0
$\begingroup$

So i have a discrete process $\xi[n]$ which is a set of independent random variables with means $m_\xi=1$ and variance $\sigma_\xi^2=1$. A new process has been formed from two samples of the initial signal $$\eta[n]=\xi[n]+2\xi[n-2]$$ I was asked to calculate the means of the new process. My solution to it is: $$E[\eta[n]]=E[\xi[n]]+2E[\xi[n-2]]=1+2E[\xi[n]]\delta[n-2]=1+2\delta[n-2]$$ Can i just replace the shifted delta function with a 1 and then my means simply equals to 3? or my solution is completely incorrect

  • 0
    Now that I look closer, I don't understand what you did with $\xi[n-2]$. What is $\delta[n-2]$? – 2017-01-29
  • 1
    The appearance of $\delta[n-2]$ is a mystery. Simply use $E[\xi[n-2]]=m_\xi$. (It is not true that $E[\xi[n-2]]=E[\xi[n]]\delta[n-2]$, actually, $E[\xi[n-2]]=E[\xi[n]]=m_\xi$ for every $n$.) – 2017-01-29
  • 0
    Thank you! i wasn't sure on that part either since i kind of mixed Discrete signal processing with random processes. Now its clear to me! – 2017-01-29

1 Answers 1

0

Your solution is correct. In fact, to compute the mean here $\xi[n]$ need not be independent, since $$\mathbb{E}(X+Y) = \mathbb{E}(X)+\mathbb{E}(Y)$$ always holds. For variance, on the other hand, $$Var(X+Y) = Var(X)+Var(Y)$$ does not generally hold. However, it holds when $X$ and $Y$ are independent (more generally uncorrelated). In your case then, both mean and variance for $\eta[n]$ are $3$.