Bounds on the conditional variance of a truncated binomial I have a binomial variable $R$ drawn from $binom(N, p)$, and I'm interested in the variance of $R$, given $R \ge Q$. The pmf of this variable $R^*$ is
$$
\phi_{R^*}(l) = \frac{\phi(l, N, p)}{P}
$$
where $P= \sum_{l=Q}^G\phi(l, N, p)$ is the probability that $R \ge Q$, and $\phi$ is the binomial pmf.
Are there any bounds on the variance of $R^*$ compared to the variance of $R$, which is $Np(1-p)$? I'm sure it's less, but can we say how much less as a function of $Q$?
My ultimate goal is to prove that
$$
Var(R^*) \le \frac{N-ER^*}{N-ER}Var(R)
$$
where $ER^*$ is the conditional mean of $R$ given that $R \ge Q$, and $ER = Np$ is the unconditional mean.

The conjecture is true for all sufficiently large $N$, namely when $N$ is large enough to use a normal approximation to the binomial, and
$$N > \frac{\pi p}{2(1-p)}\left(\sqrt{\pi}q - \frac{\exp(-q^2)}{\text{erfc}(q)}\right)^{\!-2}$$
where

*

*$q=(Q-Np)/\sqrt{2Np(1-p)}$, which is $1/\sqrt{2}$ times $Q$'s standard deviations above the mean, and

*$\text{erfc}$ is the complementary error function.

The normal distribution to the binomial has mean $Np$ and variance $Np(1-p)$. Denoting its pdf by $f$, we can approximate
\begin{align}
ER^*&\simeq\frac{M_1}{M_0}\\
Var(R^*)&\simeq\frac{M_2}{M_0}-ER^{*2}
\end{align}
using the partial moments
\begin{align}
M_0=\int_Q^\infty f(x)dx = & \frac12\text{erfc}(q)\\
M_1=\int_Q^\infty xf(x)dx = & \frac{Np}{2}\text{erfc}(q) + \sqrt{\frac{Np(1-p)}{2\pi}}\exp(-q^2)\\
M_2=\int_Q^\infty x^2f(x)dx = & \frac{Np}{2}\text{erfc}(q)(1-p+Np)\\
& +\frac{Np\exp(-q^2)}{\sqrt{\pi}}(q-pq+\sqrt{2Np(1-p)})
\end{align}
Under these approximations, the inequality in the question is equivalent to the first inequality above.
The function $\left(\sqrt{\pi}q - \frac{\exp(-q^2)}{\text{erfc}(q)}\right)^{\!-2}$ has the following graph:

which is asymptotic to $(4q^2+8)/\pi$ at $+\infty$.
