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Suppose we have $X_1,\cdots,X_n$ be a random sample with $X_i\backsim \chi^2(1)$. I'd like to show that as $n\rightarrow \infty$, $\frac{\bar{X}-1}{\sqrt{\frac{2}{n}}}$ is standard normal.

Here's my attempt.

Since $X_i\backsim \chi^2(1)$, we know that $\mathbb{E}(X_i)=1, \text{Var}(X_i)=2.$ As $n\rightarrow \infty $,the Central Limit Theorem says that, $\bar{X}$ is approximately $N(1,\frac{2}{n})$. So $ \frac{\bar{X}-1}{\sqrt{\frac{2}{n}}}\rightarrow Z\backsim N(0,1).$

Does my attempt pass as a genuine solution? Is there way of doing this using Mgf's?

Thanks.

  • 0
    Which one (question) did you answer (in an earlier comment)?2011-07-10

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