Convergence of $E[N/Q \mid Q>0]$ when $Q \sim B(N,p)$

Question

Let $Q$ be a binomial random variable with parameters $N$ and $p$.

We know, of course, that $N/Q \rightarrow_p 1/p$ when $N \rightarrow \infty$.

I was wondering whether we also have that $\mathbb{E}[N/Q \mid Q>0]\rightarrow 1/p$.

Answer 1

Yes, assuming $p>0$. Here is one way to prove.

Assume $p>0$. Let $\{Y_i\}_{i=1}^{\infty}$ be i.i.d. Bernoulli-$p$ variables. Define $Q_n = \sum_{i=1}^n Y_i$. Then $Q_n \sim Binomial(n,p)$. We want to show $E[n/Q_n|Q_n>0]\rightarrow 1/p$. Define $$ X_n = \left\{\begin{array}{ll} \left|\frac{n}{Q_n} - \frac{1}{p}\right| & \mbox{ if $Q_n>0$} \\ 0 & \mbox{ else} \end{array}\right.$$ By the law of large numbers we know that $$X_n\rightarrow 0 \quad \mbox{(with prob 1)}$$ It can be shown that there is a finite constant $M$ such that (see below) $$ E[X_n^2] \leq M \quad \forall n \in \{1, 2, 3, ...\}$$ and so $X_n$ is uniformly integrable and thus $$ \lim_{n\rightarrow\infty} E[X_n] = E[\lim_{n\rightarrow\infty} X_n] = 0 $$ Thus \begin{align} 0 &= \lim_{n\rightarrow\infty} E[X_n] \\ &= \lim_{n\rightarrow\infty} E[X_n|Q_n>0]P[Q_n>0] \\ &= \lim_{n\rightarrow\infty} E[X_n|Q_n>0](1-(1-p)^n) \end{align} Since $1-(1-p)^n\rightarrow 1$, it follows that $$ \lim_{n\rightarrow\infty} E[X_n|Q_n>0] = 0$$ and so $$ \lim_{n\rightarrow\infty} E\left[\left| \frac{n}{Q_n} - \frac{1}{p}\right| | Q_n>0\right] = 0 $$ and so by Jensen's inequality for the convex function $|x|$ we have $$ \lim_{n\rightarrow\infty}\left|E\left[\frac{n}{Q_n}|Q_n>0\right] - 1/p\right| = 0 $$ $\Box$

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Footnote: Proof that $E[X_n^2]\leq M$ for all $n$ (for some finite constant $M$): Observe that $$ X_n^2\leq \max\left\{n^2, \frac{1}{p^2}\right\}$$ So $$ E[X_n^2] = \underbrace{E[X_n^2|Q_n>n/(2p)]}_{\leq 4p^2} P[Q_n>n/(2p)] + \underbrace{E[X_n^2|Q_n\leq n/(2p)]}_{\leq \max\{n^2, 1/p^2\}}P[Q_n\leq n/(2p)]$$ Thus $$ E[X_n^2] \leq 4p^2 + \max\{n^2, 1/p^2\}P[Q_n\leq 1/(2p)]$$ By the Chernov-Hoeffding inequality: \begin{align} P[Q_n\leq n/(2p)] &\leq P[|Q_n-n/p|\geq n/(2p)]\\ &\leq 2\exp(-\frac{2(n/(2p))^2}{n}) \end{align} and so $\max\{n^2, 1/p^2\}P[Q_n\leq 1/(2p)] \rightarrow 0$. It follows that the sequence $$\max\{n^2, 1/p^2\}P[Q_n\leq 1/(2p)], \quad n \in \{1, 2, 3, ...\}$$ is upper-bounded by some positive constant $c$ for all $n$, so $E[X_n^2]\leq 4p^2+c$ for all $n$.