I want to find the covariance $K_X(t,t')$ of the following signal $X(t)$:
$X(t)=\sum\limits_{n=-\infty}^{+\infty} A_np(t-nT)$
where
$ p(t) = \begin{cases} \ 1 & \text{if } 0<t\leq T/2 \\ 0 & \mathrm{otherwise} \end{cases} $
and $A_n$ are independent random binary variables of value $-1$ or $1$ and $E[A_n]=0$.
I started with $K_X(t,t')=E[X(t)X(t')]-E[X(t)]E[X(t')]$
Is it correct that $E[X(t)]=0$?
And how can I find $E[X(t)X(t')]=E\left[\sum\limits_{n=-\infty}^{+\infty} A_np(t-nT) \sum\limits_{{n'}=-\infty}^{+\infty} A_{n'}p(t'-n'T)\right]$? Because I also found $E[X(t)X(t')]=0$ but I don't think this is correct.
For a fixed $t$, the series defining $X(t)$ has only one term which can be different from $0$ (for the potential index $n$ for which $nT\lt t\leqslant nT+T/2$). Therefore, we can switch the sum and the expectation and we find that $\mathbb E[X(t)]=0$ for each $t$ because $p(t-nT)$ is not random and $\mathbb E[A_n]=0$.
For the covariance, by the argument mentioned previously, $$\mathbb E[X(t)X(t')]=\sum_{n\in\mathbf Z}\sum_{n'\in\mathbf Z}\mathbb E[A_nA_{n'}]p(t-nT)p(t'-n'T).$$ Using independence, we obtain that $\mathbb E[A_nA_{n'}]=0$ if $n\neq n'$ and $1$ otherwise, hence $$\mathbb E[X(t)X(t')]=\sum_{n\in\mathbf Z}p(t-nT)p(t'-nT).$$ If there is an integer $n$ such that $nT\lt t,t'\leqslant nT+T/2$, then $\mathbb E[X(t)X(t')]=1$ otherwise the covariance is $0$.