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I am trying to prove that the variance of the sample lag-1 autocorrelation $$\hat{\rho}=\frac{\sum_{t=1}^n(x_t-\bar{x})(x_{t-1}-\bar{x})}{\sum_{t=1}^n(x_{t-1}-\bar{x})^2}$$ for an i.i.d. R.V is asymptotically normal with mean $0$ and variance $\frac{1}{n}$. The first two assertions are easy, but I am having trouble showing that the variance is $\frac{1}{n}$.

I have tried using a Taylor expansion on the above expression about the actual variance and covariance and then taking the variance of that expression. My taylor expantion looks like:

$$\frac{SS1}{SS2}=\frac{\sigma_1^2}{\sigma_2^2}+\frac{1}{SS2}*(SS1-\sigma_1^2)-\frac{SS1}{SS2^2}*(SS2-\sigma_2^2)$$

where SS1 is the sample autovariance, SS2 is the sample variance, $$\sigma_1^2$$ is the actual autocovarince, and $$\sigma_2^2$$ is the actual variance. Using the fact that $$\rho=\frac{SS1}{SS2}$$ I was able to get that the variance is

$$1/(T*\sigma_2^2)*(\rho^2+\rho)$$

but this is not the same as $\frac{1}{T}$. My professor has told me that there is a way to prove it using the martingale difference clt and the law of large numbers, but I have not been able to see how that approach would work.

Could someone please help? Thanks, Paul

EDIT: I figured out how to do it the other way, using the martingale difference clt. If we split the summation into $$\hat{\rho}=\frac{\sum_{t=1}^n(x_t-\bar{x})(x_{t-1}-\bar{x})}{\sigma^2}\frac{\sigma^2}{{\sum_{t=1}^n(x_{t-1}-\bar{x})^2}}$$ we can see that the right factor goes to 1 as t->inf. Now we can see that the left factor is a martingale and satisfies the criteria for the Maritingale difference CLT (all relatively easy proofs, though it takes a few step to show that the left factor is a martingale, and you need to use the fact that $$\mathbb{E}(x^2)<\inf$$, an assumption which I didn't mention before).

I am still not sure what went wrong with my Taylor series expansion idea...if the result has instead been $$1/(T*\sigma^2)*(\rho^2+\rho+1)$$ setting $$\rho=0$$ (since x is i.i.d.) would yield the desired result. I will give the answer to anyone who can do it this way.

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