Modification of the law of large numbers for Binomial random variables.

Question

Let $(p_n)_{n \geq 1}$ be a sequence of numbers in $[0,1]$ such that $p_n \to p$. Let $(X_n)_{n \geq 1}$ be a sequence of independent random variables where $X_n \sim Bin(n,p_n)$. Is it true that $X_n/n \to p$ almost surely as $n \to \infty$? In such case, how can it be proved?

Answer 1

Ok, I think I have the answer using the following:

Lemma (approximation): Let $\eta, (\eta_n)_{n \geq 1}, (\rho_k)_{k \geq 1}$ and $(\rho_{k,n})_{k,n \geq 1}$ be random elements taking values in some metric space $(S,d)$ such that $\rho_{k,n} \to \rho_k$ in distribution (as $n \to \infty$, for $k$ fixed), and $\rho_k \to \eta$ in distribution (as $k \to \infty$). Then $\eta_n \to \eta$ in distribution (as $n \to \infty$) holds under the further condition

\begin{equation} \lim_{k \to \infty} \limsup_{n \to \infty} E[d(\rho_{k,n},\eta_n)\wedge 1] = 0. \end{equation}

(The proof of the above lemma can be found in Kallenberg's Foundations of Modern Probability)

Here's my proof

Let $Z^{i}_k \sim Ber(p_k)$ independently for $i,k \geq 1$. Define $Y_{n,k} = \sum_{i=1}^{n} Z^{i}_k$ for $k,n \geq 1$ so that $X_n \overset{d}{=} Y_{n,n}$. Now by the strong law of large numbers $Y_{n,k}/n \to p_k$ a.s. (as $n \to \infty$) an by hypothesis $p_k \to p$, (notice that above limits aslo hold in distrbution).

Now,

\begin{align} E\left[\left(\frac{Y_{n,k}}{n}-\frac{Y_{n,n}}{n}\right)^2\right] & = \frac{1}{n^2}\left( E[(Y_{n,k})^2] - 2E[Y_{n,k}]E[Y_{n,n}] + E[(Y_{n,n})^2]\right)\\ & = \frac{1}{n^2}\left( np_k(1-p_k)+(np_k)^2 - 2n^2p_kp_n + np_n(1-p_n)+(np_n)^2\right)\\ & = \frac{p_k(1-p_k)+p_n(1-p_n)}{n} + (p_k-p_n)^2 \end{align}

Thus \begin{align} \lim_{k \to \infty} \limsup_{n \to \infty}E\left[\left(\frac{Y_{n,k}}{n}-\frac{Y_{n,n}}{n}\right)^2\right] = \lim_{k \to \infty} (p_k-p)^2 = 0 \end{align}

By the approximation lemma

\begin{equation} \frac{X_n}{n} \overset{d}{=} \frac{Y_{n,n}}{n} \to p \quad \text{ in distribution.} \end{equation}

Finally, since $p$ is constant, then we can conclude $X_n/n$ converges a.s. to $p$.

Can anyone tell if this is correct?