Timeline for X,Y are independent exponentially distributed then what is the distribution of X/(X+Y)
Current License: CC BY-SA 3.0
10 events
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| Sep 5, 2016 at 16:03 | history | edited | Michael Hardy | CC BY-SA 3.0 |
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| Dec 5, 2013 at 12:56 | comment | added | Albanian_EAGLE | definitely, its pretty useful :) | |
| Nov 29, 2013 at 5:36 | comment | added | Michael Hardy | You're welcome. Might you consider up-voting this answer? | |
| Nov 29, 2013 at 5:29 | comment | added | Albanian_EAGLE | very helpful, thanks. | |
| Nov 29, 2013 at 5:27 | comment | added | Michael Hardy | If $U$ and $V$ are independent exponentially distributed random variables with expected value $\alpha$, then $X=U/\alpha$ and $Y=V/\alpha$ are independent exponentially distributed random variables with expected value $1$. And $X/(X+Y)$ is equal to $U/(U+V)$. So $\alpha$ just cancels out. | |
| Nov 29, 2013 at 3:09 | comment | added | Albanian_EAGLE | but then how does the coefficient $\alpha$ simplify in the integration so that we get still just $w$? (your second method) | |
| Nov 28, 2013 at 18:51 | comment | added | Michael Hardy | It should still be true in that case. | |
| Nov 28, 2013 at 15:53 | comment | added | Albanian_EAGLE | Can we say the same for identically distributed random variables with the exponential distribution with a parameter $\alpha \neq 1$? Following the second method I think it shouldnt hold true, am I right? | |
| Jun 10, 2013 at 19:16 | history | edited | Michael Hardy | CC BY-SA 3.0 |
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| Jun 10, 2013 at 18:45 | history | answered | Michael Hardy | CC BY-SA 3.0 |