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I am trying to prove the following:

Let $X_t$ be a sequence of random variables that follows an autoregressive process; i.e. $X_t=X_{t-1}+e_t$, where $e_t$ is a zero mean i.i.d. sequence. Then $\lim\limits_{n\to \infty} \mathbf E\left(\left(\sum\limits_{i=1}^n (X_t-\bar X_t)^2\right)^{-1}\right)$ converges in probability to 0, where $\bar X_t$ denotes the sample mean.

So far, I have been able to show that $\mathbf E\left(\sum(X_t-\bar X_t)^2\right)$ converges in probability to infinity, but this alone does not imply the expectation of its reciprocal goes to 0. Are there any results I can use, along with this, to complete my proof?

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5  
Sorry, what's a Plim? –  Gerry Myerson Apr 30 '11 at 3:58
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$\mu_n = \mathbf{E}((\sum\lim_{i=1}^n (X_t-\bar X_t)^2)^{-1})$ is a sequence of numbers. There is no notion of convergence in probability in your statement. –  cardinal Apr 30 '11 at 23:32
    
Ewain: you could show the part of the proof you said you did, this could clarify the question you ask. –  Did May 20 '11 at 17:59
    
Traditionally (well, in my experience) the "autoregressive" processes are the stable ones (poles inside the unit circle). This would be more commonly called a random walk process, with continuous jumps. –  leonbloy Jul 19 '11 at 20:37

1 Answer 1

Let $Y_t=X_t-\bar X_t$ and $S_n=\displaystyle\sum_{t=1}^nY_t^2$. Then $Y_1=S_1=0$ and $(S_n)$ is a nondecreasing sequence of nonnegative random variables. If the distribution of the random variables $e_t$ has an atom at zero, then, for every $n$, $S_n=0$ with positive probability hence $1/S_n$ is not integrable. From now on we assume that the distribution of $e_t$ has no atom at zero. Then $Y_2=\frac12(X_2-X_1)=\frac12e_2$ hence $S_2=\frac14e_2^2$ and $S_n>0$ almost surely, for every $n\ge2$. Let us show that $S_n\to+\infty$ almost surely.

For every $t$, $X_t=X_0+e_1+\cdots+e_t$ hence the mean $\bar X_t$ of the values of $X_s$ from $s=1$ to $s=t$ is $\bar X_t=X_0+e_1+(t-1)e_2/t+(t-2)e_3/t+\cdots+e_t/t$, and $$ Y_t=e_2/t+2e_3/t+\cdots+(t-1)e_t/t. $$ Assume that $(S_n)$ is bounded, then $Y_t\to0$. But $Y_{t+1}=tY_t/(t+1)+te_{t+1}/(t+1)$ hence $e_{t+1}\to0$, an event which has probability zero.

Hence $S_n\to+\infty$ almost surely, $1/S_n\to0$ almost surely and $E(1/S_n)\to0$ as soon as one of the expectations $E(1/S_n)$ is finite.

To conclude that at least one random variable $1/S_n$ is integrable, some additional hypothesis on the common distribution of the random variables $e_t$ seem required.

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