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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 $|a_1|<1$.

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!

Darklink9110
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  • Did you notice that the answer you "determined" approaches the "right" answer as n tends to infinity since we have that |$a_1$|<1? – Michael R. Chernick Nov 05 '19 at 00:32
  • I did notice that, that is what makes me believe I was on the right track. But I cant seem to explain why I would need to take the limit. The only thing I can think of is that the answer determined by using the yule walker equations is a "steady state" solution. But I haven't been able to find any verification of that. – Darklink9110 Nov 05 '19 at 00:38

1 Answers1

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I think I may have figured it out. An AR process is asymptotically stationary.

Imposing the initial conditions of $w[n] = 0 $ for all $n<0$ makes the process nonstationary. Therefore when the system is stable and has poles inside the unit circle, i.e. $|a_1|<1$, after a long time the process will "forget" these initial conditions and behave as a stationary process.

In other words the formula in the question is the response of the system as it first experiences the white noise.

To come up with the answer we could either ignore the initial conditions or take the limit.

In this sense, I think it is right to assume that the Yule-Walker Equations are the "steady-state" behavior of the system.

Darklink9110
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