Suppose $X_1, X_2, X_3$ are independently and identically distributed random variables of an exponential distribution with rate $\beta$. What is the probability $P(X_1 < X_2 < X_3)$?


$$P(X_1<X_2) = P(X_2<X_3) = \frac{\beta}{2\beta}$$

Now, $P(X_1 < X_2 < X_3) = P(X_1<X_2, X_2<X_3) = P(X_1<X_2 | X_2<X_3)\times P(X_2<X_3)$

Am I moving in the right direction? How to calculate $P(X_1<X_2 | X_2<X_3)$? Is it $0.5$ due to the memoryless property?


2 Answers 2


There is no need to work so hard for this question.

Think of it this way. There are only six possible outcomes (except for those outcomes for which at least two of the three random variables are exactly equal, but these occur with probability zero):

$$X_1 < X_2 < X_3 \\ X_1 < X_3 < X_2 \\ X_2 < X_1 < X_3 \\ X_2 < X_3 < X_1 \\ X_3 < X_1 < X_2 \\ X_3 < X_2 < X_1$$ Because each $X_i$ is independent and identically distributed, the probability of any one of these is equal to the other. The underlying distribution makes no difference. Therefore, the answer is $1/6$. Had these not been IID then you'd need to do more work and the answer would depend on how they are distributed, but in this case, the variables are exchangeable.



If $X_1,X_2$ and $X_3$ are iid with distribution $f$, then

$$\mathbb P\{X_1<X_2<X_3\}=\int_{0}^\infty \int_{x_1}^\infty \int_{x_2}^\infty f(x_1)f(x_2)f(x_3)\,\mathrm d x_3\,\mathrm d x_2\,\mathrm d x_1.$$

  • $\begingroup$ If they are not independent then we need to integrate the joint pdf +1 $\endgroup$
    – Math-fun
    Commented Oct 30, 2020 at 20:37
  • $\begingroup$ @Math-fun True, iid makes the solution short and tidy $\endgroup$ Commented Oct 31, 2020 at 2:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .