# Computing conditioned probability of $X$ by $U=\min(X,Y)$

Let $X$ and $Y$ be independent random variables with $P(X\leq x)=F_x(x)$ and $P(Y\leq y)=F_y(y)$.

Let $U=\min(X,Y)$. I know that $F_u(u)=1-(1-F_x(u)(1-F_y(u))).$

By definition: $P(X \leq x |U=u)= \lim_{h\rightarrow 0}\frac{P(X \leq x,u-h<U<u+h)}{P(u-h<U<u+h)}$ if the limit exits.

I can compute $P(X<Y)=\int_{- \infty}^{+ \infty}{F_{xy}(X<Y|X=x)}f_x(x)dx=\int_{- \infty}^{+ \infty}{F_{y}(Y>x)}f_x(x)$ and solving the integral I get the desired probability.

Two questions:

1.- How can I compute $P(X<x|U=u)$ ?

2.- How can I compute $P(X<Y|U=u)$ ?

My problem is that I don't know how to compute $F_{XU}$.

• May 2, 2016 at 6:46
• Just to note that you could just as easily write $hx$ in place of $h$. In other words, the answer might depend on how you approach the condition $U=u$. May 2, 2016 at 8:14

In the context of a general, abstract $F$, the questions "how can I compute..." can be understood as a request for intuitive ways to think about these conditional probabilities. I believe many people visualize them, as I will describe here. I will illustrate the conditional probabilities involved in the two questions and then conclude with some general remarks.

By definition, the probability of an event $\mathcal{A}$ conditioned on an event of positive probability $\mathcal{E}$ is the chance of $\mathcal{A}\cap\mathcal{E}$, normalized to unity:

$$\Pr(\mathcal{A}\,|\, \mathcal{E}) = \frac{\Pr(\mathcal{A}\cap \mathcal{E})}{\Pr(\mathcal{E})}$$

Let's start with the second question, to find $\Pr(X \lt Y\,|\, U=u)$. The events are

$$\mathcal{E} = \{\omega\,|\, u-h \lt \min(X(\omega),Y(\omega)) \le u+h\}$$

and

$$\mathcal{A} = \{\omega\,|\, X(\omega) \lt Y(\omega\}.$$

We may visualize these events in a two-dimensional Cartesian space whose coordinates are the values $X(\omega)$ and $Y(\omega)$, briefly written $(X,Y)$. In the first figure, the event $\mathcal{A}$ ($X$ is smaller than $Y$) is lightly shaded in gray and the event $\mathcal{E}$ (the smaller of $X$ and $Y$ is close to $u$) is darkly colored.

The series of gray lines are contours (level sets) of the function $(X,Y)\to\min(X,Y)$. They help us see that any event of the form "$\min(X,Y)$ is close to $u$" must be a slight thickening of the contour corresponding to $u$.

The blue colors (on the vertical arm of the colored vee shape) show us the intersection $\mathcal{A}\cap\mathcal{E}$. Visually--if we ignore the tiny detail at the very tip of the vee shape--this diagram is inviting us to compute $\Pr(X \lt Y\,|\,\min(X,Y)=u)$ as the vertical probability

$$\Pr(u-h\lt X \le u+h, Y \gt u)$$

relative to the total colored probability. This immediately gives us answers in many common cases, such as when $X$ and $Y$ have continuous distributions and are identically distributed. Then, the exchangeability of $X$ and $Y$ shows that the blue and green arms have the same chances and the continuity of their distributions shows we may ignore whatever might be happening where the arms meet, implying the conditional probability must be $1/2$ in that case.

To further illustrate, suppose (as in the question) that $X$ and $Y$ are independent with differentiable distribution functions $F_X$ and $F_Y$ having derivatives $f_X$ and $f_Y$. Then, up to a vanishingly small error associated with the point of the vee, the probability of the blue vertical arm is the product of probabilities associated with its base and its height:

$$\Pr(u-h \lt X \le u+h, Y \gt u) = (F_X(u+h)-F_X(u-h))(1 - F_Y(u))$$

with a comparable formula holding for the green horizontal arm (just swap the roles of $X$ and $Y$). The Mean Value Theorem allows us to approximate this by

$$(F_X(u+h)-F_X(u-h))(1 - F_Y(u)) \approx 2h f_X(u) (1-F_Y(u)).$$

Dividing by the total probability yields

\eqalign{ \frac{\Pr(u-h \lt X \le u+h, Y \gt u)}{\Pr(u-h \lt \min(X,Y) \le u+h)} &\approx \frac{2h f_X(u) (1-F_Y(u))}{2h f_X(u) (1-F_Y(u)) + 2h f_Y(u) (1-F_X(u))} \\ &= \frac{f_X(u) (1-F_Y(u))}{f_X(u) (1-F_Y(u)) + f_Y(u) (1-F_X(u))}. }

In the limit as $h\to 0$, this becomes exact.

