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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 $Z = \max(X,Y)$.

Did
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user11697
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2 Answers2

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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 $E[\min (X,Y )]=2E[X]-E[\max (X,Y )]$

Ivan3.14
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  • 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 05 '11 at 11:43
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Hint: First find the distribution function of $Z$, then its density function, then its expectation.

Shai Covo
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  • 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 05 '11 at 08:42
  • You should maybe mention, that this simplifies to $E[Z]=1/3 a+2/3 b$ – Listing Jun 05 '11 at 09:12
  • Which some of us might prefer to write as $a+\frac23(b-a)$. – Did Jun 05 '11 at 09:25
  • What if $X \sim U(a,b)$ and $Y \sim U(c,d)$, X and Y are independent. How to find E(max(X,Y))? – Uddin May 30 '20 at 01:59