Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

up vote 4 down vote accepted

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 $[\ell, 1)$.

Edit added information as requested by OP

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.

share|improve this answer
    
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
    
@Martin I have added a few more lines to clarify what is going on. –  Dilip Sarwate Sep 24 '12 at 12:30
    
Makes sense now. Thanks for the effort! –  Martin Sep 24 '12 at 14:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.