0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $I$.

$\endgroup$
  • $\begingroup$ Thanks. I have responded to your answer in my question. $\endgroup$ – Estacionario May 22 '18 at 11:34
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Yes that is correct. $B_t$ is a Brownian motion with variance $t$. $\endgroup$ – Riccardo Sven Risuleo May 22 '18 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.