Consider the random process defined by the Ito integral $$ X_t = \int_0^t f(\tau)\, dB_\tau $$ where $f(\tau)$ is a deterministic real-valued function and $B_\tau$ denotes the canonical real-valued Wiener process.

I want to find the (presumably multivariate normal) distribution of $X_t$ for $t \in [0,T]$.


Since $X_t$ is an Ito integral and $f(\tau)$ is deterministic, I know that $\mathbb{E}(X_t) = 0$ for all $t$.


From the Ito isometry I have that $$ \mathbb{E}\left[ \left( \int_0^t f(\tau)\, dB_\tau \right)^2 \right] = \mathbb{E}\left[ \int_0^t f(\tau)^2\, d\tau \right]. $$


This is where I am stuck. How would I go about finding the autocovariance function of $X_t$ $$ \mathrm{cov}(X_t,X_s) = \mathbb{E}(X_t X_s) $$ where $s \in (0,t)$?

  • 1
    $\begingroup$ Have you considered the possibility of using $X_t=X_s+(X_t-X_s)$? $\endgroup$ – Glen_b Jun 18 '18 at 1:19
  • $\begingroup$ Using the independence of $X_s$ and $(X_t-X_s)$ I get $\mathrm{cov}(X_s,X_t)=\mathrm{Var}(X_s)$. Intuitively this seems strange as it implies a covariance which does not decay with distance between $s$ and $t$. $\endgroup$ – Estacionario Jun 18 '18 at 9:46
  • 2
    $\begingroup$ It doesn't. But the correlation does. $\endgroup$ – Glen_b Jun 18 '18 at 10:46

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