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
1 Answer
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 $(1)$.
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$\begingroup$ 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) $\endgroup$– SreejithDec 19, 2020 at 14:37
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$\begingroup$ @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
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$\begingroup$ 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) $\endgroup$– SreejithDec 19, 2020 at 15:39
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$\begingroup$ @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
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