數學 期望值與變異數題目?

(1)
∫ f(y) dy , from y = 0 to y = 2
= ∫ (y/2) dy
= [ y^2 / 4 ] , from y = 0 to y = 2
= 2^2 / 4
= 1
故 f 為機率密度函數 ... Ans

(2)
F(y)
= F( Y ≦ y )
= ∫ f(u) du , from u = 0 to u = y
= ∫ (u/2) du
= [ u^2 / 4 ] , from u = 0 to u = y
= y^2 / 4
Ans: F(y) ≡ F( Y ≦ y ) = y^2 / 4

(3)
E(Y)
= ∫ y f(y) dy , from y = 0 to y = 2
= ∫ y (y/2) dy
= ∫ (1/2)y^2 dy
= [ y^3 / 6 ] , from y = 0 to y = 2
= 8/6
= 4/3 ... Ans

(4)
E(Y^2)
= ∫ y^2 f(y) dy , from y = 0 to y = 2
= ∫ y^2 (y/2) dy
= ∫ (1/2)y^3 dy
= [ y^4 / 8 ] , from y = 0 to y = 2
= 16/8
= 2

Var (Y)
= E(Y^2) - μ^2
= E(Y^2) - [ E(Y) ]^2
= 2 - (4/3)^2
= 2 - 16/9
= 2/9 ... Ans

(5)
P( 1 ≦ Y ≦ 1.5 )
= ∫ f(y) dy , from y = 1 to y = 1.5
= ∫ (y/2) dy
= [ y^2 / 4 ] , from y = 1 to y = 1.5
= ( 1.5^2 / 4 ) - 1/4
= [ (3/2)^2 / 4 ] - 1/4
= 9/16 - 1/4
= 5/16 ... Ans

(6)
E[ (Y+1)/Y ]
= ∫ [ (y+1)/y ] f(y) dy , from y = 0 to y = 2
= ∫ [ (y+1)/y ] (y/2) dy
= ∫ (y+1)/2 dy
= ∫ ( y/2 + 1/2 ) dy
= [ (1/4)y^2 + (1/2)y ] , from y = 0 to y = 2
= 1 + 1
= 2 ... Ans