Is the mean of the left-truncated binomial distribution convex in p? The expectation of the binomial distribution of successes in $G$ trials, left-truncated at $R$, with success probability $p$, is
$$
E[X|p] = \frac{\sum_{l=R}^Gl\phi(l)}{\sum_{l=R}^G\phi(l)}
$$
where
$$
\phi(l) = \binom{G}{l}p^l(1-p)^{G-l}.
$$
Is this convex in $p$? It looks as if it is.
Update
In work so far, I've taken the first derivative as
$$
\frac{d}{dp}E[X|p] = 
\frac{\sum_{l=R}^{G}\sum_{m=l+1}^{G}(l-m)^{2}\phi(l)\phi(m)}{p(1-p)(\sum_{l=R}^G\phi(l))^2}.
$$
This used the fact (I hope!) that
$$
\frac{d}{dp} \phi(l) = \frac{l - Gp}{p(1-p)} \phi(l)
$$
and simplified a double sum $\sum_{l=R}^G \sum_{m=R}^G...$ by merging pairs of terms to produce $\sum_{l=R}^G \sum_{m=l+1}^G...$.

Since the function of interest is defined over a continuum of real values of $p$ you can check its convexity using standard calculus methods.  Consider a general ratio function:
$$f(p) = \frac{g(p)}{h(p)}.$$
The quotient rules says that:
$$f'(p) = \frac{g'(p) h(p) - g(p) h'(p)}{h(p)^2},$$
and the second-order quotient rules says that:
$$f''(p) = \frac{[g''(p) h(p) - g(p) h''(p)] h(p) - 2 [g'(p) h(p) - g(p) h'(p)] h'(p)}{h(p)^3}.$$
Convexity occurs when, for all $0<p<1$ we have:
$$[g''(p) h(p) - g(p) h''(p)] h(p) > 2 [g'(p) h(p) - g(p) h'(p)] h'(p).$$
Check this condition against the numerator and denominator functions in your example and this will tell you whether your function is convex.
