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can someone explain why this is wrong:

$\epsilon_t$ is WN~ $(0,\sigma^2)$

$X_t=\frac{\epsilon_t}{1-\phi}$

$V(X_t)=V(\frac{\epsilon_t}{1-\phi})=\frac{V(\epsilon_t)}{(1-\phi)^2}$

$V(X_t)=\frac{\sigma^2}{(1-\phi)^{2}}$ Why is this wrong? $\phi <1$ is a constant

The solution which is right is

$V(X_t)=\frac{\sigma^2}{1-\phi^2}$

Its the variance of an AR(1) process.

Thank you in advance.

mth
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1 Answers1

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Hi: You are attempting to derive the variance of an AR(1) but you are skipping some steps. The AR(1) is

$X_t = \phi X_{t-1} + \epsilon_t$ where $\epsilon_t \sim N(0,\sigma^2)$.

First, pull the first term over to the left using the lag operator.

$X_{t}(1-\phi L) = \epsilon_t$. Now when you divide by $(1 - \phi L)$, you end up with an infinite series on the right hand side.

$X_t = \frac{\epsilon_t}{1-\phi L}$.

Note that, as long as $\phi < 1$, then the right hand side is an infinite series that converges. So, we end up with:

$X_t = \sum_{t=0}^{\infty} \phi^{t} \epsilon_{t-i}$

Taking the variance of the sum results in: $var(X_t) = \sum_{t=0}^{\infty} \phi^{2t} \sigma^2 = \frac{\sigma^2}{1-\phi^2}$.

So, the issue was that the lag operator was left out which made the expression look like it wasn't an infinite series.

HERE IS AN EDIT TO PROVIDE AN ALGEBRIAC REASON WHY NORMAL VARIANCE RULES CANNOT BE USED ON A INFINITE SUM.

Let's suppose that the lag operator, $L$, is not even involved in order not to confound the issue.

Let: $ 1 + \rho + \rho^2 \ldots + \rho^n + \ldots = A $.

and Let:

$ \frac{1}{1-\rho} = B $.

I'm not 100 percent certain but I think that the reason your squaring approach doesn't work is because $A^2 \neq B^2$. Take the case of $A^2$. When one takes expectations of the individual terms of the square of the sum, all of the cross product terms are zero because of the independence of the $\epsilon_{i}$ and $\epsilon_{j}$ for $i \neq j$. Conversely, when one squares B, there is no knowledge of individual cross product terms reducing to zero.

If someone else can prove $A^2 \ne B^2$ in a less heuristic way, it's appreciated.

mark leeds
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  • Can you explain why i cant use the rules for variances? – mth Jun 26 '23 at 22:59
  • Shouldn't the solution be like this even without a lag operator?$\frac{\sigma^2}{1-\phi^2}$ because its the geometric series. – mth Jun 26 '23 at 23:01
  • Hi: I tried to explain that in the answer but it's essentially because you're initial expression: $X_t = \frac{\epsilon_t}{1-\phi}$ is not true. If it was, then you could do what you did. – mark leeds Jun 26 '23 at 23:02
  • Let me add a step so you can better see what's happening in terms of the infinite series. – mark leeds Jun 26 '23 at 23:04
  • $V(\frac{\epsilon_t^2}{1-\phi})=V(\epsilon_t^2+\epsilon_t^2\phi+\epsilon_t^2\phi^2...)=$

    $1V(\epsilon_t^2)+\phi^2V(\epsilon_t^2)+\phi^4V(\epsilon_t^2)...=1\sigma^2+\phi^2\sigma^2+\phi^4\sigma^2...=\frac{\sigma^2}{1-\phi^2}$

    Even without Lag operator its possible to derive the solution.

    But I don't understand why I can't use the rule for variances to pull out the constant.

     – mth Jun 26 '23 at 23:07
  • Hopefully now you can see how the lag operator enters into how one obtains the infinite sum. – mark leeds Jun 26 '23 at 23:13
  • you have a squared term in $V(\frac{\epsilon^2}{1-\phi})$ bt it's just a typo. In addition, I don't see how you are getting what you get because there's a $t$ subscript which has to go backwards in time. You can't just remove the lag operator because that's what causes the backwardness to happen. I think the answer is that what you wrote for V is not correct. There's an L in there so you can't use normal rules for variance because you need to expand the sum. – mark leeds Jun 26 '23 at 23:18
  • When you use "normal" rules for variance, you are assuming that the denominator is a constant but it's far from a constant. Specifically, you are using $var(ax) = a^2 var(x)$ which assumes that $a$ is fixed and constant. – mark leeds Jun 26 '23 at 23:21
  • That means a geometric series is not a constant? – mth Jun 26 '23 at 23:25
  • I'm not a math person ( why is why I'm on this list !!! ) but, if your RV has a time subscript and you want to compute the variance of the random variable, then the expression needs to be re-written as an infinite sum of random variables if that is possible. Hopefully, someone else can give a more technical answer. – mark leeds Jun 26 '23 at 23:32
  • the $(1-\phi L)$ in the denominator is not a constant like "a" is in my earlier example. – mark leeds Jun 26 '23 at 23:34
  • One more thing: Suppose that the $\epsilon_t$ variance was not constant, say $\sigma^2_t$. Then, you're geometric series expansion would not give the correct answer because you couldn't just remove the lag operator like you did. – mark leeds Jun 26 '23 at 23:44
  • @mth: Your question got me hooked ( to some degree. I'll stop and wait for a math person's feedback ) but I added something to the bottom of the answer that helped me understand things more clearly. – mark leeds Jun 27 '23 at 00:43
  • Thanks for all your help @mark leeds. I think i got it now. In a geometric series the Numerator has to be a constant as well. Thats not given here since $\epsilon$ is a random variable. Without your help i would't have get it. Thanks so much. – mth Jun 27 '23 at 12:39
  • @mth: I think that's the best way to think about it. I'm glad it helped. – mark leeds Jun 27 '23 at 15:18