# Mean and Variance of a Moving Average(1) Process

$x_t=a_0w_t +a_1w_{t-1}$
$\text{Var}(x_t)$=$\text{Var}(a_0w_t +a_1w_{t-1})$
Assume $w_t, w_{t-1}$ are independent. Then variances of sum is sum of variances.
$\text{Var}(x_t) = \text{Var}(a_0 w_t) + \text{Var}(a_1 w_{t-1})$
$Var(x_t)$ = $a_0^2\text{Var}(w_t)+a_1^2\text{Var}(w_{t-1})$
$w_t, w_{t-1}$ are both standard normal distributions. So their variance is 1:
$Var(x_t)$ = $a_0^2\text{Var}(w_t)+a_1^2\text{Var}(w_{t-1})$
$Var(x_t)$ = $a_0^2+a_1^2$

$E[x_t] =0$
Suppose $w_{t-1} = w_{low}$.
$E[x_t|w_{t-1} = w_{low}]=E[a_0w_t +a_1w_{t-1}|w_{t-1}]$
$E[w_t|w_{t-1}]=E[w_t]=0$
First term drops out so we have:
$E[x_t|w_{t-1}] = w_{low}]=E[0 +a_1w_{t-1}|w_{t-1}]$
$E[x_t|w_{t-1}=w_{low}]=a_1w_{low}$