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

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?

• What do your students know? It's pretty easy to calculate the joint distribution $(X,\min (X,Y))$ if your students know about general measures. It doesn't have density, though, as you say. – WoolierThanThou Feb 19 '20 at 10:09
• Okay, assume that I computed joint distribution and it has no density. How to compute conditional expectation? – mwrooo Feb 19 '20 at 12:31
• Well... you don't need a density, you just need a Markov kernel. If $(X,Y)\sim \mu$ and $\mu(A\times B)=\int_B \nu_y(A)\textrm{d} \gamma(y)$ for some appropriate measure $\gamma$ and family of measure $\nu_y$, then $\mathbb{E}(X|Y)=\int_{\mathbb{R}} x\textrm{d}\nu_Y$ by exactly the same argument that works in the density case. – WoolierThanThou Feb 19 '20 at 14:30
• In your case, if we denote by $\lambda^2$ the 2-dimensional Lesbegue measure and $\lambda_{\Delta}$ the Lesbegue measure on the diagonal lifted from $\mathbb{R}$. Then your measure is $\mu=1_{x\geq y} 1_{0\leq x\leq 1} \textrm{d}\lambda^2+(1-x)\textrm{d}\lambda$, which does have a Markov kernel decomposition with $\nu_y= \frac{1}{2}\delta_y+\frac{1}{2(1-y)}\cdot 1_{[y,1]}\textrm{d}\lambda$. – WoolierThanThou Feb 19 '20 at 14:45
• Okay, I agree, but the problem is how to find that $\gamma$ measure and the family of $\nu_y$? – mwrooo Feb 19 '20 at 14:46

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 $$\mathbb P(X \le x \mid \min(X,Y)=t )=\left\{ \begin{matrix}0 & \text{ for } x\lt t \\\frac12 & \text{ for } x= t \\ \frac{x+1-2t}{2-2t} & \text{ for } t \lt x\le 1 \\ 1 & \text{ for } 1 \lt x \end{matrix} \right.$$

• Very clear and transparent explanation! Thank you! – NCh Feb 19 '20 at 10:15
• Thank You very much! – mwrooo Feb 19 '20 at 10:42
• @StubbornAtom $P(X\le t \mid \min(X,Y)=t) =\frac12$ is correct, but since $P(X\lt t \mid \min(X,Y)=t) =0$ that would suggest to me $P(X= t \mid \min(X,Y)=t) =\frac12$ too – Henry Feb 19 '20 at 14:43
• @StubbornAtom On the density, you have to stretch a probability of $\frac12$ uniformly over an interval of length $1-t$, which I would have thought would make the density $\frac{1/2}{1-t} =\frac{1}{2(1-t)}$ – Henry Feb 19 '20 at 14:46
• @mwrooo - I would guess $\mathbb E[\min(X,Y)\mid \max(X,Y) =s] = \int\limits_{t=0}^s \frac{t \lambda e^{-t\lambda}}{1-e^{-s\lambda}}\, dt = \frac{1-(1+s\lambda)e^{-s \lambda}}{\lambda\left(1-e^{-s \lambda}\right)}$ – Henry Feb 20 '20 at 23:20