# Conditioning on zero probability event

I have a somewhat trivial question (no homework):

Suppose $X_1, X_2$ are i.i.d. and uniform on $[0,1]$, and the realization of their maximum is $k \in (0,1]$. What is the conditional distribution of the other random variable?

Does that make any sense? In words: given that the realization of the maximum of 2 random variables attains some value, what is the conditional distribution of the other random variable? In the example, is it uniform on $[0,k]$?

Thanks!

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No, I think you are asking for the distribution of $X_1$ given $max(X_1,...,X_n) = k$, but $X_1$ will be the max with probability $\frac 1 n$. I'm guessing that they are jointly uniform on the surface $max(x_1,...,x_n) = k$, which I think implies an atom at k and otherwise uniform. – mike Sep 11 '12 at 14:35

For two independent $U(0,1)$ random variables, the joint density of the maximum and minimum is uniformly distributed on the triangular region with vertices at $(0,0)$, $(1,0)$ and $(1,1)$. So, yes, given that the maximum has value $k \in (0,1)$, the minimum is uniformly distributed on $(0,k]$. Similarly, given that the minimum has value $\ell \in (0,1)$, the maximum is uniformly distributed on $[\ell, 1)$.
Let $W = \max\{X,Y\}$ and $Z = \min\{X,Y\}$. Then, for $a \geq b$, \begin{align} F_{W,Z}(a,b) &= P\{W \leq a, Z \leq b\}\\ &= P\left(\{W \leq a\}\cap \{Z \leq b\}\right)\\ &= P\left(\{X \leq a, Y \leq b\} \cup \{X \leq b, Y \leq a\}\right)\\ &= P\{X \leq a, Y \leq b\} + P\{X \leq b, Y \leq a\} - P\{X \leq b, Y \leq b\} )\\ &= F_{X,Y}(a,b) + F_{X,Y}(b,a) - F_{X,Y}(b,b). \end{align} If one of the steps puzzles you, think about $P(A\cup B) = P(A) + P(B) - P(A\cap B)$). On the other hand, if $a < b$, then $$F_{W,Z}(a,b) = P\{X \leq a, Y \leq a\} = F_{X,Y}(a,a).$$ For jointly continuous random variables $X$ and $Y$, this gives (upon taking partial derivatives with respect to $a$ and $b$ that $$f_{W,Z}(a,b) = \begin{cases} f_{X,Y}(a,b) + f_{X,Y}(b,a), &\text{if}~ a \geq b,\\ 0, &\text{if}~ a < b. \end{cases}$$ If you think of the joint density $f_{X,Y}$ as a solid sitting on the plane, just fold it over the diagonal line $a=b$ to get the joint density $f_{\max,\min}$ of the maximum and the minimum. For your instance, the unit cube becomes a right triangular prism of height $2$ with vertices at $(0,0)$, $(1,0)$ and $(1,1)$. Thus, the conditional density of one of the two random variables given the value of the other, being just a scaled version of the cross-section of the joint density, is a uniform density.
Could you elaborate on this some more? Specifically, if the maximum of two draws of a continuous random variable is $k$, how does one calculate the conditional probability density function of the other random variable? – Martin Sep 12 '12 at 9:17
Thanks for your detailed elaboration (and sorry for the late reply; I somehow missed your edit). Anyway, I'm not sure whether I fully understand the second line. To me it seems that $F_{W,Z}(a,b) = P\{W \leq a, Z \leq b\} = P\{W\leq a \cap Z \leq b\} = ...$ We can proceed with using $P(A \cap B) = P(A) + P(B) - P(A \cup B)$. For a two-dimensional distribution function $F_{W,Z}(a,b)$, both conditions $W \leq a$ and $Z \leq b$ must hold simultaneously, no? – Martin Sep 24 '12 at 10:02