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Assume that $X_n\sim Beta(n,n)$. Use the Delta method to show that $$ 2\sqrt{2n}(X_n-\frac{1}{2})\to^d N(0,1) $$

Since $X_n=\frac{Y_n}{Y_n+Z_n}$ where $Y_n$ and $Z_n$ are indepedndent Gamma$(n,1)$ random variables [Why?]. Each of $Y_n$ and $Z_n$ can be written as a sum of exponenetial random variables. From multivariate CLT, we have asymptotic distribution $$ \sqrt{n}([Y_n, \, Z_n]^T-[EY_1, \, EZ_1]^T)=\sqrt{n}([Y_n, \, Z_n]^T-[n, \, n]^T)\to^d N(0, \Sigma) $$ where $\Sigma=\begin{bmatrix} Var[Y_1]&Cov(Z_1, Y_1)\\ Cov(Z_1, Y_1) & Var[Z_1]\end{bmatrix}=\begin{bmatrix} n&0\\ 0 & n\end{bmatrix}$.

From the Delta method, take $f(y,z)=\frac{y}{y+z}$ with $f_y(y,z)=\frac{z}{(y+z)^2}$ and $f_z(y,z)=\frac{-y}{(y+z)^2}$ $$ 2\sqrt{2n}(X_n-\frac{1}{2})=2\sqrt{2n}(f(Y_n, Z_n)-f(n,n))\to^d [\frac{1}{4n}, -\frac{1}{4n}]\times N(0,\Sigma) $$

Question: I got the variance is $1/n$ but not 1?

$$ [\frac{1}{4n}, -\frac{1}{4n}]\begin{bmatrix} n&0\\ 0 & n\end{bmatrix}[\frac{1}{4n}, -\frac{1}{4n}]^T=1/8n $$

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  • $\begingroup$ What is the question? $\endgroup$
    – wolfies
    Commented Mar 11, 2022 at 5:24
  • $\begingroup$ @wolfies I got the limiting distribution is $N(0,1/n)$? $\endgroup$
    – Hermi
    Commented Mar 11, 2022 at 5:40

1 Answer 1

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$Y_n$ and $Z_n$ are independent $\Gamma(n,1)$ random variables, which here means that both of them can be written as the sum of $n$ independent $\text{Exp}(1)$ random variables.

So by multivariate CLT,

$$\sqrt n\left(\begin{pmatrix}\frac{Y_n}{n} \\ \frac{Z_n}{n}\end{pmatrix}-\begin{pmatrix}1 \\ 1\end{pmatrix}\right) \stackrel{d}\longrightarrow N_2\left(\begin{pmatrix}0 \\ 0\end{pmatrix}, I_2\right)$$

The $\begin{pmatrix}1 \\ 1 \end{pmatrix}$ vector and identity matrix $I_2$ comes from the mean and variance of an $\text{Exp}(1)$ variable.

Take $g(x,y)=\frac{x}{x+y}$, whence by delta method,

$$\sqrt n\left(g\left(\frac{Y_n}{n},\frac{Z_n}{n}\right)-g(1,1)\right) \stackrel{d}\longrightarrow N \left(0, (\nabla g(1,1))^T (\nabla g(1,1))\right)$$

Now the gradient vector of $g$ is

$$\nabla g(x,y)=\begin{pmatrix}\frac{y}{(x+y)^2} \\ -\frac{x}{(x+y)^2}\end{pmatrix}\,,$$

which implies

$$\nabla g(1,1)=\frac14\begin{pmatrix}1 \\ -1\end{pmatrix}$$

Hence,

$$\sqrt n\left(\frac{Y_n}{Y_n+Z_n}-\frac12\right) \stackrel{d}\longrightarrow N \left(0, \frac18\right)$$

Or,

$$2\sqrt{2n}\left(X_n-\frac12\right) \stackrel{d}\longrightarrow N \left(0, 1\right)$$

Also see If $X_n \sim \text{Beta}(n, n)$ Show that $[X_n - \text{E}(X_n)]/\sqrt{\text{Var}(X_n)} \stackrel{D}{\longrightarrow} N(0,1)$ for other methods.

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