What's the variance of the following stochastic integral?

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 $\varepsilon_t \sim \mathrm{iid}(0, \sigma^2)$

• Where you wrote $\varepsilon_t\sim\mathrm{iid}(0,\sigma^2)$, did you mean $\varepsilon_t\sim\text{iid N}(0,\sigma^2)$? $\qquad$ – Michael Hardy Jun 6 '16 at 16:33
• It means independent identically distributed with mean zero and variance $\sigma^2$, which does not necessarily normal distributed. – Tom Jun 6 '16 at 16:39

Here is my calculation, which may be wrong or not, I am not quite 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\frac{1-e^{-(\kappa+c)t}}{\kappa+c}\right) \end{align} which indicates $u_t$ exhibits heteroskedasiticity.
• Can we prove $u_t$ is weakly stationary? – Tom Jun 7 '16 at 2:52