I am trying to solve for the variance of $x[n]$, a time series process.  $$x[n] +a_1x[n-1]=w[n]$$where $w[n]$ is white noise with zero mean and variance $\sigma^2_v$. Also $|a_1|&lt;1$.  I am aware that the equation can be solved using the Yule Walker Equations, but I was trying to solve it brute force using expectation.  Here is what I did:  Take the z-transform of the process, $$X(z) + a_1X(z)z^{-1} = W(z)$$ Solving for the system function, $$ H(z) = \frac{X(z)}{W(z)} = \frac{1}{1+a_1z^{-1}}$$ Inverse z-transform, $$h[n] = (-a_1)^nu[n]$$ Using convolution, $$x[n] = \sum_{k=0}^{n-1}(-a_1)^kw[n-k]$$ Solving for the variance, $$E[x[n]x^*[n]]=E[\sum_{k=0}^{n-1}(-a_1)^kw[n-k]\sum_{l=0}^{n-1}(-a_1)^lw[n-l]]$$ $$E[x[n]x^*[n]] = \sum_{k=0}^{n-1}(-a_1)^k\sum_{l=0}^{n-1}(-a_1)^lE[w[n-k]w[n-l]]$$ Since w[n] is a white noise process, $$E[w[n-k]w[n-l]] = 0, k\neq l $$ $$E[w[n-k][w[n-l]] = \sigma^2_v, k=l$$ So the expectation becomes, $$E[x[n]x^*[n]] = \sum_{m=0}^{n-1}(a_1^2)^m \sigma^2_v$$ Therefore, $$Var[x[n]]= \sigma^2_v \frac{1-(a_1^2)^n}{1-a_1^2} $$  But this is not correct, this answer is nonstationary, and we should expect a stationary solution for the zero mean white noise.  The correct answer is $\frac{\sigma^2_v}{1-a^2_1}$  It would be very helpful if someone could point out what I did wrong. I would like to solve this problem using this method. That way I can more appreciate using Yule-Walker equations. Thank you!
Consider a stochastic process $\{X_t, t = 1, 2, \ldots\}$ following the model $$X_t = \alpha X_{t-1} + e_t,$$ where $e_t \thicksim f$.  Can I say that the distribution of the initial point, $X_1$, is the same as $f$? Can I say that the stationary marginal density, if it exists, of $\{X_t\}$ is same as $X_2 (\stackrel{D}{=}\alpha X_1 + e_2)$?  I think that the stationary marginal density, if it exists, of $\{X_t\}$ is the same as $X_2$, but not necessarily same as $X_1$.
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