I'm stuck on one of my exercises. I worked out an solution, which I think is correct, but differs from the given answer. Is my solution correct?
Suppose $\{Z_n\, n\in\mathbb{Z}\}$ is a sequence of uncorrelated r.v.'s, and $E\left[Z_n\right] = 0, Var\left[Z_n\right] = 1$. Define a autoregressive model
$$X_n = \alpha X_{n-1} + Z_n, \ |\alpha|<1$$ Prove that
$$Cov(X_n,X_{n+m}) = \frac{\alpha^{|m|}}{1-\alpha^2}, n, m\in \mathbb{Z}$$
My solution:
First suppose that $m > 0$. Observe that $$X_n = \sum_{i=0}^{n-1} \alpha^{i} Z_{n-i}$$ We have $$ Cov(X_n,X_{n+m}) = \sum_{i=0}^{n-1}\sum_{j=0}^{n+m-1}\alpha^{i+j} Cov(Z_{n-i},Z_{n+m-j}) $$
Note that $Cov(Z_{n-i}, Z_{n+m-j})=1$ if and only if $j = m+i$, i.e. given each $i = 0,\cdots n-1$, there corresponds only one $j = m+i$, which ranges between $m, n+m-1$, s.t. the term of the above sum does not vanish. Therefore,
$$ Cov(X_n,X_{n+m}) = \sum_{i=0}^{n-1}\alpha^{2i+m} = \frac{\alpha^m(1-\alpha^{2n})}{1-\alpha^2} $$
Which definitely differs from the provided formula.
However, I noticed that the result I've got equates the provided formula, if $n \rightarrow \infty$. Therefore,
- Is the exercise wrong, or missing some prerequisites?
- Is my understanding of the exercise, especially of $X_n$, correct? For I in my solution supposed $X_1 = Z_1$.
- Is my deduction wrong? What should the exercise be correctly solved?
P.S. I'm still working on stochastic process and am not familiar of autoregressive model, actually. All I know is from the definition given by the exercise.
You have assumed that $X_0=0$, which violates the autoregressive equation at $n=0$. Instead, the system starts from 0 in the far distant past, and the correction term in your formula disappears.
More precisely, the formula for $X_n$ becomes $$ X_n=\sum_{i=0}^{\infty}\alpha^i Z_{n-i}, $$ and therefore for all $m\geq 0$ by the same argument as yours we obtain that $$ \text{Cov}(X_n,X_{n+m})=\sum_{i=0}^{\infty}\alpha^{2i+m}=\frac{\alpha^m}{1-\alpha^2}. $$