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?

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

  • $\begingroup$ 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


$Y_n$ 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

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


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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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