Where is Lag operator in MA(q) process variance?

Question

MA(q) process with intercept looks like: $x_{t}= c_0+w_t+ \Theta_1w_{t-1}+ \dotsb +\Theta_qw_{t-q}$

Equation may be represented in terms of the lag operator:

$x_{t}=c_{0}+(1+\Theta_1L+ \dotsb + \Theta_qL^q)w_t$

I want to derive the variance for MA(q) process. So i get:

$Var[x_{t}]=(1+\Theta_1L+ \dotsb+\Theta_qL^q)^2\sigma^2$

But the correct answer is:

$Var[x_{t}]=(1+\Theta_1^2+ \dotsb+\Theta_q^2)\sigma^2$

I have 2 questions:

1. Where is lag operator in variance?

2. Why there is square on each $\Theta$? According to $Var$ properties: $Var(aX)=a^2Var(X)$.

Answer 1

1. Lag operator applies only to sequences, not to constants like $\sigma^2$. Moreover, since $(w_t)$ is a white noise, we know that $\text{Var}(w_t)$ doesn't depend on $t$, thus $\forall k\in\Bbb N,\;L^kw_t. = w_{t-k}$ has the same variance as $w_t$, i.e. $\text{Var}(L^kw_t)=\sigma^2$.

2. Exactly because of the property you recalled.