# How to prove $P(X_2 < X_3\mid X_1 = \min(X_1, X_2, X_3)) = P(X_2 < X_3)$?

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

that.

• huh..what was the downvote for...? Commented Nov 25, 2019 at 13:08
• What are $X_1,X_2,X_3$? And please use MathJax for formatting math. Commented Nov 25, 2019 at 18:11
• 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. Commented Nov 25, 2019 at 18:20
• Sorry! It is my fault. I add the information of variables Xi. Commented Nov 26, 2019 at 15:59

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