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

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

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Your idea is fine, however, there are a few mistakes. Here is another solution. Note that, \begin{align*} u_t &= \int_{t-1}^{t} \int_{0}^{s}e^{-\kappa(t-s)-c(s-r)}dW(r) \,ds \\ &= \int_{0}^{t-1} \int_{t-1}^{t}e^{-\kappa(t-s)-c(s-r)}ds\,dW(r) + \int_{t-1}^{t} \int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}ds\,dW(r)\\ &= e^{-\kappa t} \int_{t-1}^t e^{(\kappa -c)s}ds \int_0^{t-1} e^{cr}dW(r) + \frac{e^{-\kappa t}}{\kappa -c}\int_{t-1}^t e^{cr} \left[e^{(\kappa -c)t}-e^{(\kappa-c)r} \right] dW(r)\\ &= e^{-\kappa t} \int_{t-1}^t e^{(\kappa -c)s}ds \int_0^{t-1} e^{cr}dW(r) + \frac{e^{-c t}}{\kappa -c}\int_{t-1}^t e^{cr} dW(r) - \frac{e^{-\kappa t}}{\kappa -c}\int_{t-1}^t e^{\kappa r} dW(r). \end{align*} Then, \begin{align*} E(u_t^2) &=\left(e^{-\kappa t} \int_{t-1}^t e^{(\kappa -c)s}ds\right)^2 \int_0^{t-1} e^{2cr} dr \\ &\quad + \frac{e^{-2c t}}{(\kappa -c)^2}\int_{t-1}^t e^{2cr} dr + \frac{e^{-2\kappa t}}{(\kappa -c)^2}\int_{t-1}^t e^{2\kappa r} dr - 2\frac{e^{-(\kappa+c) t}}{(\kappa -c)^2}\int_{t-1}^t e^{(\kappa + c)r} dr. \end{align*}

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  • $\begingroup$ Can we prove $u_t$ is weakly stationary? $\endgroup$ – Tom Jun 7 '16 at 2:52
  • $\begingroup$ @Tom: what do you mean weakly stationary? You may ask this as another question. $\endgroup$ – Gordon Jun 7 '16 at 12:38

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