3
$\begingroup$

Integral of a Real white gaussian noise

In this question, is the answer not equal to infinity ? Answer is mentioned as 6. But my doubt is cant we think of it like a linear combination of many independent random variables each having infinite variance, so the resulting random variable also has infinite variance. This was a question asked in GATE exam conducted in India for Electronics and Communication stream

Improve this question
$\endgroup$
2
  • $\begingroup$ But is it not an integration in time domain ? $\endgroup$
    – Sreejith
    Dec 19, 2020 at 13:50
  • $\begingroup$ But is it not a time domain integration from t=5 to t =7 $\endgroup$
    – Sreejith
    Dec 19, 2020 at 13:55

1 Answer 1

Reset to default
3
$\begingroup$

Since $W(t)$ is assumed to be zero-mean, also the RV $Y$ is zero-mean. Hence, the variance of $Y$ is given by

$$\begin{align}\sigma_Y^2&=E\left\{Y^2\right\}\\&=E\left\{\int_{-\infty}^{\infty}W(t_1)\phi(t_1)dt_1\int_{-\infty}^{\infty}W(t_2)\phi(t_2)dt_2\right\}\\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi(t_1)\phi(t_2)E\big\{W(t_1)W(t_2)\big\}dt_1dt_2\tag{1}\end{align}$$

where $E\big\{W(t_1)W(t_2)\big\}$ is the auto-correlation function $R_W(t_2-t_1)$ of $W(t)$. Now you just have to figure out the expression for $R_W(\tau)$ and solve the integral $(1)$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Isnt this integral also evaluates to infinity, because E[ W(t1) W(t2) ] is non zero only for t1 = t2 and E[W^2 (t) ] evaluates infinity , and the remaining is energy of phi (t) $\endgroup$
    – Sreejith
    Dec 19, 2020 at 14:37
  • $\begingroup$ @Sreejith: No, you get a Dirac delta impulse inside the integral and you can use the sifting property to solve the integral. $\endgroup$
    – Matt L.
    Dec 19, 2020 at 15:23
  • $\begingroup$ Its delta ( t1 - t2 ) and is zero except for t1 = t2, when t1 = t2 we get integral of delta (0). phi^2(t) ,which is different from integral of delta ( t ) phi ^2 ( t ). how do we apply sifting property for integral of delta (0) . phi^2 ( t) $\endgroup$
    – Sreejith
    Dec 19, 2020 at 15:39
  • $\begingroup$ @Sreejith: You don't get an integral over $\delta(0)$, because $\delta(0)$ is meaningless. I guess you have to review the sifting property: $\int_{-\infty}^{\infty}\phi(t_2)\delta(t_2-t_1)dt_2=\ldots$. $\endgroup$
    – Matt L.
    Dec 19, 2020 at 15:47
  • $\begingroup$ Thank you very much, I got it now $\endgroup$
    – Sreejith
    Dec 19, 2020 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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