# Conditional Expected Value- maximum and minimum

I have to compute $$E(max(X,Y)|min(X,Y)=t)$$ where X and Y are independent and have uniform distribution on $$[0,1]$$.

I made this task:

$$E(max(X,Y)|min(X,Y)=t)=f(t)$$

$$E(max(X,Y)\mathbf{1}_{min(X,Y)>s})=Ef(min(X,Y))\mathbf{1}_{min(X,Y)>s}$$

$$\int_{s}^{1}x*2xdx=\int_{s}^{1}f(x)*(2-2x)dx$$

$$2s^2=f(s)*(2-2s)$$

$$f(s)=(2-2s)/2s^2$$

But the answer is diffrent. Where did I mistake? Thanks in advance.

• – StubbornAtom Feb 18 at 12:25

## 1 Answer

I think you computed your left-hand side incorrectly.

$$E[\max(X,Y) 1_{\min(X,Y) > s}] = \int_s^1 \int_x^1 y \, dy \, dx + \int_s^1 \int_y^1 x \, dx \, dy = \int_s^1 (1-t^2) \, dt$$

Setting the integrand equal to $$f(t) 2(1-t)$$ yields $$f(t) = \frac{1+t}{2}$$.

Alternate approach:

\begin{align}E[\max(X,Y) \mid \min(X,Y) = t] &= \int_0^\infty P(\max(X,Y) \ge u \mid \min(X,Y) = t) \, du \\ &= \int_0^t 1 \, du + \int_t^1 \frac{2(1-u)}{2(1-t)} \, du \\ &= t + \frac{(1-t)^2 / 2}{1-t} = t + \frac{1-t}{2} = \frac{1 + t}{2}. \end{align}