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

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____

$(A)1$

$(B)2.31$

$(C)4.54$

$(D)0$

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

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

\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.

• There doesn't exist any such process unless "autocorrelation" is replaced by "autocovariance." That's because the size of the autocorrelation cannot exceed $1$, whereas $R$ obviously gets much larger than that. Another clue that $R$ cannot be autocorrelation is that the mean and autocorrelation alone do not determine the variance of any function of $X$ that depends on its scale--and that obviously includes $Y$.
– whuber
Commented Nov 15, 2017 at 18:51

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}

• @Eldioo Can you help me how did you change from this integration $$\int_0^2\int_0^2R(t-s)dt~ds$$ to this integration please. $$\int_{-2}^2(2-|x|)R(x)dx$$ Commented Nov 15, 2017 at 10:27
• @Xi'an did'nt get you ? should I ask another question for this? Commented Nov 15, 2017 at 16:33
• @Rohit If you don't know how to calculate the double integral, check my edit. Does that clear things up or do you want another hint? Commented Nov 15, 2017 at 18:00
• @Eldioo can you do it with the double integral without the identity I am not getting how to do at the last part. Commented Nov 15, 2017 at 18:12
• @Eldioo Can you provide me any link for that identity of integration please Commented Nov 15, 2017 at 19:36

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.