In the first question, the event in question becomes

$$\mathcal{A} = \{\omega\,|\, X(\omega) \lt x\}.$$

The second figure repeats the scheme of the first, shading $\mathcal{A}$ (the event $X\lt x$) gray, coloring $\mathcal{E}$ (the smaller of $X$ and $Y$ is close to $u$), and depicting contours of $U=\min(X,Y)$ for reference:

Again the color distinguishes the event $\mathcal{A}\cap\mathcal{E}$ (in blue) from its complement $\mathcal{E}\setminus\mathcal{A}$ (in green). It serves as a guide to writing down regions of integration (when $X$ and $Y$ have densities) or limits of sums (when $X$ and $Y$ are discrete) and for reasoning about these probabilities. As before, the conditional probability will equal the total probability in the blue region divided by the total probability in the colored region.

Such diagrams allow us quickly to draw useful conclusions. For instance, although you could have shown this algebraically before, it is now instantly obvious that this conditional probability must be zero whenever $x \lt u$: for sufficiently small $h$, the gray region and colored region would not intersect. It is just as obvious that the distribution function

$$x\to \Pr(X \lt x\,|\, U=u)$$

will likely jump from zero to a positive value at $x=u$, because within the arbitrarily small range from $u-h$ to $u+h$ it includes all the (relative) probability of the vertical arm of the vee.

To say more, in any detail, would require additional information about the variables. Usually the principal technical concerns are whether $X$ and $Y$ have density functions and--if they do not--how the distinctions between $\lt$ and $\le$ are going to matter in computing the probabilities.

Note that this visualization works even when $X$ and $Y$ are not independent.

• Could you show me how to compute this probabilities with the following two examples?: 1st example: $X \sim Uniform(0,5)$ $Y \sim Uniform(0,3)$ and $u= \frac{1}{2}$ and 2nd example $X \sim Exponential(\lambda=5)$ $Y \sim Exponential(\lambda=3)$ and $u=\frac{1}{3}$. In the second example I suppose that P(X<Y|U=u)=5/(5+3) because of the memoryless of the exponential, but I want to understand the process to get it. May 2, 2016 at 15:43
• I added an explicit formula for the second question. You may work out a comparable formula for the first question using the second figure.
– whuber
May 2, 2016 at 16:29
• Before I could see your edited post I was getting the expression for my second question. To answer my first question I suppose that I can get it by conditioning so $P(X<x|U=u)=P(X<x|U=u, X<Y)*P(X<Y|U=t)+ P(X<x|U=u, Y<X)*(1-P(X<Y|U=t))=1*P(X<Y|U=t) + P(X<x|U=u, Y<X)*(1-P(X<Y|U=t))$ when $u\leq x$ and $P(X<x|U=u, Y<X)=\frac{P(u<X \leq x)}{P(X>u)}$. Is it right. May 2, 2016 at 19:45
• I didn't check your whole calculation, but just looking at the very last equality suggests there's a problem: your formula does not appear to account for the vertical (blue) part of the probability at all (the section where $X\approx u$ and $Y \gt u$). If you use the figures as a guide to writing the probabilities, you should be able to see the parts of the figure even in the final answer.
– whuber
May 2, 2016 at 20:09
1. The easier definition of $P(X<x|U=u)$ is

$$P(X<x|U=u)=\int_{-\infty}^xf_{X|U}(t|u)dt,$$

where

$$f_{X|U}(t,u)=\frac{f_{XU}(t,u)}{f_U(u)},$$

where $f_{X,U}(t,u)$ is the probability density of the $(X,U)$ and $f_U(u)$ is the probability density of $U$.

1. Also

$$P(X<Y)=\int_{(x,y)\in\mathbb{R}^2:x<y}f_X(x)f_Y(y)dxdy.$$

Your formula with the cdf inside the integral is probably wrong.

1. We have

$$F_{XU}(x,u)=P(X<x, U<u) = P(X<x, \min(X,Y)<u) = P(X<x \cap \min(X,Y) < u).$$

Now

$$\min(X,Y)>u = X>u \cap Y>u,$$

so you can use de Morgan's law to get the expression for $\min(X,Y)<u$ (you should be aware of it, because you have stated the formula for $F_U(y)$) and then use the distributivity property of union and intersection together with the fact that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ to arrive at the final expression for $F_{XU}(x,u)$.

This is not a complete solution, but it may clarify things a bit.

• I think your 3. is not right. $X <Y$ does mean $U=X$, but this isn't the condition, $U=u$ is. I don't think you can interchange $X$ and $U$ in the conditional part. You have $\frac {Pr (X=U=u)}{Pr (U=u)}$ May 2, 2016 at 11:18
• My problem is to compute $f_{X,U}$ once I got it, I can use it to compute $f(x|u)=\frac{f(x,u)}{f(u)}$ but How I get $f(x,u)$? May 2, 2016 at 14:28
• To see it with an example supose that $X \sim Uniform(0,5)$ and $Y \sim Uniform(0,3)$ and $U=\min(X,Y)=\frac{1}{3}$ How can I compute $P(X<x|u=\frac{1}{3})$ ? May 2, 2016 at 14:31