I have been trying to solve the following problem, but have not been successful yet. I was hoping anyone could nudge me in the correct direction.
Let $\{ w_t; t = 0,1,\ldots \}$ be a white noice process with variance $\sigma^2$ and let $|\phi| < 1$ be a constant. Consider $x_0 = \frac{w_o}{\sqrt{1-\phi^2}}$ and $$ x_t = \phi x_{t-1} + w_t,\ \ \ \ t=1,2,\ldots. $$ Show that $var(x_t)$ is constant in $t$.
MY ATTEMPT:
Firstly, by recursively writing out $x_t$ it can be seen that: $$ x_t = \phi^t x_0 + \sum_{k=0}^t \phi^k w_{t-k} = \phi^t \frac{w_0}{\sqrt{1-\phi^2}} + \sum_{k=0}^t \phi^k w_{t-k}. $$ By linearity of the expectation, it follows that $E(x_t) = 0$. Thus $var(x_t) = E(x_t^2)$. \begin{align} var(x_t) &= E(x_t^2) = E([\phi^t \frac{w_0}{\sqrt{1-\phi^2}} + \sum_{k=0}^t \phi^k w_{t-k}]^2) \\ &= \phi^{2t}\frac{1}{1-\phi^2} E(w_0^2) + 2 \frac{\phi^t}{\sqrt{1-\phi^2}}E(w_0\sum_{k=0}^t \phi^k w_{t-k}) + E([\sum_{k=0}^t \phi^k w_{t-k}]^2) \\ &= \phi^{2t}\frac{\sigma^2}{1-\phi^2} + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 + \sigma^2 \sum_{k=0}^t \phi^{2k} \\ &= \phi^{2t}\frac{\sigma^2}{1-\phi^2} + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 + \sigma^2 \frac{1-\phi^{2(t+1)}}{1-\phi^2} \\ &= \frac{\sigma^2}{1-\phi^2}( \phi^{2t} + 1 - \phi^{2t} \phi^2) + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 \\ &= \frac{\sigma^2}{1-\phi^2} + \frac{\sigma^2}{1-\phi^2}(\phi^{2t}(1-\phi^2)) + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 \\ &= \frac{\sigma^2}{1-\phi^2} + \sigma^2 \phi^{2t} + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2. \end{align} This is where I am stuck: I believe the second and third term should somehow cause the dependence on $t$ to cancel out, but can't seem to figure it out. Is this the way to go or am I missing something else? In any case, thanks for your time!
Trivially it can be seen that $\text{Var}(x_0)$ is constant.
Observe $$\begin{align} &x_1 = \dfrac{\phi w_0}{\sqrt{1 - \phi^2}} + w_1 \\ &x_2 = \dfrac{\phi^2w_0}{\sqrt{ 1- \phi^2}} + \phi w_1 + w_2 \\ &x_3 = \dfrac{\phi^3w_0}{\sqrt{1 - \phi^2}} + \phi^2 w_1 + \phi w_2 + w_3 \\ &\vdots \\ &x_t = \dfrac{\phi^tw_0}{\sqrt{ 1- \phi^2}} + \sum_{k=1}^{t}\phi^{t-k} w_k\text{.} \end{align}$$ Using the fact that $\{w_t: t \geq 0\}$ are uncorrelated random variables, we have that their pairwise covariances are $0$, thus $$\text{Var}(x_t) = \dfrac{\phi^{2t}\sigma^2}{1-\phi^2} + \sum_{k=1}^{t}\phi^{2(t-k)}\sigma^2 = \sigma^2\left(\dfrac{\phi^{2t}}{1-\phi^2} + \phi^{2t}\sum_{k=1}^{t} \phi^{-2k} \right)\text{.}$$ We have that $$\begin{align} \sum_{k=1}^{t} \phi^{-2k} &= \phi^{-2}+\phi^{-4}+\cdots+\phi^{-2t} \\ &= \phi^{-2}\left[1+\phi^{-2}+\cdots+\phi^{-2(t-1)}\right] \\ &= \phi^{-2}\left(\dfrac{1-\phi^{-2t}}{1-\phi^{-2}} \right) \\ &= \dfrac{1-\phi^{-2t}}{\phi^2 - 1} \\ &= \dfrac{\phi^{-2t} - 1}{1-\phi^2} \end{align}$$ thus $$\phi^{2t}\sum_{k=1}^{t}\phi^{-2k} = \dfrac{1 - \phi^{2t}}{1-\phi^2}$$ and $$\text{Var}(x_t) = \dfrac{\sigma^2}{1-\phi^2}\text{.}$$