# Assume that $X_n\sim Beta(n,n)$, how to drive the limiting distribution of $2\sqrt{2n}(X_n-\frac{1}{2})\to^d N(0,1)$

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?

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

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

$$Y_n$$ 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,

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