0
$\begingroup$

Let $(X_n)_{n \in \Bbb N}$ be i.i.d. ; $\Bbb E [X_1] = 0$; $0 \lt \mathrm{Var}[X_1]\lt \infty$. Now i want to show that $$Y_n := \frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}} \xrightarrow[]{d} \mathcal N(0,1)$$ I know that convergence in distribution is equivalent to $$P(Y_n \le x)\xrightarrow[]{n\to \infty}P(Z \le x)$$ with $Z$ having a $\mathcal N(0,1)$ distribution.
I don't really know how to start, can someone give me a hint on how to show this? Thanks in advance!

$\endgroup$
  • 3
    $\begingroup$ This follows by mixing 3 theorems: classical Central Limit Theorem, Law of Large Numbers, and the Slutsky's theorem. $\endgroup$ – Sangchul Lee Jul 18 '17 at 10:47
2
$\begingroup$

Let $\sigma^2=Var(X_1)$ ... then just a little hint: \begin{align} \frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}}&=\frac{\sigma\sqrt n}{\sqrt{\sum_{k=1}^n X_k^2}}\frac{\sum_{k=1}^n X_k}{\sigma\sqrt n}\\ &=\frac{\sigma}{\sqrt{\frac1n\sum_{k=1}^n X_k^2}}\frac{\sum_{k=1}^n X_k}{\sigma\sqrt n}\\ \end{align}

Now show using an LLN that $\frac{\sigma}{\sqrt{\frac1n\sum_{k=1}^n X_k^2}}\to1$ and use a classic CLT to establish $\frac{\sum_{k=1}^n X_k}{\sigma\sqrt n}\to \mathcal{N}(0,1)$.

$\endgroup$
1
$\begingroup$

$$ \frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}} = \frac{\frac{1}{\sqrt{n}}\sum_{k=1}^n X_k}{\sqrt{\frac{1}{n}\sum_{k=1}^n X_k^2}} = \frac{\sqrt{n}\bar{X}_n}{\sqrt{\frac{1}{n}\sum_{k=1}^n X_k^2}}. $$ Where by WLLN and the continuous mapping theorem $\sqrt{\frac{1}{n}\sum_{k=1}^n X_k^2} \xrightarrow{p} \sqrt{EX^2}=\sqrt{Var(X)}$, and by CLT $$ \sqrt{n}\bar{X}_n \xrightarrow{D}N(0,Var(X)), $$ hence by Slutsky $$ \frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}} \xrightarrow{D} \frac{N(0, Var(X))}{\sqrt{Var(X)}} = N(0,1). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.