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Let $X_t=\int_0^t \left(\frac{1-t}{1-u}\right)^k dW_u$. Assume $0\lt s \lt t\lt T$. Is the following the right way to compute the covariation of $X_s$ and $X_t$?

$$ \begin{align} \text{Cov}(X_t, X_s) &= \mathbb{E}\left[\left( X_t - \mathbb{E}[X_t] \right)\left( X_s - \mathbb{E}[X_s] \right)\right]\\ &=\mathbb{E}[X_t X_s]\\ &=\mathbb{E}\left[ \int_0^t \left(\frac{1-t}{1-u}\right)^k dW_u\int_0^s \left(\frac{1-s}{1-u}\right)^k dW_u\right]\\ &=\mathbb{E}\left[ \left(\int_s^t \left(\frac{1-t}{1-u}\right)^k dW_u+\int_0^s \left(\frac{1-t}{1-u}\right)^k dW_u\right)\int_0^s \left(\frac{1-s}{1-u}\right)^k dW_u\right]\\ &=\mathbb{E}\left[ \int_0^s \left(\frac{1-t}{1-u}\right)^k dW_u\int_0^s \left(\frac{1-s}{1-u}\right)^k dW_u\right]\\ &=\mathbb{E}\left[ \int_0^s \left(\frac{1-t}{1-u}\right)^k\left( \frac{1-s}{1-u}\right)^k du\right]\\ \end{align} $$

Danny
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    For a straightforward approach, write $X_t$ and $X_s$ as stochastic integrals on $(0,T)$ and apply the general formula. – Did Jun 18 '15 at 07:02
  • do you mean something like $X_t=\int_0^T f(t).\mathbb{1}_{s<t}dW_s$? – Danny Jun 18 '15 at 07:13
  • Indeed, but this can onnly be done if $f(t)\mathbf 1_{s<t}$ in your comment is actually some $f(t,s)\mathbf 1_{s<t}$ hence the covariance of $X_t$ and $X_s$ is by definition $$\int_0^T(f(t,u)\mathbf 1_{u<t})\cdot(f(s,u)\mathbf 1_{u<s})\cdot du=\cdots$$ – Did Jun 18 '15 at 09:24

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