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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 = X 1 + X 2 . Since X 3 is independent of S, P(X 1 + X 2 < X 3 ) = E[P(X 3 > S | S)]. Because X 3 ~ U(0, 1), P(X 3 > S) = 1 − S, for 0 ≤ S ≤ 1 0, for S > 1 The sum S = X 1 + X 2 has the triangular density f S (s) = s, for 0 ≤ s ≤ 1 f S (s) = 2 − s, for 1 ≤ s ≤ 2 Hence, P(X 1 + X 2 < X 3 ) = ∫ 0 1 (1 − s)s ds = ∫ 0 1 (s − s²) ds = [s²/2 − s³/3] 0 1 = 1/2 − 1/3 = 1/6 . Final Answer P(X 1 + X 2 < X 3 ) = 1/6