Timeline for 2 discrete distributions with equal mean and variance
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16, 2015 at 8:21 | comment | added | hanna | Thank you! I get a variance of 70.6 for case B. I think there is something wrong...With one distribution where this holds, I could compute other ones, which would be very great! | |
| Feb 15, 2015 at 4:53 | comment | added | soakley | Yes, it is possible. Let $A$ take on values $-6$ and $+6$ with probability ${{1} \over {2}}$ each. Let $B$ take on values $-14$ with probability ${{1} \over {7}},$ zero with probability ${{5} \over {14}},$ and $4$ with probability ${{1} \over {2}}.$ Then both $A$ and $B$ have a mean of zero and a variance of $36.$ | |
| Feb 13, 2015 at 8:19 | comment | added | hanna | Ok, that makes sense. But if I change the highest point in the 2-point distribution to increase variance it should be possible, shouldn't it? | |
| Feb 11, 2015 at 23:33 | comment | added | soakley | No formula that I know about. I don't think it is possible to do what you want even under only the main constraint. Think of it this way. Since you are leaving the highest point the same for both cases, you need to take the lower point of the 2-point distribution and divide it into 2 pieces (to make the 3-point distribution). You can keep the mean the same by dividing that point symmetrically. But the variance will then have to increase since you are including a point further to the left. That's far from rigorous, but experiment around a bit to convince yourself. | |
| Feb 11, 2015 at 13:32 | comment | added | hanna | Daer Soakley, is there any sort of formula that tells me under which constraints it is possible? The main constraint I want to implement is that the probability for the highest outcome should be the same in both cases. | |
| Feb 10, 2015 at 23:01 | comment | added | soakley | Note that this does not say that you cannot have a 2-point distribution and a 3-point distribution with equal mean and variance. Just not under your constraints. So, for example, consider that $A$ takes on $-6$ with probability ${1 \over 2}$ and $+6$ with probability ${1 \over 2}.$ Then let $B$ take on $-12$ with probability ${1 \over 8},$ zero with probability ${3 \over 4},$ and $+12$ with probability ${1 \over 8}.$ These have identical means and variances. | |
| Feb 10, 2015 at 15:46 | vote | accept | hanna | ||
| Feb 9, 2015 at 18:41 | history | edited | soakley | CC BY-SA 3.0 |
Reviised answer to fit the questioner's new problem definition.
|
| Feb 7, 2015 at 13:52 | comment | added | soakley | What it shows is that there was no solution that had positive probabilities for all three outcomes of the random variable $B.$ But now you have changed the problem definition again. | |
| Feb 7, 2015 at 13:49 | history | edited | soakley | CC BY-SA 3.0 |
deleted 1 character in body
|
| Feb 7, 2015 at 9:03 | comment | added | hanna | Thank you for your comment. But Ithink this misses out that the second distribution should take 3 different values, one takes value 0, one takes value 6000 and one is unknown. This woud change the variance of the second distribution. | |
| Feb 7, 2015 at 1:27 | history | answered | soakley | CC BY-SA 3.0 |