Timeline for Difference of hypoexponential distributions
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2017 at 10:12 | vote | accept | Muriel | ||
| Oct 14, 2017 at 10:12 | |||||
| Oct 14, 2017 at 10:12 | comment | added | Muriel | As I have added before, $X$ and $Y$ are dependent. I am not sure I see where this solution is assuming independence. | |
| S Oct 3, 2017 at 15:58 | history | suggested | Max | CC BY-SA 3.0 |
corrected spelling
|
| Oct 3, 2017 at 15:57 | comment | added | Mateusz Kwaśnicki | This answer is as unclear as the question: in the first part, $X$ and $Y$ appear to be coupled (for $X>Y$ with probability one), while in the other part the are apparently assumed to be independent (although even in this case we have $f_{(X,Y)}(x,y)=(p_1 f_{\alpha_1}(x)+\ldots+p_n f_{\alpha_n}(x))(q_1 f_{\beta_1}(y)+\ldots+q_m f_{\beta_m}(y))$ rather than the expression given in the answer). | |
| Oct 3, 2017 at 15:40 | review | Suggested edits | |||
| S Oct 3, 2017 at 15:58 | |||||
| May 6, 2017 at 12:23 | history | answered | Henry.L | CC BY-SA 3.0 |