Timeline for How do you calculate the expected value of a discrete distribution without replacement?
Current License: CC BY-SA 4.0
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| Apr 12, 2022 at 22:21 | comment | added | Jarle Tufto | @knrumsey +1 But there is an easier and more general way of deriving the variance. | |
| Apr 12, 2022 at 20:46 | history | edited | knrumsey | CC BY-SA 4.0 |
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| Apr 12, 2022 at 20:45 | comment | added | knrumsey | @BruceET, Maybe the confusion lies in the fact that you are looking at $(X_1+X_2+X_3)/3$, while the OP is asking about the sum $X_1+X_2+X_3$. They are of course highly related, but I am just trying to focus on how the question was asked. // I have edited the code so that it now computes both the expected value and the variance of $X_1+X_2+X_3$. | |
| Apr 12, 2022 at 20:39 | history | edited | knrumsey | CC BY-SA 4.0 |
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| Apr 12, 2022 at 20:33 | history | edited | knrumsey | CC BY-SA 4.0 |
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| Apr 12, 2022 at 20:31 | comment | added | knrumsey | @BruceET I did not mean to suggest that $E(\bar X) \neq \mu$. I have scanned my answer, and I can't seem to find where I said this. Please point out (or feel free to edit) what you are referring to, so that I can edit to make the answer more clear. | |
| Apr 12, 2022 at 3:28 | vote | accept | brubsby | ||
| Apr 12, 2022 at 0:29 | comment | added | BruceET | Sorry about the mistake--which should have been obvious. (+1) I agree that the expectation of the sample mean of three is the same with and and without replacement, as shown in my simulation within simulation error. But why do you say $E(\bar X_3) \ne \mu,$ the population mean? That seems another obvious contradiction. // Does some modification of your code find the standard deviations? | |
| Apr 11, 2022 at 23:53 | history | answered | knrumsey | CC BY-SA 4.0 |