Let $X_1$ and $X_2$ be two random variables uniformly distributed on $(0, 1)$. It is easy to calculate the distribution of minimum and maximum of these two numbers:
$$
P[\max(X_1, X_2)<x] = x^2
$$
$$
P[\min(X_1, X_2)<x] = 1 – P[\min(X_1, X_2)>x] \\
= 1 – (1-x)^2 = 2x-x^2
$$
But what would be the distribution of the minimum, given that the maximum is given and equal to some number $z$, i.e.
$$
P[\min(X_1, X_2)<x | \max(X_1, X_2)=z] = ?
$$
Best Answer
Good question, this had me stumped at first. The minimum is actually the uniform distribution on [0,z].
Without loss of generality, let $X_1$ be at the maximum value z.
Then $P(X_2 > z) = 0$ since z is the max, and $P(X_2 \le z) = 1$.
Let Y be the event that $X_2 \le z$. Then by Bayes rule. $$ f_{X_2}(x|Y) = P(Y|X_2=x)f_{X_2}(x)/P(Y) \\ = 0 \quad for \quad x > z \\ = 1(1)/z \quad for \quad x \le z \\ $$ Therefore the minimum is uniform on [0,z]