When $X$ is an exponential random varaible, the memoryless property is stated as

$$P(X>s+t\mid X>s)=P(X>t)$$

But, I am not sure how to prove

$$P(X_2 < X_3\mid X_1 = \min(X_1, X_2, X_3)) = P(X_2 < X_3)$$

$X_1$~ $exp$ ($λ1$), $X_2$~ $exp$ ($λ2$), $X_3$~ $exp$ ($λ3$), and $X_1$,$X_2$,$X_3$ are independent variables.

Here is my proof:

$$P(X_2 < X_3\mid X_1 = \min(X_1, X_2, X_3)) = P(X_1< X_2 < X_3)/ P(X_1 = \min(X_1, X_2, X_3))$$

And, prove $$P(X_1< X_2 < X_3)/ P(X_1 = \min(X_1, X_2, X_3)) = P(X_2 < X_3)$$

{I have known how to calculate $P(X_1< X_2 < X_3)$ and $P(X_1 = \min(X_1, X_2, X_3))$ }

But, this procedure is too complicated , so I am wondering if there is much easier way too prove


  • $\begingroup$ huh..what was the downvote for...? $\endgroup$
    – Amy Chang
    Commented Nov 25, 2019 at 13:08
  • $\begingroup$ What are $X_1,X_2,X_3$? And please use MathJax for formatting math. $\endgroup$ Commented Nov 25, 2019 at 18:11
  • 2
    $\begingroup$ You didn't introduce the $X_i$. In case they're independent samples of $X$, both sides are $\frac12$ simply by symmetry; that has nothing to do with the exponential distribution. $\endgroup$
    – joriki
    Commented Nov 25, 2019 at 18:20
  • $\begingroup$ Sorry! It is my fault. I add the information of variables Xi. $\endgroup$
    – Amy Chang
    Commented Nov 26, 2019 at 15:59

1 Answer 1


Does this work? First, prove that for any $s$, we have $$P(X_2<X_3 \mid s < X_2, s < X_3) = P(X_2<X_3)$$ using memorylessness. Then note that

$$P(X_2<X_3 \mid X_1 < X_2, X_1 < X_3) = \\ \int_x P(X_2<X_3 \mid X_1 =x, X_1 < X_2, X_1 < X_3)p(X_1=x \mid X_1 < X_2, X_1 < X_3)dx = \\ \int_x P(X_2<X_3 \mid x < X_2, x < X_3)p(X_1=x \mid X_1 < X_2, X_1 < X_3)dx = \\P(X_2<X_3) \int_x p(X_1=x \mid X_1 < X_2, X_1 < X_3)dx = P(X_2<X_3)$$

Here, lowercase $p$ denotes the conditional density function. (If this works then it seems it's not necessary for $X_1$ to follow any particular distribution.)


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