Find the Variance of random variable $Y=\int_{0}^{2}X(t)dt $

Question

If $X(t)$ is stationary process having mean value $E[X(t)]=3$ and autocorrelation function $R_{xx}(\tau)=9+2e^{-|\tau|}$.The variance of random variable $Y=\int_{0}^{2}X(t)dt $ will be____

$(A)1$

$(B)2.31$

$(C)4.54$

$(D)0$

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I have no idea how to solve this but I tried like this

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$$E[X^2]=R_{xx}(0)=11$$ $$E[X]^2=9$$

\begin{align} Var(Y)&=\int_{0}^{2}Var(X(t))dt\\ &=\int_{0}^{2}E[X^2(t)]-E[X]^2dt\\ &=\int_{0}^{2}2dt\\ &=4 \end{align}

But its not in the options.

Answer 1

First of all: The error of your approach was to pull out the integral from the Variance function. While this works (under some conditions) for the expected value, it doesn't for $$E\left[\left(\int_0^2X(t)dt\right)^2\right],$$ due to the square term.

Note that:

• You can calculate $E[Y]$ by interchanging integration and expected value.
• $E[Y]^2$ can be calculated by using \begin{align}E[Y^2]&=E\left[\left(\int_0^2X(t)dt\right)\left(\int_0^2X(s)ds\right)\right]\\ &=\int_0^2\int_0^2E[X(t)X(s)]dt~ds\\ &=\int_0^2\int_0^2R(t-s)dt~ds\\ &\left(=\int_{-2}^2(2-|x|)R(x)dx\right) \end{align}

Altogether, this should yield answer $(C)$.

EDIT: I used the identity $$\int_0^a\int_0^a f(x-y) dx~dy =\int_{-a}^a (a-|z|)f(z) dz$$ for the last part (Proof), but it's not needed. You can also directly calculate the double integral. Just split it into a positive and negative part, in order to get rid of the absolute term in the exponent. After that it's plain calculation.

EDIT2: Note that \begin{align} \int_0^2\int_0^2R(t-s)dt~ds&=\int_0^2\int_0^2 9+2\exp^{-|t-s|}dt~ds\\ &=36+2\int_0^2\int_0^2\exp^{-|t-s|}dt~ds\\ &=36+2\left(\int_0^2\int_s^2\exp^{-(t-s)}dt~ds+\int_0^2\int_0^s\exp^{t-s}dt~ds\right). \end{align}

Answer 2

Hint #1: The equality $$Var(Y)=\int_{0}^{2}Var(X(t))dt$$ is incorrect

Hint #2: If you are unfamiliar with continuous time process, try to use first a (Riemann) discretisation of the integral.

Hint #3: this question has already been asked on X validated.