Let $X(t)$ be a Gaussian white noise process with mean zero and variance $\sigma^2 \in \mathbb{R}$. Let $\tau \in \mathbb{R}$ be a constant. How would I identify the distribution of $$I = \int_0^t X(u) e^{u/\tau} \, du$$ where $t \in \mathbb{R}$ denotes time?

EDIT: Having read What is meant by a continuous-time white noise process?, it looks like it is more correct to write $$I = \int_0^t \sigma e^{u/\tau} \, dB_u$$ where $B_u$ denotes Brownian motion.

EDIT 2: Using the Ito Isometry to evaluate the variance of $I$ I get $$\mathbb{E}\left[\left(\int_0^t \sigma e^{u/\tau} \, dB_u\right)^2\right] = \mathbb{E}\left[\int_0^t \sigma^2 e^{2u/\tau} \, du\right]= \dfrac{\sigma^2 \tau}{2}\left(e^{2t/\tau}-1\right).$$ I would guess that the expectation of $I$ is zero since each of the normal random variables in the series has mean zero, but how does this follow from the Ito isometry?


I will assume your integral is an Ito integral. You can see $I$ as the stochastic limit of a series of normal random variables; so $I$ has a Normal distribution.

In addition, it is a Martingale and $$ \mathbf{E}[I_t] = \mathbf{E}[I_0] = 0$$

Then, you can apply Ito isometry to find the the variance of $I$.

  • $\begingroup$ Thanks. I have responded to your answer in my question. $\endgroup$ – Estacionario May 22 '18 at 11:34
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    $\begingroup$ You're welcome! Sorry I didn't point it out. Ito integrals have zero mean. $\endgroup$ – Riccardo Sven Risuleo May 22 '18 at 11:43
  • $\begingroup$ Does the $dB_u$ notation imply that the variance of the white noise process is equal to one? If so, am I correct in multiplying the integrand by sigma? $\endgroup$ – Estacionario May 22 '18 at 12:01
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    $\begingroup$ Yes that is correct. $B_t$ is a Brownian motion with variance $t$. $\endgroup$ – Riccardo Sven Risuleo May 22 '18 at 12:23

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