Show that $\frac{X_n-nm}{\sqrt{nm}} \xrightarrow{d}N(0,1) , X_n \in Bin(n^2,m/n),m>0$

Question

Let $X_n \in Bin(n^2,m/n),m>0$. I want to show that $$\frac{X_n-nm}{\sqrt{nm}} \xrightarrow{d}N(0,1) $$ as $n \to \infty$

Im stuck in this problem. I started by checking the limit distribution of $X_n$ and hope to be able to use the CLT, but cant quite get event that.

$$P(X_n=k)=\frac{(n^2)!}{k!(n^2-k)!}(\frac{m}{n})^k(1-\frac{m}{n})^{n^2-k}=$$ $$\frac{m^k}{k!}\frac{n^2!}{(n^2-k)!}(1-\frac{m}{n})^{n^2}$$

But what is $$\lim_{n\to \infty} \frac{n^2!}{(n^2-k)!}(1-\frac{m}{n})^{n^2}$$

I had a feeling that it converges to $e^{-m}$ and tried to reach that conclusion but with no success. Any tips? Thanks

Answer 1

CLT tells you that if $Z_i$ are i.i.d, then $$\sqrt{n}(\frac{1}{n}\sum_{i=1}^n Z_i - \mu) \rightarrow \mathcal{N}(0,\sigma^2)$$ where $\mu = E(Z_i)$ and $\sigma^2 = var(Z_i)$. Now, notice that a sum of Bernoulli is Binomial, i.e. $$X_n = \sum_{i=1}^{n^2} Z_i$$ where $Z_i$ is Bernoulli with rate $p = \frac{m}{n}$. So this means that by CLT $$\sqrt{n^2}(\frac{1}{n^2}\sum_{i=1}^{n^2} Z_i - \mu) \rightarrow \mathcal{N}(0,\sigma^2)$$ where $\mu = p = \frac{m}{n}$ and $\sigma^2 = p(1-p) = \frac{m}{n}(1 - \frac{m}{n}) = \frac{m(n-m)}{n^2}$. Hence $$n(\frac{1}{n^2}X_n- \frac{m}{n}) \rightarrow \mathcal{N}(0,\frac{m(n-m)}{n^2})$$ i.e. $$n(\frac{1}{n^2}X_n- \frac{m}{n}) \rightarrow \sqrt{\frac{m(n-m)}{n^2}}\mathcal{N}(0,1)$$ or $$\sqrt{\frac{n^2}{m(n-m)}}n(\frac{1}{n^2}X_n- \frac{m}{n}) \rightarrow \mathcal{N}(0,1)$$ In the large regime $n >> m$ (since $n \rightarrow \infty$), then $$\sqrt{\frac{n^2}{nm}}n(\frac{1}{n^2}X_n- \frac{m}{n}) \rightarrow \mathcal{N}(0,1)$$ or $$n\sqrt{\frac{1}{nm}}(\frac{1}{n}X_n- m) \rightarrow \mathcal{N}(0,1)$$ which concludes the proof. $$\sqrt{\frac{1}{nm}}(X_n- nm) \rightarrow \mathcal{N}(0,1)$$

Answer 2

First note that since the Binomial r.v. is obtained through a sum of independent Bernoulli random variables, we can write $X_n$ as \begin{align*} X_n = \sum_{i=1}^{n^2} Y_i, \end{align*} where $Y_i \sim \text{Bernoulli}(m/n)$. Now define \begin{align*} Z_{i,n} \doteq \frac{Y_i - m/n}{\sqrt{nm}}. \end{align*} To show that $\sum_{i=1}^{n^2} Z_{i,n} \Rightarrow N(0,1)$, it suffices to check the Lindeberg conditions in order to apply the CLT. These are,

(i) $\sum_{i=1}^{n^2} EZ_{i,n}^2 \to 1$,

(ii) $\sum_{i=1}^{n^2} EZ_{i,n}^2\textbf{1}_{\{|Z_{i,n}| > \varepsilon\}} \to 0$ for any $\varepsilon > 0$.

Indeed, \begin{align*} \sum_{i=1}^{n^2} EZ_{i,n}^2 &= \sum_{i=1}^{n^2} \frac{1}{nm} E(Y_i - m/n)^2 \\ &= \sum_{i=1}^{n^2} \frac{1}{nm} \cdot m/n(1-m/n) \\ &= \sum_{i=1}^{n^2} \frac{(1-m/n)}{n^2} \\ &= 1-m/n \\ &\to 1, \end{align*} as $n\to\infty$, since $m$ is fixed. For the second condition, note that \begin{align*} |Z_{i,n}| > \varepsilon \iff |Y_i - m/n| > \varepsilon \sqrt{nm}. \end{align*} But since $m$ is fixed, $m/n$ is eventually always less than 1. Hence $|Y_i - m/n| < 2$ almost always. This means that after some $n$, we will have $\varepsilon \sqrt{nm} > 2$ and $|Y_i - m/n| < 2$ almost always, so the indicator will be $0$ for every term in the sum. Hence the entire sum is equal to $0$, almost always (there exists an $N$ after which the statement is always true).

Hence, by the Lindeberg CLT, we have the desired convergence in distribution.