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.
 A: I need to change the notation a little to make it a little more consistent with stats references and easier to follow.  I'll switch to the orginal poster's notation at the end.
$$X \sim binomial(n, p)$$
$$f(x,n,p) = {n \choose p} p^x (1-p)^{n-x}$$
$$F(x,n,p) = P(X \le x) = \sum_{k=0}^{x} f(k,n,p)$$
For the truncated distribution, you can find many references to show this:
$$f(x,n,p|a < X \le b) = \frac{g(x,n,p)}{F(b) - F(a)}$$
where $g(x,n,p) = f(x,n,p)$ if $a < x \le b$ and $g(x,n,p) = 0$ otherwise.

$$E(X | a < X \le b) = \sum_{a+1}^b xf(x,n,p|a<X\le b) \\
= \frac{1}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b x {n \choose p} p^x (1-p)^{n-x} \\
= \frac{np}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b \frac{(n-1)!}{(n-x)!(x-1)!} p^{x-1} (1-p)^{n-x} \\ 
= \frac{np}{F(b,n,p)-F(a,n,p)} \sum_{a}^{b-1} \frac{(m)!}{(m-y)!(y)!} p^{y} (1-p)^{m-y}  \ \ where\ \ y=x-1\ \ and\ \  m=n-1 \\
= \frac{np}{F(b,n,p)-F(a,n,p)} [F(b-1,n-1,p) - F(a-1,n-1,p)]$$

$$E(X(X-1) | a<X\le b) = \sum_{a+1}^b x(x-1)f(x,n,p|a<X\le b) \\
= \frac{1}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b x(x-1) {n \choose p} p^x (1-p)^{n-x} \\
= \frac{n(n-1)p^2}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b \frac{(n-2)!}{(n-x)!(x-2)!} p^{x-2} (1-p)^{n-x} \\
= \frac{n(n-1)p^2}{F(b,n,p)-F(a,n,p)} \sum_{a-1}^{b-2} \frac{(m)!}{(m-y)!(y)!} p^{y} (1-p)^{m-y}   \ \ where\ \ y=x-2\ \ and\ \  m=n-2 \\
= \frac{n(n-1)p^2}{F(b,n,p)-F(a,n,p)} [F(b-2,n-2,p) - F(a-2,n-2,p)]$$

$$V(X | a<X\le b) = E(X^2| a<X\le b) - E(X| a<X\le b)^2 \\
= E(X(X-1)| a<X\le b) + E(X| a<X\le b) - E(X| a<X\le b)^2$$

$$V(X | a<X\le b) = \frac{n(n-1)p^2[F(b-2,n-2,p) - F(a-2,n-2,p)]}{F(b,n,p)-F(a,n,p)} + \frac{np[F(b-1,n-1,p) - F(a-1,n-1,p)]}{F(b,n,p)-F(a,n,p)}  - \left[ \frac{np[F(b-1,n-1,p) - F(a-1,n-1,p)]}{F(b,n,p)-F(a,n,p)} \right]^2$$

Now, switching to the original notation and situation, I can get the ratio of variances as a function of $q$, but not quite the nice inequality you were looking for.
$$\frac{V(R|q-1<R\le n)}{V(R)} = \frac{1}{np(1-p)} \left( \frac{n(n-1)p^2[1 - F(q-3,n-2,p)]}{1-F(q-1,n,p)} + \frac{np[1 - F(q-2,n-1,p)]}{1-F(q-1,n,p)}  - \left[ \frac{np[1 - F(q-2,n-1,p)]}{1-F(q-1,n,p)} \right]^2 \right)$$
The next step would be to show by a series arguments and operations to the right hand side, that
$$\frac{V(R|q-1<R\le n)}{V(R)} \le \frac{n-\frac{np(1-F(q-2,n-1,p))}{(1-F(q-1,n,p))}}{n-np} = \frac{N-ER^*}{N-ER}$$
Update on the Normal Approximation
Matt F's post using the normal approximation inspired me to continue, but I'm not sure that path reaches the ultimate goal of proving the inequality under all conditions.
See Wikipedia's Truncated Normal Distribution page for these facts
Let $\alpha = (a-\mu)/\sigma$ and $\beta = (b-\mu)/\sigma$ and $\phi$ is the standard normal PDF and $\Phi$ is the normal CDF.
$$E(X|a<X<b) = \mu - \sigma \frac{\phi(\beta)-\phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)}$$
$$V(X|a<X<b) = \sigma^2 \left[ 1-\frac{\beta \phi(\beta)-\alpha \phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)} - \left(\frac{\phi(\beta)- \phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)}\right)^2\right]$$
Now, substituting in for this specific situation: with $W = 1 - \Phi(\alpha)$ and $\alpha = (q-np)/\sqrt{np(1-p)}$
$$\frac{V(R|q<R\le n)}{V(R)} = \frac{\sigma^2 \left[1 + \frac{\alpha \phi(\alpha)}{W} - \frac{\phi(\alpha)^2}{W^2}\right]}{\sigma^2} \le \frac{n-ER^*}{n-ER} = \frac{n-\left( np + \frac{\sqrt{np(1-p)} \phi(\alpha)}{W} \right)}{n - np}$$
$$ 1+\frac{(q-np)\phi(\alpha)}{W \sqrt{np(1-p)}} - \frac{\phi(\alpha)^2}{W^2} \le 1 - \frac{\phi(\alpha)\sqrt{np(1-p)}}{(n-np)W} $$
cancelling terms
$$\frac{q-np}{\sqrt{n(1-p)}} - \frac{\sqrt{p}\phi(\alpha)}{W} \le -\frac{p}{\sqrt{n(1-p)}}$$
To guarantee the inequality of interest:
$$q \le (n-1)p + \frac{\sqrt{np(1-p)}\ \ \phi\left( \frac{(q-np)}{\sqrt{np(1-p)}}\right) }{1-\Phi\left( \frac{(q-np)}{\sqrt{np(1-p)}} \right)}$$
A: 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 $+\infty$.
