# Probability that $X_1 > X_2 > X_3$ if all $X_1$, $X_2$, $X_3$ are uniform

Let $$X_1$$, $$X_2$$, and $$X_3$$ be independent and identically distributed random variables with uniform distribution [0, 1]. What is the probability that $$X_1 > X_2 > X_3$$?

I know that this is easily solved using the fact that there are $$3! = 6$$ possible permutations of $$X_1, X_2, X_3$$ so the answer is $$\frac{1}{6}$$. But I am trying to solve it using integrals and I'm not getting the right answer. Here is what I have so far:

$$P(X_1 > X_2 > X_3) = \int_{0}^{1} P(X_1 > X_2 > X_3 | X_3 = x_3)dx_3 = \int_{0}^{1} P(X_1 > X_2 \cap X_2 > X_3 | X_3 = x_3)dx_3.$$

Using the definition of conditional probability, this can be split into

$$\int_{0}^{1} P(X_1 > X_2 | X_2 > X_3, X_3 = x_3)P(X_2 > X_3 | X_3 = x_3)dx_3 = \int_{0}^{1} P(X_1 > X_2 | X_2 > x_3)P(X_2 > X_3 | X_3 = x_3)dx_3.$$

I then find $$P(X_1 > X_2 | X_2 > x_3) = \int_{x_3}^{1} P(X_1 > X_2 | X_2 = x_2)dx_2 = \int_{x_3}^{1} (1 - x_2)dx_2 = \frac{1}{2} - x_3 + \frac{x_3^2}{2}.$$

Substituting this into the original expression yields $$\int_{0}^{1} P(X_1 > X_2 | X_2 > x_3)P(X_2 > X_3 | X_3 = x_3)dx_3 = \int_{0}^{1} (\frac{1}{2} - x_3 + \frac{x_3^2}{2})(1 - x_3)dx_3 = \frac{1}{8}.$$

Clearly I'm making a mistake here but I am unable to figure out what it is. I can use another method and find that $$P(X_1 > X_2 > X_3) = \int_{0}^{1} \int_{x_3}^{1} P(X_1 > X_2 | X_2 = x_2) dx_2dx_3 = \int_{0}^{1} \int_{y}^{1} (1 - x) dx_2dx_3 = \frac{1}{6}$$, but why is the calculation above incorrect? Thanks.

The mistake is in the line $$P(X_1 > X_2 | X_2 > x_3) = \int_{x_3}^1 P(X_1 > X_2|X_2=x_2)dx_2.$$ You forgot to divide by $$P(X_2 > x_3) = 1-x_3$$, so the RHS is $$P(X_1 > X_2\cap X_2 > x_3)$$ rather than $$P(X_1 > X_2 | X_2 > x_3)$$.

• Makes sense, thanks! Seems like resolving that error gives the right answer Dec 8, 2022 at 19:35
• I took another look and I'm a little confused. Why is $P(X_1 > X_2 | X_2 > x_3) = \int_{x_3}^{1} P(X_1 > X_2 | X_2 = x_2)dx_2$ not true? Isn't $P(X_1 > X_2 | X_2 > x_3) = P(X_1 > X_2 | X_2 = x_3) + P(X_1 > X_2 | X_2 = x_3 + dx_2) + . + P(X_1 > X_2 | X_2 = 1)$, which is simply the integral? How is this integral equal to the probability that $(X_1 > X_2) \cap (X_2 > x_3)$? Dec 8, 2022 at 19:54
• No, you can't break up a conditional probability like that. Even if $B$ and $C$ are disjoint, $P(A|B \cup C) \ne P(A|B)+P(A|C)$. That sort of derivation would work for $P((X_1 > X_2) \cap (X_2 > x_3))$, though, because we do have that $P(B \cup C) = P(B) + P(C)$ for disjoint sets $B,C$. Of course, this is slightly glossing over the technicalities of trying to write this out as an infinite sum, but it works as an informal argument. Dec 8, 2022 at 20:11

Here is an easier way to solve the problem.

It is straightforward to see that:

$$\mathcal{S} = \{(X_1,X_2,X_3) : 1 > X_1 > X_2 > X_3 > 0\}$$

can be rewritten as:

$$\mathcal{S} = \{(X_1,X_2,X_3) : X_3 \in (0, X_2) \wedge X_1 \in (1, X_2) \wedge X_2 \in (0,1)\}.$$

As a consequence:

$$P(\mathcal{S}) = \int_0^1 \left(\int_{X_2}^1 \left(\int_0^{X_2} dX_3\right) dX_1\right) dX_2 = \\ =\int_0^1 \left(\int_{X_2}^1 X_2 dX_1\right) dX_2 = \\ =\int_0^1 (X_2-X_2^2) dX_2 = \\ =\frac{1}{2}-\frac{1}{3} = \frac{1}{6}.$$