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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.

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    This is correct. If you are not supposed to use the Central Limit Theorem, you can use the Characteristic Function.2011-06-23
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    @leonbloy:The question didn't specify. Thanks for verifying.:)2011-06-23
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    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 $N(1,\frac2n)$?).2011-06-24
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    @Didier: Please what can be done to correct that step?2011-06-24
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    What did you want to say when you wrote that $\bar X$ was approximately $N(1,\frac2n)$?2011-06-24
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    @Didier:that $\bar{X}$ was approximately Normally distributed with mean $1$ and variace $\frac{1}{n}$ by the CLT.2011-06-24
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    Your last comment is not a mathematical statement. The fact that some random variables $Y_n$ are approximately $N(1,7)$ (or better said, that $(Y_n)$ converges in distribution to $N(1,7)$) has a precise mathematical meaning. But what is the meaning of what you write? Forget about probability theory for a moment and consider the statement that some sequence $x_n$ converges to $x$ in a given metric space. This means that $d(x_n,x)\to0$, right? But you say something similar to *$x_n$ is approximately $y_n$*. Does this mean that $d(x_n,y_n)\to0$? Then $x_n=1/n$ is approximately $y_n=3/n$...2011-06-24
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    So what I should have was that as $n\rightarrow \infty$, $\bar{X}$ converges in distribution to $N(1,\frac{2}{n})$...right?2011-06-24
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    @Nana let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/617/discussion-between-didier-piau-and-nana)2011-06-24
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    @Nana: Any progress on your problem?2011-07-10
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    @Didier: I'm afraid not...:)2011-07-10
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    @Nana: In that case... any progress on the answer to any of the (specific) questions asked in my comments?2011-07-10
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    @Didier: I think I answered one of the questions in an earlier comment. If you don't mind, would you tell me how to approach it better.2011-07-10
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    Which one (question) did you answer (in an earlier comment)?2011-07-10

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