I have the following question at hand:

Suppose $U,V$ are iid random variables following Unif$(0,1)$. what is the conditional distribution of $U$ given $Z:=\max(U,V)$ ?

I tried writing $Z=\Bbb{I}\cdot V+(1-\Bbb{I})\cdot U$ where $\Bbb{I}=\begin{cases}1&U<V\\0&U>V\end{cases}$

But I am not getting anywhere.

  • 3
    $\begingroup$ This may be wrong, but here goes. If $U$ is the max, then $U=Z$. Otherwise $U<V=Z$, so $U$ would be uniform on $[0,Z]$. The two cases should have equal probability, so then $U$ has a mix of the two distributions? $\endgroup$
    – GeoMatt22
    Aug 27 '16 at 20:24
  • $\begingroup$ Cross-posted at math.stackexchange.com/questions/1905481/…. $\endgroup$ Jan 4 '20 at 6:51

A picture might help. Independent uniform distributions on the interval $[0,1]$ may be considered a uniform distribution on the unit square $I^2 = [0,1]\times [0,1]$. Events are regions in the square and their probabilities are their areas.


Let $z$ be any possible value of $\max(U,V)$. The set of coordinates $(U,V)$ where $\max(U,V)=z$ forms the top and right edges of a square of side $z$. Let $dz$ be a small positive number. The set of coordinates $(U,V)$ whose maximum lies between $z$ and $z+dz$ forms a narrow thickening of that square, as shaded in the figure. Its area is the difference of the areas of two squares, one of side $z+dz$ and the other of side $z$, whence

$$\Pr(z \le Z \le z+dz) = (z+dz)^2 - z^2 = 2z\,dz + (dz)^2.\tag{1}$$

Let $u$ be any possible value of $U$: it is marked with a vertical dashed line in the figures.

The left panel shows a case where $u \le z$: The chance that $U\le u$ would be the area to the left of that line (equal to $u$); but the event that $U\le u$ and $Z$ lies between $z$ and $z+dz$ is just the brown shaded area. It's a rectangle, so its area is its width $u$ times its height $dz$. Thus,

$$\Pr(U \le u, z \le Z \le z+dz) = u\,dz.\tag{2}$$

The right panel shows a case where $z \lt u \le z+dz$. Now the chance that $U \le u$ and $z \lt Z \le z+dz$ consists of two rectangles. The top one has base $u$ and height $dz$; the right one has base $(u-z)$ and height $z+dz$. Therefore

$$\Pr(U \le u, z \le Z \le z+dz) = u\, dz + (u-z)(z+dz).\tag{3}$$

By definition, the conditional probabilities are these chances divided by the total chance that $z \le Z \le z+dz$, given in $(1)$ above. Divide $(2)$ and $(3)$ by this value. Letting $dz$ be infinitesimal, and retaining the standard part of the result, gives the chances conditional on $Z=z$. Thus, when $0 \le u \le z$,

$$\Pr(U \le u\,|\, Z=z) = \frac{u\,dz}{2z\,dz + (dz)^2} = \frac{u}{2z + dz} \approx \frac{u}{2z}.$$

When $z \lt u \le z+dz$, write $u = z + \lambda dz$ for $0 \lt \lambda \le 1$ and compute

$$\Pr(U \le u|Z=z) = \frac{u\, dz + (u-z)(z+dz)}{2z\,dz + (dz)^2} = \frac{(z + \lambda dz)dz + (\lambda dz)(z+dz)}{2z\,dz+(dz)^2}\approx\frac{1+\lambda}{2}.$$

Finally, for $u \gt z+dz$, the brown area in the right panel has grown to equal the gray area, whence their ratio is $1$.

These results show that the conditional probability grows linearly from $0$ to $z/(2z)=1/2$ as $u$ grows from $0$ to $z$, then shoots up linearly from $1/2$ to $1$ in the infinitesimal interval between $z$ and $z+dz$, then stays at $1$ for all larger $u$. Here's a graph:

Figure 2

Because $dz$ is infinitesimal, it is no longer possible to distinguish $z$ from $z+dz$ visually: the plot jumps from a height of $1/2$ to $1$.

Putting the foregoing together into a single formula to be applied to any $z$ for which $0 \lt z \le 1$, we could write the conditional distribution function as

$$F_{U|Z=z}(u) = \left\{\begin{array}{ll} 0 & u \le 0 \\ \frac{u}{2z} & 0 \lt u\le z \\ 1 & u \gt z. \end{array} \right.$$

This is a complete and rigorous answer. The jump shows that a probability density function will not adequately describe the conditional distribution at the value $U=z$. At all other points, though, there is a density $f_{U|Z=z}(u)$. It is equal to $0$ for $u\le 0$, $1/(2z)$ for $0 \le u \lt z$ (the derivative of $u/(2z)$ with respect to $u$), and $0$ for $u \gt z$. You could use a "generalized function" to write this in a density-like form. Let $\delta_z$ be the "generalized density" giving a jump of magnitude $1$ at $z$: that is, it's the "density" of an atom of unit probability located at $z$. Then the generalized density at $z$ can be written $\frac{1}{2}\delta_z$ to express the fact that a probability of $1/2$ is concentrated at $z$. In full, we could write

$$f_{U|Z=z}(u) = \left\{\begin{array}{ll} 0 & u \le 0 \\ \frac{1}{2z} & 0 \lt u\lt z \\ \frac{1}{2}\delta_z(u) & u=z \\ 0 & u \gt z. \end{array} \right.$$


First consider the distribution of the maximum $Z$ conditional on $U=u$. The maximum $Z$ becomes equal to $u$ in the event that $V<u$ with conditional probability $u$. Otherwise, $Z$ takes some value greater than $u$ equal to $V$. The overall conditional distribution will thus be a mixture between a point mass at $u$ (of size u) and a uniform density on $(u,1)$ (integrating to $1-u$). Representing the point mass by the Dirac delta function, the generalised probability density function (gpdf) of this conditional distribution is $$ f_{Z|U=u}(z) = u\delta(z-u) + \begin{cases} 1 & \text{for }u<z<1 \\ 0 &\text{otherwise}. \end{cases} $$ The joint gpdf of $Z$ and $U$ is then \begin{align} f_{Z,U}(z,u) &= f_{Z|U=u}(z)f_U(u) \\ &= u\delta(z-u) + \begin{cases} 1 & \text{for }0<u<z<1 \\ 0 &\text{otherwise}. \end{cases} \end{align} The pdf of the maximum is $f_Z(z)=2z$. Hence, the conditional gpdf of $U$ given the maximum $Z$ becomes \begin{align} f_{U|Z=z}(u) &= \frac{f_{Z,U}(z,u)}{f_Z(z)} \\ &= \frac12\delta(z-u) + \begin{cases} \frac1{2z} & \text{for }0<u<z \\ 0 &\text{otherwise}, \end{cases} \end{align} a mixture between a point mass at $z$ with probability 1/2 and a uniform density on $(0,z)$ integrating to 1/2.

  • $\begingroup$ Well , +1 for your great help!! But i have a problem.. I dont know DIRAC DELTA function. ... So can it be done without it? $\endgroup$
    – Qwerty
    Aug 27 '16 at 21:44
  • 1
    $\begingroup$ I don't know. It seems like a convenient way of representing a distribution which is part discrete and part continuous. A thread at math.stackexchange has some further discussion. $\endgroup$ Aug 27 '16 at 21:52
  • 1
    $\begingroup$ @Qwerty: a way to work without delta function is to avoid densities, using instead cdfs. $\endgroup$
    – Xi'an
    Jul 8 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.