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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.

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    $\begingroup$ 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 $Y$. $\endgroup$ – whuber Nov 15 '17 at 18:51
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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}

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  • $\begingroup$ @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. $$\int_{-2}^2(2-|x|)R(x)dx$$ $\endgroup$ – Rohit Nov 15 '17 at 10:27
  • $\begingroup$ @Xi'an did'nt get you ? should I ask another question for this? $\endgroup$ – Rohit Nov 15 '17 at 16:33
  • $\begingroup$ @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? $\endgroup$ – Eldioo Nov 15 '17 at 18:00
  • $\begingroup$ @Eldioo can you do it with the double integral without the identity I am not getting how to do at the last part. $\endgroup$ – Rohit Nov 15 '17 at 18:12
  • $\begingroup$ @Eldioo Can you provide me any link for that identity of integration please $\endgroup$ – Rohit Nov 15 '17 at 19:36
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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.

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