Convergence of $E[N/Q \mid Q>0]$ when $Q \sim B(N,p)$ 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 $\mathbb{E}[N/Q \mid Q>0]\rightarrow 1/p$.
 A: 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
$$ \lim_{n\rightarrow\infty}\left|E\left[\frac{n}{Q_n}|Q_n>0\right] - 1/p\right| = 0 $$
$\Box$

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 $n$.
