Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Could anybody help me with solving such problem:

You are given two independent random variables $X$ and $Y$ with continuous uniform distribution. You are to find expected value $E$ of random variable $Z = \max(X,Y)$.

share|cite|improve this question

As Shai Covo said find the distribution function for this. Notice that:

P($\max (X,Y )\leq x$)=P($X\leq x$)P($Y\leq x$) as they are equaly distributed then you already have the distribution of the max then derive for density , multiply by x and integrate and that's it

in general I got $\displaystyle E[\max (X,Y )]=xF(x)\big|_{-\infty}^{\infty}-\int_{-\infty}^{\infty} F^2(x)dx$, for X, Y independent r.v both with distribution equal to F(x) and density f(x) and $E[\min (X,Y )]=2E[X]-E[\max (X,Y )]$

share|cite|improve this answer
Indeed, integration by parts, as your answer suggests, is useful here (assuming that $X$ and $Y$ are i.i.d. uniform$[a,b]$ rv's). Specifically, letting $F_Z$ denote the distribution function of $Z$, we have $$ {\rm E}(Z) = \int_a^b {xf_Z (x)\,dx} = xF_Z (x)\big|_a^b - \int_a^b {F_Z (x)\,dx} = b - (b-a)/3, $$ which is equal to $ \frac{{2b^2 - ab - a^2 }}{{3(b - a)}}$ (that I found by calculating $\int_a^b {xf_Z (x)\,dx} $). – Shai Covo Jun 5 '11 at 11:43

Hint: First find the distribution function of $Z$, then its density function, then its expectation.

share|cite|improve this answer
To check yourself, in the case where $X$ and $Y$ are independent ${\rm uniform}[a,b]$ random variables, ${\rm E}[Z] = \frac{{2b^2 - ab - a^2 }}{{3(b - a)}}$. – Shai Covo Jun 5 '11 at 8:42
You should maybe mention, that this simplifies to $E[Z]=1/3 a+2/3 b$ – Listing Jun 5 '11 at 9:12
Which some of us might prefer to write as $a+\frac23(b-a)$. – Did Jun 5 '11 at 9:25

To consult this page could be profitable.

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.