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I have independent $X_i$ with mean $0$ and variance $1$. They are normally distributed.

From the $X_i$ I construct $Y_i$ = $X_i$ + $X_{i+1}$. The $Y_i$ aren’t independent.

I’m curious to know the distribution of $\frac{\sum_{i=1}^n Y_i}{\sqrt{n}}$. Since the $Y_i$ aren’t independent I don’t think I can use CLT. If I break down the sum into a sum of $X_i$, then I can’t get rid of the n term in the variance. Is there a better solution?

Anonymous
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  • there are extensions of the CLT for dependent variables. https://link.springer.com/article/10.1007/BF01025872 https://link.springer.com/chapter/10.1007/978-1-4612-0865-5_9 – herb steinberg Jun 11 '18 at 21:21

1 Answers1

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Notice that $$\sum_{i=1}^n Y_i = \sum_{i=i}^{n} X_i + X_{i+1} = X_1 + X_{n+1} +\sum_{i=2}^{n} 2X_i.$$

It follows that $$\frac{\sum_{i=1}^n Y_i}{\sqrt{n}} = \frac{\sum_{i=1}^n 2 X_i}{\sqrt{n}} + \frac{- X_1 +X_{+1}}{\sqrt{n}}.$$ It's easy to see that $\frac{- X_1 +X_{+1}}{\sqrt{n}} \rightarrow 0$ almost surely. Also, by the central limit theorem, $\frac{\sum_{i=1}^n 2 X_i}{\sqrt{n}} \rightarrow \mathcal{N}(0, 4)$ in distribution.

By using slutzky's theorem, one gets $$\frac{\sum_{i=1}^n 2 X_i}{\sqrt{n}} + \frac{- X_1 +X_{+1}}{\sqrt{n}} \rightarrow Z + 0$$ in distribution, where $Z \sim \mathcal{N}(0, 4)$.

cdwe
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