# Find the limiting distribution of Bernoulli random variables

If $X_1, X_2,\ldots$ are i.i.d Bernoulli random variables with mean $\frac{1}{2}$. Define $T_n = \sqrt{n}(\frac{4\sum_{i=1}^{n} X_i - 2n}{\sum_{i=1}^{n} X_i^2})$. Find the pmf or pdf of the limiting distribution of the sequence $T_1, T_2,\ldots$

My thought: We consider the function $g(x) = 4x$. Since this function and its derivative are continuous, and both are nonzero at $u_x = \frac{1}{2}$, we could apply the theorem saying $\sqrt{n}[\frac{g(\overline{X_n})- g(\frac{1}{2})}{|g'(\frac{1}{2})\sigma_{x}}]\rightarrow N(0,1)$ in distribution, where $X_n = \frac{X_1+\ldots + X_n}{n}$ and $\sigma_{x} = \frac{1}{4}$ (since $X$ is Bernoulli). Now, my plan is to figure out the convergence in probability of $\sum_{i=1}^{n} X_i^2$, but I am stucked here.

Could anyone please help with this last piece of puzzle?

## 1 Answer

You can use the law of large numbers to talk about what happens to $\frac{1}{n} \sum_{i=1}^n X_i^2$. Note that you divided the numerator of $T_n$ by $n$, so you should also do that for the denominator.

• I'm so dumb:( We only need to show $E(X_i^2) < \infty$ which is obvious since it equals to $\frac{3}{4}$, and $Var(X_i^2) < \infty$. Since $X_i^2$ are i.i.d because $X_i$ are i.i.d, $\frac{1}{n} \sum_{i=1}^{n} X_i^2\rightarrow \frac{3}{4}$ in probability. Thus the final result is convergence in distribution to $\frac{3}{4}N(0,1)$, so the limiting distribution has pdf $\frac{3}{4}\frac{1}{\sqrt{2\pi}}e^{-x^2}$. Is this correct? – user177196 Mar 2 '17 at 7:07
• $E[X_1^2]=1/2$. Also, the pdf you have written does not integrate to $1$. [More specifically, if $X$ has pdf $p_X(x)$, then the pdf of $cX$ is not $cp_X(x)$.] – angryavian Mar 2 '17 at 7:12
• Thank you for your patience. Is the pdf equal to $N(0,\frac{3}{4})$? – user177196 Mar 2 '17 at 7:22
• @user177196 If $Z$ is standard normal, then $cZ \sim N(0,c^2)$. – angryavian Mar 2 '17 at 7:24
• @user177196 Confirm: this is true. – NCh Mar 2 '17 at 10:49