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 $\sum_{l=R}^G \sum_{m=l+1}^G...$ .

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