^2(t-\tau') $$
when,
$$ t-\tau' < t-\tau $$ $$ \tau < \tau' $$
And consequently, my integration should have been:
$$ \int_{0}^{t} \Biggl(\int_{0}^{\tau'} \sigma^2(t,\tau')d\tau + \int_{\tau'}^{t} \sigma^2(t,\tau)d\tau\Biggr)d\tau'$$
The subtle difference being integrating of the variance of the function of tau' below tau' and tau above rather than tau below and tau' above in my original attempt.