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Let's say that a stochastic process is defined as $$X(t) = \sum_{n = -\infty}^{+\infty} X_n * f(t- nT)$$ where $X_n$ is the $n$-th symbol among independent random symbols. I have proved that $\mathbb E[X(t)] = 0$. But, I want to prove that this stochastic process is Wide Sense Stationary(WSS) which means by definition that its autocorrelation function $R_{xx}(t+k,t) = \mathbb E[X(t+k)*X(t)] = R_x(k)$ i.e. the auto-correlation function of $X(t)$ depends only on $k$, which is the time difference between $t+k$ and $k$. What I have proved is that the auto-correlation function of $X(t)$ is zero but I am not sure that this proves that this stochastic process is WSS. I have attached a picture of the problem and what I have done so far in order to be clearer.

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    $\begingroup$ Hint: In the double sum that you write, it is not always true that $E[X_nX_m] = E[X_n]E[X_m]$. When $n$ happens to equal $n$ (which happens infinitely often as $m$ and $n$ range from $-\infty$ to $\infty$). In this special case, $E[X_n^2] > 0$ and so the sum cannot equal $0$. Whether it has the desired WSS property that you seek is up to you to determine. $\endgroup$
    – Dilip Sarwate
    Commented Nov 18, 2013 at 2:38
  • $\begingroup$ Thanks for the answer. Basically, I have 2 questions. First, why E[XnXm]=E[Xn]*E[Xm] is not always true given that those symbols are independent? But let's suppose that I undestand this. Then, what the double sum on the 3rd line of the proof equals to given that E[Xn^2]>0 infinitely often? $\endgroup$
    – mgus
    Commented Nov 18, 2013 at 8:00
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    $\begingroup$ Try it with just two variables $X_1$ and $X_2$. $X_mX_n$ can be one of four different things: $$X_1X_1 = X_1^2,\\ X_1X_2,\\ X_2X_1,\\ X_2X_2 = X_2^2.$$ Do all of these four random variables have mean $0$? In particular, is $E[X_1X_1] = E[X_1]E[X_1]$? Is $X_1$ independent of $X_1$? $\endgroup$
    – Dilip Sarwate
    Commented Nov 18, 2013 at 13:28
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    $\begingroup$ Corrected first sentence of my comment: Hint: In the double sum that you write, it is not always true that $E[X_nX_m]=E[X_n]E[X_m]$. When $m$ happens to equal $n$ (which happens infinitely often as $m$ and $n$ range from $-\infty$ to $\infty$), $E[X_mX_n] = E[X_nX_n] = E[X_n^2] > 0 \neq E[X_n]E[X_n]$. $\endgroup$
    – Dilip Sarwate
    Commented Nov 18, 2013 at 14:01

1 Answer 1

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Your proof is incorrect. the step indicated below is not true, i.e.: $$ E[X_n X_m] \not = E[X_n] E[X_m] $$ in general.

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  • $\begingroup$ Maybe you could highlight the area where you point out the error with thicker or darker lines so that tired and colorblind eyes like mine can see more clearly what you are pointing out? I would not have posted my redundant comment on the main question if I had seen what you were indicating. Oh, and +1 of course..... $\endgroup$
    – Dilip Sarwate
    Commented Nov 18, 2013 at 13:34
  • $\begingroup$ @DilipSarwate: Done! $\endgroup$
    – Peter K. ♦
    Commented Nov 18, 2013 at 13:44

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