Question

Say we have X ~ Uniform(0, 1) and Y ~ Uniform(0, 1). What is the expected value of the minimum of X and Y?

Solution

The questions has asked to find the expected value of the minimum value of the joint probability of random variables X and Y, which are both distributed as the standard uniform distribution.

Let \(Z\) be the minimum value of \(X\) and \(Y\) so, \(Z = min(X, Y)\). To find the expected value of \(Z\), we need the Probability density function (PDF) of \(Z\).

Then, we have

\[P(Z\le z) = P(X\le z, Y\le z) = 1 - P(X>z, Y>z)\]

Since, \(X\sim Unif(0, 1)\) and \(Y\sim Unif(0, 1)\), so,

\[P(X>z) = 1 - z\] \[P(Y>z) = 1 - z\]

Also, \(X\) and \(Y\) are independent and identically distributed (i.i.d) as they both are distributed as the standard uniform distribution and knowing about \(X\) doesn’t help in determining \(Y\).

Thus,

\[P(X>z, Y>z) = P(X>z)\cdot P(Y>z) = (1 - z)\cdot (1 - z) = (1 - z)^2\]

Then, we get \(P(Z\le z)\) as,

\[F_Z = P(Z\le z) = 1 - (1 - z)^2\]

\(F_Z\) is the Cumulative distribution function (CDF) of Z. Now, to find the PDF of Z, we just need to differentiate the CDF of Z with respect to z. Thus,

\[P_Z = \frac{d(F_Z)}{dz} = 2(1 - z)\]

This is the PDF of the random variable Z. Now, to find the expected value of Z, which is also distributed as standard uniform, we integrate the PDF.

\[E(Z) = \int_0^1 z P_Z dz = 2\int_0^1 z(1 - z) dz = 2 (\frac{1}{2} - \frac{1}{3}) = \frac{1}{3}\]

Therefore, the expected value of the minimum of \(X\) and \(Y\) is \(\frac{1}{3}\).