# Autocorrelation of a stochastic process which is a sum

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. • 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. – Dilip Sarwate Nov 18 '13 at 2:38
• 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? – mgus Nov 18 '13 at 8:00
• 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 $X_1$? – Dilip Sarwate Nov 18 '13 at 13:28
• 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 $\infty$), $E[X_mX_n] = E[X_nX_n] = E[X_n^2] > 0 \neq E[X_n]E[X_n]$. – Dilip Sarwate Nov 18 '13 at 14:01

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. 