Solution
Square Beats Product
Show the problem again
X, Y and Z are independent and uniform on [0, 1]. What is the probability that Z² > XY? Give a reduced fraction.
Worked solution
The answer is 5/9. Taking square roots, Z² > XY is equivalent to Z > √X·√Y. For a fixed value c in [0, 1], P(Z > c) = 1 − c, so P(Z² > XY) = 1 − E[√X·√Y] = 1 − E[√X]·E[√Y] = 1 − (2/3)² = 5/9, using independence and E[√X] = ∫₀¹ √x dx = 2/3. In words: Z beats the geometric mean of X and Y with probability 5/9, even though Z beats each of X and Y separately only half the time. The geometric mean is pulled downward, since it never exceeds the arithmetic mean.
Source: Original problem written for AxiomIQ.