Expected Value of the Minimum of Two Uniforms
Let $A \sim U(0, 5)$ and $B \sim U(0, 10)$ be independent uniform random variables. Compute $E[\min(A, B)]$.
Hints
- For the expected value of a non-negative random variable, recall the survival function identity: $E[X] = \int_0^\infty P(X > t)\,dt$.
- For independent variables, $P(\min(A,B) > t) = P(A > t) \cdot P(B > t)$. Write out each survival function piecewise.
- Since $P(A > t) = 0$ for $t > 5$, you only need to integrate from $0$ to $5$: $\int_0^5 \frac{(5-t)(10-t)}{50}\,dt$.
Worked Solution
How to Think About It: When you see $E[\min(\cdot)]$ for independent continuous variables, your first instinct should be the survival function method: $E[X] = \int_0^\infty P(X > t)\,dt$ for non-negative $X$. For a minimum of independent variables, $P(\min(A,B) > t) = P(A > t)P(B > t)$, which factors nicely. This avoids the messier double integral approach.
Quick Estimate: The minimum is bounded above by $A$, so $E[\min(A,B)] < E[A] = 2.5$. The minimum is also bounded below by 0. Since $B$ has a wider support, it often exceeds $A$, so the minimum is often just $A$. But $B$ pulls the minimum down when $B < A$. Rough guess: somewhere around $2$. Let's compute exactly.
Approach: Use the survival function identity $E[\min(A,B)] = \int_0^\infty P(A > t)P(B > t)\,dt$.
Formal Solution: The survival functions are:
- $P(A > t) = \frac{5 - t}{5}$ for $t \in [0, 5]$, and $0$ for $t > 5$
- $P(B > t) = \frac{10 - t}{10}$ for $t \in [0, 10]$, and $0$ for $t > 10$
Since $P(A > t) = 0$ for $t > 5$, the integral is:
$$E[\min(A,B)] = \int_0^5 \frac{(5-t)(10-t)}{50}\,dt$$
Expand the numerator:
$$(5-t)(10-t) = 50 - 15t + t^2$$
Integrate:
$$E[\min(A,B)] = \frac{1}{50}\int_0^5 (50 - 15t + t^2)\,dt = \frac{1}{50}\left[50t - \frac{15t^2}{2} + \frac{t^3}{3}\right]_0^5$$
$$= \frac{1}{50}\left(250 - \frac{375}{2} + \frac{125}{3}\right) = \frac{1}{50}\left(\frac{1500 - 1125 + 125}{6}\right) \cdot \frac{6}{6}$$
Computing the bracket: $250 - 187.5 + 41.\overline{6} = 104.1\overline{6} = \frac{625}{6}$.
$$E[\min(A,B)] = \frac{625}{6 \times 50} = \frac{625}{300} = \frac{25}{12}$$
Answer: $E[\min(A,B)] = \dfrac{25}{12} \approx 2.083$.
Intuition
The survival function method is the standard tool for computing expectations of order statistics. The identity $E[X] = \int_0^\infty P(X > t)\,dt$ transforms the problem from integrating a complicated density (which requires convolution or conditioning for the minimum) into integrating a simple product of survival functions. For independent variables, this product factors immediately.
This technique appears everywhere in quant work -- from pricing first-to-default baskets (the minimum of correlated default times) to computing expected first-arrival times in queuing models. The key habit: whenever you see "expected value of a minimum (or maximum)," reach for the survival function before attempting a joint density approach.