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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ and $Y$ be independent random variables with uniform distribution on $[0,1]$, in notation:

$X$~$Unif(0,1)$, and $Y$~$Unif(0,1)$.

Derive (a) $E(min(X,Y))$, (b) $E(|X - Y|)$, (c) $E((X+Y)^2)$.

I've been getting seriously frustrated with this question because I can't seem to find any examples in the books I have or my class notes. :( Any help is very much appreciated! Even links to online examples :'(

share|cite|improve this question
up vote 3 down vote accepted


It is a matter of setting up integrals defining these expectation: $$ \mathbb{E}\left(\min(X,Y)\right) = \int_0^1 \int_0^1 \min(x,y) \mathrm{d}x \mathrm{d}y = \int_0^1 \left(\int_0^y x \mathrm{d}x + \int_y^1 y \mathrm{d} x \right) \mathrm{d}y $$ $$ \mathbb{E}\left(|X-Y|\right) = \int_0^1 \int_0^1 |x-y| \mathrm{d}x \mathrm{d}y = \int_0^1 \left(\int_0^y (y-x) \mathrm{d}x + \int_y^1 (x-y) \mathrm{d} x \right) \mathrm{d}y $$ $$ \mathbb{E}\left((X-Y)^2\right) = \int_0^1 \int_0^1 (x-y)^2 \mathrm{d}x \mathrm{d}y $$ You should be able to finish these off by evaluating them.

share|cite|improve this answer
Thanks so much! :) – Fred Feb 17 '13 at 16:55
Just wondering if this problem can be set up similarly: Let $X$~$Exp(\lambda)$, that is, $f_X (x) = \lambda*exp(-\lambda)$ and $Y$~$Exp(\mu)$, that is, $f_Y (y) = \mu*exp(-\mu)$ Find the density of $Z=min(X,Y)$ when X,Y are independent. – Fred Feb 17 '13 at 17:34
@Panda To solve the latter problem it is easier to first find $1-F_Z(z) = \mathbb{P}\left(Z > z \right) = \mathbb{P}\left(\min(X,Y) > z \right) = \mathbb{P}\left(X > z, Y > z \right) = (1-F_X(z)) (1-F_Y(z))$. It now remains to differentiate $f_Z(z) = F_Z^\prime(z)$. – Sasha Feb 17 '13 at 17:45
@Panda You can either use the law of the total probability $\mathbb{P}\left(X<Y\right) = \int_0^\infty F_X(y) f_Y(y) \mathrm{d}y$, or set up the integral for the probability: $\mathbb{P}\left(X<Y\right) = \int_0^\infty \int_0^\infty [x<y] f_X(x) f_Y(y) \mathrm{d}x\mathrm{d} y = \int_0^\infty \left(\int_0^y f_X(x) \mathrm{d}x \right) f_Y(y)\mathrm{d} y$. The inner integral is just $F_X(y)$. – Sasha Feb 17 '13 at 19:16
Thanks so much! You've been a huge help! :) – Fred Feb 17 '13 at 19:32

Seems I'm late, nevertheless will try.

There's a slower way, but it may give you a few more insights. Define $W=\min (X,Y)$. I'll start with the discrete case, continuous one is similar. Also $P(X=j)= P(Y=j)=p_j.$

If $X, Y$ are iid and defined on $\mathbb{N}$, to get $W=0$ you have two identical cases: either $X=0 \cap Y \geq1$ or the other way around. For $W=1 \ X=1 \ \text{and} \ Y \geq 2$ or the other way around. And so on. Putting it together, you get the CDF of $W$: $$ \mathbf{P}(W \leq k)= \sum_{j=0}^{k}p_j \mathbf{P}(Y \geq j+1) + \sum_{j=0}^{k}p_j \mathbf{P}(X \geq j+1) $$ For the case in your problem:

$$ F_{W}(w)=\mathbf{P}(W \leq w)=\int_{0}^{w}f_{X}(w')\mathbf{P}(Y \geq w')dw' + \int_{0}^{w}f_{Y}(w')\mathbf{P}(X \geq w')dw'\\ = 2 \int_{0}^{w}(1-w')dw'=2w \bigg( 1-\frac{w}{2}\bigg) $$ Taking the derivative we get $$ f(w)=2(1-w) $$

Now for the actual problem: $$ \mathbf{E}W = \int_{0}^{1}w f(w) dw=2 \int_{0}^{1}w(1-w)dw=\frac{1}{3} $$

share|cite|improve this answer

Your Answer


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.