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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.

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  • $\begingroup$ Start with the definition of a convex function, and then see if this definition applies here. Have you tried this already? $\endgroup$
    – Mahmoud
    yesterday
  • $\begingroup$ I have written it out, but proving the relevant inequality looks hard! $aE[X|p] +(1-a)E[X|q] > E[X|ap+(1-a)q]$... it's not easy to see how this will simplify $\endgroup$
    – dash2
    19 hours ago

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