#2. Expected Value by conditioning Problems - Solutions
1. E(drive time)=.2(55)+.4(50)+.4(45)=49
45
2. E(X)=E(E(X|Ω))=∫4 E(X|Ω = t)𝑓Ω (𝑡)𝑑𝑡, we know that Ω is uniform on the interval [5,45]. So
𝑓Ω (𝑡) = 1/40 for 5<t<45. We also know that X is exponential with mean Ω, so when Ω=t, the
mean of X is t. That is E(X|Ω = t) = 𝑡. Putting this together:
45
45 𝑡
𝑑𝑡
40
E(X)=E(E(X|Ω))=∫4 E(X|Ω = t)𝑓Ω (𝑡)𝑑𝑡 = ∫5
= 25
I could also say: E(X|Ω)= Ω so
E(X)=E(E(X|Ω))= E(Ω)=25.
3.
Game
probability
Cubs
.4
Steelers
.3
Red Sox
.2
Curling
.1
E(X)=.4(50)+.3(70)+.2(65)+.1(24)
distribution
N(50,16)
Unif (20,120)
Expo. Mean 65
24 ith prob. 1
mean
50
70
65
24
4.
Prob.
.04
.02
.94
damage
1000X
15000
0
payment
1000(X-1)+
14000
0
Expected payment
3290.57
14000
0
We need to compute: 𝐸((𝑋 − 1)+ ) = ∫(𝑥 − 1)+ 𝑓(𝑥)𝑑𝑥. Because (X-1)+=0 if X<1, the integral starts at
15
x=1: : 𝐸((𝑋 − 1)+ ) = ∫1 (𝑥 − 1). 5003𝑒 −𝑥/2 𝑑𝑥. Integrating by parts we get=3.290566
Finally, E(payment)=.04(3290.57)+.02(14000)
Problem 7, section 4.7: f(x,y)=x+y, 0<x,y<1.
1
𝐸(𝑌|𝑋 = 𝑥) = ∫0 𝑦𝑓𝑌|𝑋 (𝑦|𝑥)𝑑𝑦, so we need to compute the conditional density
1
1
𝑓𝑋 (𝑥) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 = ∫ (𝑥 + 𝑦)𝑑𝑦 = 𝑥 + 1/2
0
0
𝑓𝑌|𝑋 (𝑦|𝑥) =
1
1
𝑓(𝑥, 𝑦)
𝑥+𝑦
=
𝑓𝑋 (𝑥)
𝑥 + 1/2
𝑥+𝑦
1
1
𝑥
+1/3
2
Finally: 𝐸(𝑌|𝑋 = 𝑥) = ∫0 𝑦𝑓𝑌|𝑋 (𝑦|𝑥)𝑑𝑦 = ∫0 𝑦 𝑥+1/2 𝑑𝑦 = 𝑥+1/2 ∫0 (𝑥𝑦 + 𝑦 2 )𝑑𝑦 = 𝑥+1/2