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Consider exponential random variables $X$, $Y$, and $Z$ with $\lambda_x$, $\lambda_y$, and $\lambda_z$, respectively. Now I want to calculate the following integration:

$$E[X1_{\{X<Y\}}1_{\{X<Z\}}]=\int_{0}^{\infty}X1_{\{X<Y\}}1_{\{X<Z\}}f_X(x)=?$$

any hint, please

I know that there is a very simple solution for this expectation, but I want to calculate such integration.

Optimize Life
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  • Notice that $\mathbb{I}_{X – Fiodor1234 Sep 14 '21 at 13:53
  • Thanks, I actually know this, but I want to solve this directly by specifying integral intervals. – Optimize Life Sep 14 '21 at 14:17
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    The expectation is incorrectly expressed. The expectation is a *triple* integral, $$E\left[X1_{X\lt Y}1_{X\lt Z}\right] = \int_0^\infty \int_0^\infty \int_0^\infty x 1_{x\lt y} 1_{x \lt z} f_{X,Y,Z}(x,y,z)\,\mathrm{d}x\mathrm{d}y\mathrm{d}z.$$ You can easily evaluate it by assuming $X,Y,Z$ are *independent.* – whuber Sep 14 '21 at 15:34
  • @whuber, Thanks, can you explain to me in more detail? If it is in the form of an answer, I would be very grateful. – Optimize Life Sep 14 '21 at 15:47

1 Answers1

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To continue Whubers answer, based on independence assumption between the random variables $X, Y, Z$.

$$\mathbb{E}[X\mathbb{I}_{X<Y}\mathbb{I}_{X<Z}] = \int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}x \mathbb{I}_{x<y} \mathbb{I}_{x<z}f_{X}(x)f_{Y}(y)f_{Z}(z)dzdydx$$

$$=\int_{0}^{\infty}\int_{0}^{\infty}x\mathbb{I}_{x<y}f_{X}(x)f_{Y}(y)\int_{x}^{\infty}f_{Z}(z)dz \ dydx$$

For exponential distributions with parameter $\lambda_{z}$ it holds that $F(Z>x) = \int_{x}^{\infty}f_{Z}(z)dz = e^{-\lambda_{z}x}$

$$=\int_{0}^{\infty}\int_{0}^{\infty}x\mathbb{I}_{x<y}f_{X}(x)f_{Y}(y) e^{-\lambda_{z}x} \ dydx$$

$$= \int_{0}^{\infty}xe^{-\lambda_{z}x}f_{X}(x) \int_{x}^{\infty}f_{Y}(y)dy \ dx$$ $$ = \int_{0}^{\infty}xe^{-(\lambda_{z}+\lambda_{y})x} f_{X}(x)dx$$

Now using Integration by parts we have

$$= -\frac{\lambda_{x}}{\lambda_{z}+\lambda_{y}+\lambda_{x}}\int_{0}^{\infty}x(e^{-(\lambda_{z}+\lambda_{y}+\lambda_{x})x})^{'}dx = -\frac{\lambda_{x}}{\lambda_{z}+\lambda_{y}+\lambda_{x}} (0 - \int_{0}^{\infty}e^{-(\lambda_{z}+\lambda_{y}+\lambda_{x})x}dx ) = \frac{\lambda_{x}}{(\lambda_{z}+\lambda_{y}+\lambda_{x})^{2}}$$

Fiodor1234
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    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. – whuber Sep 14 '21 at 18:57
  • @whuber ooops I'll correct it right away! – Fiodor1234 Sep 14 '21 at 19:00
  • @whuber Does the square on the denominator should disappear because based on my calculations it doesnt? – Fiodor1234 Sep 14 '21 at 19:12
  • Yes, now it looks good. Notice that your new answer includes $\lambda_x$ and is inversely proportional to it. – whuber Sep 14 '21 at 19:14
  • @whuber Is it possible to intuitively interpret this result?? The expected value of $X$ when it is less from both $Y$ and $Z$ ? – Fiodor1234 Sep 14 '21 at 19:18
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    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 https://stats.stackexchange.com/a/429589/919 for intuition. – whuber Sep 14 '21 at 19:21
  • 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 – Fiodor1234 Sep 14 '21 at 19:36
  • Thanks a lot for your clear answer – Optimize Life Sep 15 '21 at 05:02