Help with probability please?

Question

Let A, B, and C be independent uniform random variables on the interval (0,1). Show that the probability that AB is greater than C^2 is 5/9.

Answer 1

Actually, the probability that AB is greater than C^2 is 4/9, not 5/9. Observe:

Let B and C be fixed, then P(AB > C²) = P(A > C²/B) = {0 if C²/B ≥ 1, 1-C²/B if C²/B < 1}

Now, let C be fixed. P(AB > C²) = [0, 1]∫P(AB > C² | B=x) dx = [0, 1]∫{0 if C²/x ≥ 1, 1-C²/x if C²/x<1} dx. Obviously, C²/x ≥ 1 ⇔ x≤C², so this is [C², 1]∫1-C²/x dx. But this is:

[C², 1]∫1-C²/x dx
x - C² ln x |[C², 1]
1 - C² - C² ln (1/C²)
1 - C² + 2C² ln C

Finally, consider that the probability that AB > C² is:

[0, 1]∫P(AB>C² | C² = x) dx
[0, 1]∫1 - x² + 2x² ln x dx
x - x³/3 | [0, 1] + [0, 1]∫2x² ln x dx
1 - 1/3 + [0, 1]∫2x² ln x dx
2/3 + [0, 1]∫2x² ln x dx

Integrating by parts:

2/3 + 2x³/3 ln x | [0, 1] - [0, 1]∫2x²/3 dx

2(1)³/3 ln 1 = 0, but the left limit is not strictly defined and so must be evaluated as a limit:

2/3 - [x→0]lim (2x³/3 ln x) - [0, 1]∫2x²/3 dx
2/3 - 2/3 [x→0]lim (ln x/(1/x³)) - [0, 1]∫2x²/3 dx
2/3 - 2/3 [x→0]lim ((1/x)/(-3/x⁴)) - [0, 1]∫2x²/3 dx
2/3 - 2/3 [x→0]lim (-x³/3) - [0, 1]∫2x²/3 dx
2/3 - [0, 1]∫2x²/3 dx

Finally:

2/3 - [0, 1]∫2x²/3 dx
2/3 - 2x³/9 | [0, 1]
2/3 - 2/9
4/9

So in fact, the probability is 4/9, not 5/9. This makes sense -- since A and B are independent, E(AB) = [0, 1]×[0, 1]∬xy dA = [0, 1]∫[0, 1]∫xy dy dx = [0, 1]∫x [0, 1]∫y dy dx = [0, 1]∫x dx [0, 1]∫y dy = 1/2 * 1/2 = 1/4, but E(C²) = [0, 1]∫x² dx = 1/3. Since AB has a lower expected value than C², it makes sense that P(AB>C²) < 1/2 (although note that in general E(X) < E(Y) does NOT imply that P(X>Y) < 1/2, it merely makes it less surprising).