Three shrinking uniform draws on the unit interval

Question

Draw X uniformly on [0, 1]. Then draw Y uniformly on [0, X]. Then draw Z uniformly on [0, Y]. What is the expected value of Z?

Accepted Answer

The draws are nested: each one's upper limit is the previous draw. Unwind with the tower rule, using the fact that a uniform draw on [0, m] averages \frac{m}{2}. Innermost, condition on Y. Given Y, the draw Z is uniform on [0, Y], so E[Z \mid Y] = \frac{Y}{2}. Taking expectations, E[Z] = \frac{E[Y]}{2}. Middle, condition on X. Given X, the draw Y is uniform on [0, X], so E[Y \mid X] = \frac{X}{2}, giving E[Y] = \frac{E[X]}{2}. Outermost, X is uniform on [0, 1], so E[X] = \frac{1}{2}. Now climb back out, halving at each stage: E[Y] = \frac{1/2}{2} = \frac14, \qquad E[Z] = \frac{1/4}{2} = \frac…

Three shrinking uniform draws on the unit interval

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