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Let $X_1, X_2,...$ be IID rv with $E[X_1] = 5$ and $Var[X_1] = 9$ Find the limiting distribution of

$\sqrt{n} \frac{(X_1 + X_2 + ... + X_n - 5n)}{(X_1^2 + X_2^2 + ... + X_n^2)}$

as $n \rightarrow \infty$. Explicate your argument.

So we can rewrite

$\sqrt{n} \frac{(X_1 + X_2 + ... + X_n - 5n)}{(X_1^2 + X_2^2 + ... + X_n^2)}$ = $\frac{\frac{1}{\sqrt{n}}(X_1 + X_2 + ... + X_n - 5n)}{\frac{1}{n}(X_1^2 + X_2^2 + ... + X_n^2)}$

And use the central limit theorem on the nominator. However can anyone explain why the upper part becomes

$3 \cdot \frac{1}{3\sqrt{n}}(X_1 + X_2 + ... + X_n - 5n) \rightarrow N(0, 9)$

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  • $\begingroup$ Because $\sigma^2=9$ and $\frac{1}{\sqrt{n\sigma^2}}(X_1 + X_2 + ... + X_n - 5n) \rightarrow N(0, 1)$. $\endgroup$ – Did Dec 30 '15 at 17:08
  • $\begingroup$ Yes but why do I have to multiply by $3$ and not only use $N(0,1)$? $\endgroup$ – Ptru Dec 30 '15 at 17:48
  • $\begingroup$ This is the same statement: $Z_n\to N(0,1)\iff 3Z_n\to N(0,9)$. $\endgroup$ – Did Dec 31 '15 at 7:23

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