Timeline for Integrating with considering two indicator function
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 15, 2021 at 5:02 | comment | added | Optimized Life | Thanks a lot for your clear answer | |
| Sep 15, 2021 at 5:00 | vote | accept | Optimized Life | ||
| Sep 14, 2021 at 19:36 | comment | added | Fiodor1234 | Interesting point! Also, the results of the integral is like $\frac{Var(min(X,Y,Z))}{\mathbb{E}[X]}$ restricting that $X$ has to be the minimum. Then I assume restricting that any of the other two random variables is the minimum we would have $\frac{Var(min(X,Y,Z))}{\mathbb{E}[Y]}$, $\frac{Var(min(X,Y,Z))}{\mathbb{E}[Z]}$ respectively | |
| Sep 14, 2021 at 19:21 | comment | added | whuber♦ | Yes: the result looks like it must be related to the distribution of $\min(X,Y,Z).$ That distribution is easy to work out. Compare the answer to yours--or better, compare it to what you obtain upon permuting the roles of $X,$ $Y,$ and $Z.$ That will bring to your attention the perhaps not so obvious relation $$\min(X,Y,Z) = X1_{X\lt Y}1_{X\lt Z}+Y1_{Y\lt Z1}1_{Y\lt X}+Z1_{Z\lt X}1_{Z\lt X}.$$ Then, remembering the relationship between Poisson processes and exponential random variables, ponder the illustrations I posted at stats.stackexchange.com/a/429589/919 for intuition. | |
| Sep 14, 2021 at 19:18 | comment | added | Fiodor1234 | @whuber Is it possible to intuitively interpret this result?? The expected value of $X$ when it is less from both $Y$ and $Z$ ? | |
| Sep 14, 2021 at 19:14 | comment | added | whuber♦ | Yes, now it looks good. Notice that your new answer includes $\lambda_x$ and is inversely proportional to it. | |
| Sep 14, 2021 at 19:12 | comment | added | Fiodor1234 | @whuber Does the square on the denominator should disappear because based on my calculations it doesnt? | |
| Sep 14, 2021 at 19:11 | history | edited | Fiodor1234 | CC BY-SA 4.0 |
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| Sep 14, 2021 at 19:04 | history | edited | Fiodor1234 | CC BY-SA 4.0 |
added 9 characters in body
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| Sep 14, 2021 at 19:00 | comment | added | Fiodor1234 | @whuber ooops I'll correct it right away! | |
| Sep 14, 2021 at 18:57 | comment | added | whuber♦ | Yes, this is one way to do it. However, you lost $\lambda_x$ near the end of your calculations and the integration was incorrectly performed. This is easy to see without any calculation: since you have taken the $\lambda_{*}$ to be rates, the result should be inversely proportional to them--but not inversely proportional to their squares. | |
| Sep 14, 2021 at 18:08 | history | answered | Fiodor1234 | CC BY-SA 4.0 |