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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.

$$ P[\min(X_1, X_2)<x | \max(X_1, X_2)=z] = ? $$

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  • $\begingroup$ why is the first result $2x - x^2$? I keep getting $x(1-x)+x(1-x) = 2x - 2x^2$ $\endgroup$ – Alex Jun 10 '14 at 15:20
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    $\begingroup$ @Alex, I added the derivation that I think is correct. In your case, didn't you miss the case that both of the random variables are smaller than $x$? $\endgroup$ – Grzenio Jun 10 '14 at 15:39
  • $\begingroup$ OK thanks I got it $\endgroup$ – Alex Jun 10 '14 at 15:47
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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]

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  • $\begingroup$ Thanks for that! Does it hold in the more general case where we have $N$ random variables $X_1$, ..., $X_N$, and we condition the distribution of one of them (which is not maximal) on the maximum of all these random variables being $z$? $\endgroup$ – Grzenio Jun 12 '14 at 7:50
  • $\begingroup$ New problem $\implies$ New question. $\endgroup$ – Did Jul 14 '14 at 18:20

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