Timeline for Expectation of cross product of normal distributed variables
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2022 at 21:14 | vote | accept | granular_bastard | ||
| Jan 11, 2021 at 20:26 | comment | added | JimB | OK. I see now. The expectation of the square of a random variable is the sum of the variance and the square of the mean (assuming the variance and mean exist). Rather than performing the known integration I just used a replacement rule. | |
| Jan 11, 2021 at 20:01 | comment | added | granular_bastard | no, just try to understand the step. Can also be answered here: mathematica.stackexchange.com/questions/238037 | |
| Jan 11, 2021 at 19:49 | comment | added | JimB | Do you have an alternative in mind? | |
| Jan 11, 2021 at 19:18 | comment | added | granular_bastard | Why the replacement must be performed in the given way? mean = taylor //. z[i_]^2 -> [Sigma]^2 + [Mu][i]^2; | |
| Jan 28, 2020 at 14:41 | comment | added | granular_bastard | A computer simulation might help for a few cases but not in general. | |
| Jan 28, 2020 at 14:37 | comment | added | JimB | It might just be that estimating the mean through random samples is the way to go which doesn't take too much computer time if you just have to do a few of these. | |
| Jan 27, 2020 at 4:46 | comment | added | granular_bastard | The solution diverges if vectors become parallel. It cannot be applied in the case that is described here: math.stackexchange.com/questions/3520115/… | |
| Jan 27, 2020 at 3:56 | history | edited | JimB | CC BY-SA 4.0 |
Added in comment about also estimating the mean through random samples.
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| Jan 27, 2020 at 3:45 | history | answered | JimB | CC BY-SA 4.0 |