Timeline for How to express the expectation and variance of a truncated binomial distribution without summation?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 20 at 17:21 | comment | added | Iosif Pinelis | At this point, I don't have a proof of this conjecture. | |
| Sep 20 at 1:40 | comment | added | GodsDusk | @IosifPinelis how do you get it? same result as my code simulation | |
| Sep 19 at 20:51 | comment | added | Iosif Pinelis | It seems that the sign of your displayed expression is the same as that of $1/2-p$. | |
| Sep 16 at 14:22 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
|
| Sep 16 at 13:22 | comment | added | GodsDusk | @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. | |
| Sep 16 at 13:04 | comment | added | Iosif Pinelis | Specifically, how do you truncate? There are a number of ways to truncate, at the same level. Please define the truncated distribution formally. | |
| Sep 16 at 12:02 | history | edited | gmvh |
edited tags
|
|
| S Sep 16 at 11:16 | review | First questions | |||
| Sep 16 at 12:02 | |||||
| S Sep 16 at 11:16 | history | asked | GodsDusk | CC BY-SA 4.0 |