Noise added to a Random Process if we have a discrete random process
\begin{equation}
x\left(n\right)\:=\:0.2x\left(n-1\right)+w\left(n\right)+w\left(n-1\right)
\end{equation}
where $ w\left(n\right)$ is a noise with a mean
$
m_w=0.5$ and variance $\sigma ^2_w\:=\:1$.
I  am trying to find the value of $E\left\{x^2\left(n\right)\right\}$.
Here's my work below:
\begin{equation}
E\left\{x^2\left(n\right)\right\} = E\bigg\{\big[0.2x\left(n-1\right)+w\left(n\right)+w\left(n-1\right)\big]^2\bigg\}
\end{equation}
for
$$E\big\{w\left(n-i\right)^2\big\}=\sigma _w^2+m_w^2=1+0.5\:=\:1.5.$$
After that, I get the following
\begin{align}
E\left\{x^2\left(n\right)\right\} =&amp;\phantom{+} 0.04 E\left\{x^2\left(n-1\right)\right\}\\
&amp;+0.4E\left\{w\left(n\right)x\left(n-1\right)\right\}\\
&amp;+0.4E\left\{w\left(n-1\right)x\left(n-1\right)\right\}\\
&amp;+E\left\{w^2\left(n\right)\right\}\\
&amp;+2E\left\{w\left(n\right)w\left(n-1\right)\right\}\\
&amp;+E\left\{w^2\left(n-1\right)\right\}
\end{align}
Am not sure what to do from here? Please help.
Hi: I don't know whether white noise can have a non-zero mean ( Dilipe can tell us that )  but, if it can't, then just call it whatever you call white noise with a non-zero mean.
But, assuming it's zero, this is how you can do it. ( If it's not zero, then the  solution is similar but the last step would be a little different. Use
$$\operatorname{Var}\left\{X\right\} = E\left\{X^2\right\} - E\left\{X\right\}^2$$
Note also I wrote it with &quot;t's&quot; and &quot;epsilon's&quot;  just because I'm more used to those but you can think of $t$ as $n$ and $\epsilon$ as $w$.
\begin{align}
x_t &amp;= 0.2 x_{t-1} + \epsilon_{t} + \epsilon_{t-1}\\
\implies x_{t}(1 - 0.2L) &amp;= \epsilon_{t} + \epsilon_{t-1} \tag{$\scriptsize{\text{using lag operator}}$}\\
\implies x_{t} &amp;=  \frac{\epsilon_{t}}{1 - 0.2L} + \frac{\epsilon_{t-1}}{1 - 0.2L}\\
\implies x_{t} &amp;= \sum_{i = 0}^{\infty} 0.2^{i} \epsilon_{t-i} + \sum_{i = 0}^{\infty} 0.2^{i} \epsilon_{t-1-i}
\end{align}
So, since the mean is zero, the $E\left\{x^2_{t}\right\}$ is the variance of the RHS.
So, it is
\begin{align}
\sum_{i = 0}^{\infty} \left(0.2^{i}\right)^2 \cdot \sigma^2_\epsilon + \sum_{i = 0}^{\infty} \left(0.2^{i}\right)^2 \cdot \sigma^2_\epsilon&amp;= \sum_{i = 0}^{\infty} 0.04^{i} \cdot \sigma^2_\epsilon + \sum_{i = 0}^{\infty} 0.04^{i} \cdot \sigma^2_\epsilon\\
 &amp;=  \frac{2 \cdot \sigma^2_{\epsilon}}{(1- 0.04)}
\end{align}