Assume $X_n$ is an iid Gaussian random process with zero mean and variance $\sigma^2$, and $U_n$ be an iid binary random process with $P_r\{ U_{n}=1\}=P_r\{U_n=-1\}=0.5$, and $\{U_n\}$ is independent of $\{X_n\}$, now let $Q_n=X_n+U_n$,and $B_n=X_n+U_0$,what is the mean of $Q_n$ and $B_n$ ?

In my opinion

$E[Q_n]=E[X_n+U_n]=E[X_n]+E[U_n]=0+(1 \times 0.5+-1 \times 0.5)=0$

$E[B_n]=E[X_n+U_0]=E[X_n]+E[U_0]=0+(\frac{1}{2} \times 1 \times 0.5+\frac{1}{2} \times -1 \times 0.5)=0$

Actually,i am not sure about that the $E[U_0]$ is $0$ ,if wrong,can anyone tell me how to calculate it?

  • $\begingroup$ It's unconventional to refer to $\{X_n\}$ as a random process if $n$ takes value on $\{0, 1, 2, \ldots\}$ (as it is exaggerated). A more suitable name would be a sequence of random variables. $\endgroup$
    – Zhanxiong
    Jun 13, 2018 at 1:29
  • $\begingroup$ ^Not in the slightest. AFAIK, the name comes from an analogy with dynamical processes, and discrete time dynamical processes are known as such. Sure, they're sequences, but by that coin all dynamical processes should be called functions. What is certainly true is that if the random variables are all independent, one tends to refer to sequences, and if they are dependent in some nice way, processes. $\endgroup$ Jun 13, 2018 at 1:35

1 Answer 1


$U_n$ are iid. It implies that $U_1$, $U_2$, ... have all the same expectation.

So actually, $E[Q_n]=E[B_n]=E[Q_1]=E[B_1]= 0$. The expectations do not depend on $n$ at all. Your calculations are correct.

  • $\begingroup$ Is my calculation of $E[U_0]$ right? $E[U_0]=(\frac{1}{2} \times 1 \times 0.5+\frac{1}{2} \times -1 \times 0.5)$ i wonder that $\times \frac{1}{2}$ is right or wrong? $\frac{1}{2}$ means the probability that $U_0$ is $1$ or $-1$ $\endgroup$
    – Shine Sun
    Jun 13, 2018 at 1:38
  • $\begingroup$ $E[X] = \sum\limits_x x P(X=x) = 1\times 0.5 - 1 \times 0.5 =0$ $\endgroup$
    – W. Volante
    Jun 13, 2018 at 11:35

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