# What's the variance of the following stochastic integral and is it weakly stationary?

The stochastic integral is defined as $$u_t = \int_{t-1}^t e^{-\kappa(t-s)}\int_0^s e^{-c(s-r)} \, dW(r) \, ds.$$ where $W(t)$ is a standard Brownian motion, $\kappa$ and $c$ are both positive.

I know this integral can be viewed as $$u_t = \int_{t-1}^t e^{-\kappa(t-s)}J_c(s) \, ds,$$ where $J_c(s) = \int_0^s e^{-c(s-r)} \, dW(r)$, is an Ornstein-Uhlenbeck process. But how do we handle this double integral and use Ito's isometry to get the variance of it?

Further, does this integral admit a Wold representation, that is, $$u_t = \sum_{j = 0}^\infty F_j \varepsilon_{t-j},$$ where $\varepsilon_t \sim \mathrm{iid}(0, \sigma^2)$.

Here is my calculations, which may be wrong, I am not so sure...

\begin{align} Var(u_t) &= E(u_t^2) = E\left(\int_{t-1}^{t} \int_{0}^{s}e^{-\kappa(t-s)-c(s-r)}dW(r) ds\right)^2 \\ &= E\left(\int_{0}^{t} \int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}ds dW(r) \right)^2 \\ &= \int_{0}^{t} \left(\int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}ds\right)^2 dr \\ &= \int_{0}^{t} e^{-2\kappa t + 2cr} \left(\int_{r}^{t}e^{(\kappa-c)s}ds\right)^2 dr \\ &= \dfrac{1}{(\kappa-c)^2}\int_{0}^{t} e^{-2\kappa t + 2cr} \left(e^{(\kappa-c)t}- e^{(\kappa-c)r} \right)^2 dr \\ &= \dfrac{1}{(\kappa-c)^2}\left(\int_{0}^{t} e^{-2c(t-r)} dr + \int_{0}^{t} e^{-2\kappa(t-r)} dr - 2\int_{0}^{t} e^{-(\kappa+c)(t-r)} dr\right)\\ &= \dfrac{1}{(\kappa-c)^2}\left(\dfrac{1-e^{-2ct}}{2c} + \dfrac{1-e^{-2\kappa t}}{2\kappa} - 2\dfrac{1-e^{-(\kappa+c)t}}{\kappa+c}\right) \end{align} which indicates $u_t$ exhibits heteroskedasiticity. But still, I don't know if it is weakly stationary.

It seems that you consider $u_t$ as being a discrete time (DT) process for $t=1$, $2$, $\dots$ yet the integral makes sense for any real positive $t$, then defining a continuous time (CT) process $u(t)$ for $t \geq 0$. Roughly speaking, $u(t)$ results from applying two linear filters to a CT Gaussian white noise, the two impulse responses $f(t)$ and $g(t)$ being $$f(t) := e^{-ct} \,1_{[0,\,\infty)}(t), \quad g(t) := e^{-Kt} \,1_{[0,\,1]}(t).$$ The composition of the two linear filters is equivalent to a linear filter with its impulse response $h(t)$ given by the convolution of $f(t)$ and $g(t)$ $$h(t) = \int f(s) g(t-s) \, \text{d}s.$$ However, some complications are due to the initial conditions and the related transient effects.
The Ornstein-Uhlenbeck (OU) process $J(s)=\int_0^s f(s-r) \, \text{d}W(r)$ is not weakly stationary due to its initial condition $J(0) = 0$, but it has a stationary distribution and can be considered as stationary for large $t$. Indeed, a stationary OU, say $J^\star(s)$, would be obtained by starting from the stationary distribution at $s=0$ or equivalently by starting from any distribution in the remote past. This could be written as $$J^\star(s) := \int_{-\infty}^s e^{-c(s-r)} \, \text{d}W(r) = \int_{0}^\infty e^{-cr'} \, \text{d}W(s-r'),$$ and $J(s) - J^\star(s) = e^{-cs} X$ for some Gaussian r.v. $X$, so $J(s)$ is nearly equal to $J^\star(s)$ unless $s$ is small. In other words, $J(s)$ and $J^\star(s)$ are equal up to an initial warm-up period.
Your derivation of the variance of $u(t)$ is correct and, as you noted, the variance is not constant but tends to a constant with exponential speed. Thus none of the DT and CT processes $u_t$ and $u(t)$ can be weakly stationary, hence none has a (DT or CT) Wold representation. However, except for a warm-up period, $u(t)$ is close to the stationary processes $u^\star(t)$ with a CT Wold representation $$u^\star(t) = \int_{0}^\infty h(s) \, \text{d}W(t-s),$$ so $u^\star(t)$ results from applying the linear filter with impulse response $h(t)$ to a white noise process. The variance of $u^\star(t)$ is given by the isometry $$\text{Var}[u^\star(t)] = \int_0^\infty h(t)^2 \,\text{d}t.$$ After a warm-up period, the DT process $u_t$ is equal to the DT process $u^\star_t$ which is stationary and hence has a DT Wold representation. Finding this DT Wold representation does not seem to be straightforward, but the CT representation can be used instead.