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Given a binomial distribution with parameters $ n $ and $ p $, where $ n $ is an odd integer greater than or equal to 3, I am interested in the truncated binomial distribution where we truncate at $ k = \frac{n+1}{2} $ (i.e., $ X \geq k $). I would like to express the expectation and variance of this truncated distribution using only $ n $ and $ p $, and without involving summation or complicated series expressions.

My ultimate goal is to determine the sign of the following expression: $$ (E - np)(2p - 1) + D - np(1 - p) + (E - np)^2 $$ where $ E $ is the expectation and $ D $ is the variance of the truncated distribution.

Is there a way to express the expectation and variance of this truncated distribution in terms of $ n $ and $ p $ without using summation symbols or recursive relationships? I am open to using standard special functions, but would prefer to avoid summation signs if possible.

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  • $\begingroup$ Specifically, how do you truncate? There are a number of ways to truncate, at the same level. Please define the truncated distribution formally. $\endgroup$
    – Iosif Pinelis
    Commented Sep 16 at 13:04
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    $\begingroup$ @IosifPinelis The truncated distribution normalizes the binomial distribution over the range $ x \geq k $ by dividing by the cumulative probability $ P(X \geq k) $, which is: $$ P(X \geq k) = \sum_{x=k}^{n} \binom{n}{x} p^x (1 - p)^{n - x} $$ I am particularly interested in this left-truncated binomial distribution and its expectation $ E[X|X \geq k] $ and variance $ \text{Var}(X|X \geq k) $, as well as the sign of the expression I provided earlier. $\endgroup$
    – GodsDusk
    Commented Sep 16 at 13:22
  • $\begingroup$ It seems that the sign of your displayed expression is the same as that of $1/2-p$. $\endgroup$
    – Iosif Pinelis
    Commented Sep 19 at 20:51
  • $\begingroup$ @IosifPinelis how do you get it? same result as my code simulation $\endgroup$
    – GodsDusk
    Commented Sep 20 at 1:40
  • $\begingroup$ At this point, I don't have a proof of this conjecture. $\endgroup$
    – Iosif Pinelis
    Commented Sep 20 at 17:21

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