隨機變數機率問題。

1. the PDF of X is dF(x)/dx, so,
f(x)= 0, if x<-1
0.2 if -1<=X<0
0.3 if 0<=X<1
0 if X>=1
(a) P(X = -1) = ∫ f(x)dx [from -1 to -1] =0
-----or-----
P(X = -1) =F(-1) -F(-1) =0

(b) P(-0.6 < X < 0)
=F(0) - F(-0.6)
= [0.3+0.2(0+1)] - [0.3+0.2(-0.6+1)]
=0.12
---------or-----------
P(-0.6 < X < 0)
= ∫ f(x)dx [from -0.6 to 0]
= ∫ 0.2 dx [from -0.6 to 0]
= 0 - (0.2)(-0.6) =0.12

(c) similar to (a), the answer is 0
(d) E(X) = ∫ x f(x)dx [from -1 to 1]
=∫ 0.2x dx [from -1 to 0] + ∫ 0.3x dx [from 0 to 1]
=0.05

2(a)
f(x) = λ^x e^(-λ)/x!
f(x + 1)
= λ^(x+1) e^(-λ)/(x+1)!
= λ * λ^x * e^(-λ)/ (x+1)x!
= (λ/x+1) λ^x e^(-λ)/x!
=λ/x+1 f(X)

(b) f(1+1)
= (λ/1+1) f(1)
= (λ/2) f(0+1)
= (λ/2) (λ/0+1) f(0) = 2f(0)
For f(0) not equal 0, (λ/2) (λ) = 2
λ = 2 or -2 (rejected since λ should be +ve)
P(x = 5) = 2^5*e^(-2)/5! = 4/(15 e^2) = 0.0360894...

(1a) F(-1 -)=0; F(-1 +) = 0.3
So P(-1) = 0.3 - 0 = 0.3
(1d) P(0) = F(0+) - F(0-) = 0.2
E(x) = 0.3 * (-1) + 0.2 * (0) + ∫ 0.2x dx [from -1 to 0] + ∫ 0.3x dx [from 0 to 1]
= -0.25