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.
Best Answer
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 $P(X_1 > X_2 | X_2 > x_3)$.