Finding $E(X\mid \min(X,Y))$ where $X,Y$ are independent $U(0,1)$ variables

Question

Let $X,Y$ be two independent random variables, uniformly distributed on $(0,1)$. I have to compute $E(X\mid\min(X,Y)=t)$, but in very educational way, I mean, I have to explain it to my students and I would rather prefer do not use very advanced methods. Is it possible? I was trying to compute common distribution function and then take its density, but I think that density does not exist. So, is there any easy, FORMAL way?

Answer 1

You could try drawing a square something like this as the joint distribution is uniform on the square

The condition $\min(X,Y)=t$ corresponds to points on the orange and red lines (orange for $t=X \le Y$ and red for $t=Y \le X$

Conditioned on this, the probability $P(X=t \mid \min(X,Y)=t) =\frac12$ by symmetry, and if not then $X$ is uniformly distributed between $t$ and $1$ so with density $f(x)=\frac{1}{2(1-t)}\mathbf 1_{t \lt x \lt 1}$

This makes the conditional expected value $\mathbb E[X \mid \min(X,Y)=t ]=\frac12 t + \frac12\left(\frac{t +1}{2}\right) = \frac 34 t +\frac14$

This turns out to be formal, with the corresponding conditional CDF $$\mathbb P(X \le x \mid \min(X,Y)=t )=\left\{ \begin{matrix}0 & \text{ for } x\lt t \\\frac12 & \text{ for } x= t \\ \frac{x+1-2t}{2-2t} & \text{ for } t \lt x\le 1 \\ 1 & \text{ for } 1 \lt x \end{matrix} \right.$$