Random variables $A_1,A_2,A_3$ are drawn independently from the exponential distribution with parameters $\lambda_1,\lambda_2,\lambda_3$. Find the probability that $A_1<\min(A_2,A_3,1)$.

If the question were just to find the probability that $A_1<\min(A_2,A_3)$, the answer would be $\frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}$ by some properties of exponential random variables (although I'm not sure how to derive it -- we cannot separate it into $\text{Pr}(A_1<A_2)\cdot\text{Pr}(A_1<A_3)$.) Here there is the extra condition $A_1<1$, which complicates things further.


The probability sought for is

$$p=\lambda_1\lambda_2\lambda_3\iiint_A e^{-(\lambda_1x+\lambda_2y+\lambda_3z)}\ dxdydz$$ where

$$A=\{(x,y,z):x,y,z\ge 0,x<1,x<y,x<z \}.$$


$$p=\lambda_1\lambda_2\lambda_3\int_0^1 e^{-\lambda_1x}\left[\int_x^{\infty}e^{-\lambda_2y}\ dy\int_x^{\infty}e^{-\lambda_3z}\ dz\right]\ dx=$$

$$=\lambda_1\int_0^1 e^{-(\lambda_1+\lambda_2+\lambda_3)x}\ dx=\frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}\left(1-e^{-(\lambda_1+\lambda_2+\lambda_3)}\right).$$

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    $\begingroup$ Double check your integration. $\endgroup$ – John Dawkins Oct 4 '15 at 16:42
  • $\begingroup$ @JohnDawkins: Thanks for warning. $\endgroup$ – zoli Oct 4 '15 at 16:52
  • $\begingroup$ This (and several analogous calculations) show that the minimum of $A_1,A_2,A_3$ is independent of their ranks (i.e, which is smallest, middle, largest). One wonders if the complete set of order statistics is independent of the ranks. $\endgroup$ – John Dawkins Oct 4 '15 at 16:59
  • $\begingroup$ @JohnDawkins: Me dont know. I am only integrating : ) $\endgroup$ – zoli Oct 4 '15 at 17:02

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