I apologise in advance - this is heavily related to another question, but I don't have enough reputation to post a comment on the relevant question to ask for clarification.
I was looking at this question, and the answer provided by Sesame. I'm stuck on calculating the variance of the stochastic integral. Here's my line of reasoning:
From the answer, we know that $\mathrm{E}[x_t] = x_0\exp(-at)$, and that $$ \begin{align} \mathrm{Var}[x_t] &= \int_0^t \exp(-2a(t - s)) \textrm d s \\ &= \exp(-2at)\int_0^t \exp(2as) \textrm d s \\ &= \frac1{2a}(1 - \exp(-2at)) \end{align} $$ Also from the answer, we know that $$ \begin{align} \mathrm E[y_t] &= y_0\exp(-bt) + \int_0^t \exp (-b(t - s)) \mathrm E[\exp(- x_s)] \mathrm ds \\ &= y_0\exp(-bt) + \underbrace{\int_0^t \exp (-b(t - s)) \exp\left(- x_0\exp(-as) + \frac1{4a}(1 - \exp(-2as))\right) \mathrm ds}_{\equiv I} \end{align} $$ using the properties of a log-normally distributed random variable.
To work out the variance of $y_t$, $$ \begin{align} \mathrm{Var}[y_t] &= \mathrm E[y_t^2] - (\mathrm E[y_t])^2 \\ &= \mathrm E\left[ \left( \int_0^t \exp (-b(t - s) - x_s) \mathrm ds\right)^2\right] - (\mathrm E[y_t])^2 \end{align} $$ Firstly, I'm not sure where the leading factor of $y_0 \exp(-bt)$ has gone in $\mathrm E[y_t^2]$. I think the line should read $$ \begin{align} \mathrm{Var}[y_t] &= \mathrm E\left[ \left( \int_0^t \exp (-b(t - s) - x_s) \mathrm ds\right)^2\right] - I^2 \\ &= \mathrm E\left[ \int_0^t \int_0^t \exp (-b(t - u) - x_u) \exp (-b(t - v) - x_v) \mathrm du \mathrm dv\right] - I^2 \\ &= \int_0^t \int_0^t \exp (-b(2t - u - v)) \mathrm E\left[ \exp(- (x_u + x_v) )\right]\mathrm du \mathrm dv - I^2 \end{align} $$ The sum of two Gaussian random variables is also Gaussian, therefore, $$ x_u + x_v \sim \mathcal N \left(x_0\left(e^{-au} + e^{-av}\right), \frac1{2a}(2 - \exp(-2au) - \exp(-2av))\right) $$ Substituting this back into the expression for the $\mathrm{Var}[y_t]$, $$ \begin{align} \mathrm{Var}[y_t] &= \int_0^t \int_0^t \exp (-b(2t - u - v)) \\ &\hspace{1cm}\exp\left(-x_0\left(e^{-au} + e^{-av}\right) + \frac1{4a}(2 - \exp(-2au) - \exp(-2av))\right)\mathrm du \mathrm dv - I^2 \\ &= \int_0^t \exp \left(-b(t - u) -x_0 \exp(-au) + \frac1{4a}(1 - \exp(2au))\right) \mathrm du \\ &\hspace{1cm}\int_0^t \exp \left(-b(t - v) -x_0 \exp(-av) + \frac1{4a}(1 - \exp(2av))\right)\mathrm dv - I^2 \\ &= I^2 - I^2 \\ &= 0 \end{align} $$ I'm fairly sure that the variance is non-zero, however, I'm not sure where I've gone wrong - if I had to guess, it would either be when exchanging the expectation value with the double integral, or summing the two Gaussian random variables (which implicitly assumes that they're independent).
My question is, therefore: how does one solve for the variance of this integral?
Edit: I've realised that I misread the previous question, in which $b$ represents Brownian motion - I suspect this would invalidate both the answer I refer to and the approach I take in my question, as $b$ becomes a random variable; how would one solve the system given the $x_t$ and $y_t$ I have defined, i.e. $$ \begin{align} x_t &= x_0\exp(-at) + \int_0^t \exp(-a(t-s)) \textrm dW_s \\ y_t &= y_0\exp(-bt) + \int_0^t \exp(-b(t-s) - x_s) \textrm ds \end{align} $$ and treating $b$ as a constant?