✔ 最佳答案
(a) E(Y)
= ∫ yf(y) dy (from 0 -> +∞)
= ∫ y d[-(1 - F(y)]
= y[-(1 - F(y)] | [0, +∞] + ∫ (1 - F(y) dy...(*)
Consider b[-(1 - F(b)] where b is a positive constant
b[(1 - F(b)]
= ∫ b f(y) dy (from b -> +∞)
< ∫ y f(y) dy
So, since E(Y) < +∞, the last integral will tend to 0 when b -> + ∞
Thus, y[-(1 - F(y)] | [0, +∞] = 0 and
(*) becomes E(Y)
= ∫ (1 - F(y) dy
= P(Y > y) dy
(b) ∫ P(Y < -y) dy (from 0 -> +∞)
= ∫ F(-y) dy
= yF(-y) | [0,+∞] + ∫ yf(-y) dy
= ∫ yf(-y) dy
= ∫ xf(x) dx (from 0 to -∞ , after letting x = -y)
= -∫ xf(x) dx (from -∞ to 0)
E(Y)
= ∫ yf(y) dy (from -∞ -> +∞)
= ∫ yf(y) dy (from 0 -> +∞) + ∫ yf(y) dy (from -∞ -> 0)
= ∫ P(Y > y) dy (from 0 -> +∞) - ∫ P(Y < -y) dy (from 0 -> +∞)
