Timeline for How to compute the variances of quotients of normal variables?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2018 at 14:56 | comment | added | whuber♦ | @Dilip It is unnecessary to define the mean to define a variance, because the variance of a random variable $X$ can be expressed in terms of the expectation of $(X_1-X_2)^2$ where the $X_i$ are independent copies of $X.$ | |
| Dec 8, 2018 at 5:06 | comment | added | Dilip Sarwate | Isn't it the case that if $A$ and $B$ are a random variable whose joint density is nonzero everywhere in a neighborhood of the origin $0$, then $\frac AB$ does not have an expectation and so in the simplest viewpoint, one cannot even begin to define the variance of $\frac AB$? There exists no mean of $\frac AB$ from which we can compute deviations whose mean-square value is desired. | |
| Dec 7, 2018 at 21:00 | history | edited | whuber♦ | CC BY-SA 4.0 |
added 126 characters in body
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| Dec 7, 2018 at 20:54 | history | answered | whuber♦ | CC BY-SA 4.0 |