# If $X_n \sim \text{Beta}(n, n)$, show that $[X_n - \text{E}(X_n)]/\sqrt{\text{Var}(X_n)} \stackrel{D}{\longrightarrow} N(0,1)$

Let $$X_n \sim \mathbf{B}(n,n)$$ (Beta distribution), with pdf

$$f_n(x) = \frac{1}{\text{B}(n,n)}x^{n-1}(1 - x)^{n-1},~~ x \in (0,1).$$

Knowing that $$\text{E}(X_n) = 1/2$$ and that $$\text{Var}(X_n) = 1/[4(2n+1)]$$, prove that

$$2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \stackrel{D}{\longrightarrow} N(0,1).$$

I thought about doing it by the definition of convergence in distribution, but the cdf of $$2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}})$$ is obscene. I wouldn't know how to calculate the limit $$\text{lim}_{n \to \infty} F_{Y_n}(x)$$ where $$Y_n = 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}})$$.

Then I thought about proving convergence in probability, since converge in probability $$\Rightarrow$$ convergence in distribution. The problem is that it may not even converge in probability so it would be wasted work.

Edit:

I did some work and this is where I'm at:

Definition. A sequence of random variables $$X_1, X_2, ...$$, converges in distribution to a random variable X if

$$\text{lim}_{n \to \infty} F_{X_n}(x) = F_X(x)$$

So we have to prove that

$$\text{lim}_{n \to \infty} F_{Y_n}(x) = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy$$

Where $$Y_n = 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}})$$.

Now,

\begin{align} P(Y_n \leq x) & = P(2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \leq x) \\ & = P(X_n - 1/2 \leq \frac{x}{2\sqrt{2n+1}} \\ & = P(X_n \leq \frac{x}{2\sqrt{2n+1}} + 1/2) \\ & = F_{X_n} \Bigl( \frac{x}{2\sqrt{2n+1}} + \frac{1}{2} \Bigr) \\ & = \frac{1}{B(n,n)}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt \end{align}

We use Stirling's approximation to $$\text{B}(n,n)$$:

$$B(a, b) \approx \sqrt{2\pi} \frac{a^{a - 1/2}b^{b - 1/2}}{(a + b)^{a + b - 1/2}}$$

So $$\text{B}(n, n) \approx \frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}$$, after simplification.

Substituting the Stirling approximation (we do this because it converges asymptotically and we're taking the limit), we obtain

$$\frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt.$$

So what's left to do is prove that

$$\text{lim}_{n \to \infty} \frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy.$$

Edit 2: I asked my professor for guidance on how to finish the last step. All he said was "apply the limit theorem to solve directly".

• Does the proof given in math.wm.edu/~leemis/chart/UDR/PDFs/BetaNormal.pdf help? It seems easier to work with the probability density function and apply Scheffé’s lemma. Sep 30, 2020 at 11:02
• I'm not at all familiar with Scheffé's lemma. It's not material I've seen in my Statitical Inference course so I wouldn't be allowed to use it. But I will take a look at the proof, maybe it helps in something. Sep 30, 2020 at 11:11
• math.stackexchange.com/questions/1982321/…
– user140541
Sep 30, 2020 at 19:03
• Also asked at stats.stackexchange.com/q/489814/119261. Oct 3, 2020 at 5:33
• @StubbornAtom I really needed an answer and here on MSE I wasn’t getting any. Oct 3, 2020 at 10:27

The answer is in portuguese because I'm a native portuguese speaker.

O último cálculo na demonstração acima é um problema computacional excessivamente difícil. Aqui a ideia é apresentarmos uma demonstração alternativa, que se dá pelos seguintes passos:\

1º: Mostramos que a densidade de $$Y_n$$ converge para a densidade de $$Z$$, onde $$Z \sim N(0, 1).$$\

2º: Invocamos o \textit{Lema de Scheffé} para terminar a demonstração. O Lema de Scheffé é um resultado em Teoria da Medida que, no nosso caso, implica que se $$f_{Y_n}(x) \longrightarrow f_Z(x)$$, então $$F_{Y_n}(x) \longrightarrow F_Z(x)$$, provando a definição Convergência em Distribuição. Em resumo, temos um trabalho facilitado por causa de um resultado mais forte e sofisticado.\

Muito bem, ao diferenciar as equações (1) e (4), obtemos $$f_{Y_n}(x) = f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}}.$$

Agora temos que demonstrar que

$$\text{lim}_{n \to \infty} f_{Y_n}(x) = \text{lim}_{n \to \infty} f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}} = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$$

De fato,

\begin{align*} f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}} & = \frac{1}{2\sqrt{2n+1}} \frac{1}{B(n,n)} (\frac{x}{2\sqrt{2n+1}} + \frac{1}{2})^{n-1} (\frac{1}{2} - \frac{x}{2\sqrt{2n+1}})^{n-1} \\ & = \frac{1}{2\sqrt{2n+1}} \frac{1}{B(n,n)}(\frac{1}{4} - \frac{x^2}{4(2n+1)})^{n-1} \\ & = \frac{1}{2\sqrt{2n+1}} \frac{2^{2n - 1} \sqrt{n}}{\sqrt{\pi}} (\frac{1}{4} - \frac{x^2}{4(2n+1)})^{n-1} \\ & = \frac{1}{2\sqrt{2n+1}} \frac{2^{2n - 1} \sqrt{n}}{\sqrt{\pi}} (\frac{1}{4})^{n-1}(1 - \frac{x^2}{2n-1})^{n-1} \\ & = \frac{1}{\sqrt{\pi}} \sqrt{\frac{n}{2n + 1}}(1 - \frac{x^2}{2n-1})^{n-1}. \end{align*}\

Aplicando o limite $$\text{lim}_{n \to \infty}$$ na úlima expressão acima, obtemos

\begin{align*} \text{lim}_{n \to \infty} \frac{1}{\sqrt{\pi}} \sqrt{\frac{n}{2n + 1}}(1 - \frac{x^2}{2n-1})^{n-1} &= \frac{1}{\sqrt{\pi}} \frac{1}{\sqrt{2}} \text{lim}_{n \to \infty} (1 + \frac{(-x^2/2)}{n - \frac{1}{2}})^{n-1} \\ &= \frac{1}{\sqrt{2\pi}}e^{-x/2}. \end{align*}

Mostramos que $$\text{lim}_{n \to \infty} f_{Y_n}(x) = f_Z(x)$$. Agora, pelo \textit{Lema de Schéffe}, temos que $$\text{lim}_{n \to \infty} F_{Y_n}(x) = F_Z(x)$$, o que prova a convergência em distribuição desejada e termina a demonstração.