Conditional Uniform Selection: Expected Value
Adam draws a number $A$ uniformly at random from $(0, 1)$. After seeing Adam's draw $a$, Eve draws a number $V$ uniformly at random from $(a, 1)$.
Find $E[V]$.
Hints
- Eve's distribution depends on Adam's draw, so think about conditioning -- what tool lets you compute an expectation by averaging over a conditioning variable?
- Apply the Law of Total Expectation: $E[V] = E[E[V \mid A]]$. First find $E[V \mid A = a]$ for a fixed $a$.
- Since $V \mid A = a \sim U(a, 1)$, its mean is $(a + 1)/2$. Now take the expectation of $(A + 1)/2$ over $A \sim U(0, 1)$.
Worked Solution
How to Think About It: This is a nested randomness problem -- Eve's distribution depends on Adam's outcome. The right reflex is the Law of Total Expectation: condition on Adam's draw first, compute Eve's expected value given that draw, then average over Adam. Before any calculation, you can gut-check the answer. Adam averages $1/2$, so Eve -- who always picks above Adam -- should average somewhere above $1/2$ and below $1$. A value around $3/4$ seems plausible.
Quick Estimate: If Adam drew $a = 1/2$ (his average), Eve would draw uniformly from $(1/2, 1)$, giving $E[V \mid A = 1/2] = 3/4$. That is the plug-in estimate, and it turns out to be exactly right -- the answer is $3/4$.
Approach: Apply the Law of Total Expectation $E[V] = E[E[V \mid A]]$, using the fact that $V \mid A = a$ is $U(a, 1)$.
Formal Solution:
Set up the distributions: $$A \sim U(0, 1), \qquad V \mid A = a \sim U(a, 1)$$
The conditional expectation of Eve's draw given Adam's: $$E[V \mid A = a] = \frac{a + 1}{2}$$
Apply the Law of Total Expectation: $$E[V] = E\!\left[E[V \mid A]\right] = E\!\left[\frac{A + 1}{2}\right] = \frac{1}{2}\left(E[A] + 1\right) = \frac{1}{2}\left(\frac{1}{2} + 1\right) = \frac{3}{4}$$
Answer: $E[V] = \dfrac{3}{4}$
Intuition
The Law of Total Expectation is one of the most useful tools in probability -- it says you can always break a hard expectation into a conditional expectation (easy to compute for a fixed given value) and then average that over the conditioning variable. This two-step decomposition turns a problem with nested randomness into two simple calculations.
In practice, this structure shows up constantly in quantitative finance: a payoff that depends on a random parameter (think: a stochastic volatility model where you condition on the vol path, or a Bayesian market-making model where you condition on the true value). The plug-in intuition is also worth internalizing -- when conditioning is linear in the conditioning variable, plugging in the mean of the conditioning variable gives you the exact answer, not just an approximation. Here, $E[V \mid A = a] = (a+1)/2$ is linear in $a$, so the plug-in estimate at $a = E[A] = 1/2$ is exact.