Let X1, X2, and X3 be independent and identically distributed random variables with uniform distribute on [0, 1]. The probability P(X1+X2 < X3) is ____
Solution
Let S = X1 + X2. Since X3 is independent of S,
P(X1 + X2 < X3) = E[P(X3 > S | S)].
Because X3 ~ U(0, 1),
P(X3 > S) =
- 1 − S, for 0 ≤ S ≤ 1
- 0, for S > 1
The sum S = X1 + X2 has the triangular density
- fS(s) = s, for 0 ≤ s ≤ 1
- fS(s) = 2 − s, for 1 ≤ s ≤ 2
Hence,
P(X1 + X2 < X3) = ∫01 (1 − s)s ds = ∫01 (s − s²) ds = [s²/2 − s³/3]01 = 1/2 − 1/3 = 1/6.
Final Answer
P(X1 + X2 < X3) = 1/6