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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.

share|improve this question
    
This is correct. If you are not supposed to use the Central Limit Theorem, you can use the Characteristic Function. –  leonbloy Jun 23 '11 at 1:46
    
@leonbloy:The question didn't specify. Thanks for verifying.:) –  Nana Jun 23 '11 at 2:12
    
This is of course the rough idea. However the step asserting that *$\bar X$ is approximately $N(1,\frac2n)$* cannot be considered as correct (what does that mean to say that some random variables $Y_n$ are approximately $N(1,\frac2n)$?). –  Did Jun 24 '11 at 6:00
    
@Didier: Please what can be done to correct that step? –  Nana Jun 24 '11 at 6:03
    
What did you want to say when you wrote that $\bar X$ was approximately $N(1,\frac2n)$? –  Did Jun 24 '11 at 6:05

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