Maths

1. C (2,0,-3,5,10) v(0,1)=1.25 v(1,2)=1.25 ; v(2,1) = 1/1.25 = 0.8 v(2,3)=1+1=2 ; v(3,2)=1/(1+1)=0.5 v(3,4)=1+1=2 ; v(4,3)=1/(1+1)=0.5
The smallest payment at time 1 = 2v(0,1) + (-3) v(2,1) + 5v(3,2)v(2,1) + 10v(4,3)v(3,2)v(2,1)
= 2(1.25) + (-3)(0.8) + 5 (0.5)(0.8) + 10(0.5)(0.5)(0.8) = 2.5-2.4+2+2 = 4.1
The smallest payment you would accept is 4.1.

2. Let K be the initial payment.
a)9980.89
b)10117.4

3. E[X]=1, Var(X)=5
V(X) = E(X2) – [E(X)] 2
5=E(X2)-1
E(X2) = 6
a) E[(2+X) 2] = E(4+4X+ X2) = 4 + 4 E(X) + E(X2) = 4+4+6 = 14
b) Var (4+3X) = E [(4+3X) 2] – [E(4+3X)] 2 = E (16+6X+9 X2) – [4+3E(X) ] 2 = 16+6 E(X)+9 E(X2) – 16 - 24 E(X) - 9 [E(X) ] 2 = 16 + 6 +9(6) -16 -24 -9 = 27
4. C (2,0,-3,5,10) v(0,1)=1.25 v(1,2)=1.25 ; v(2,1) = 1/1.25 = 0.8 v(2,3)=1+1=2 ; v(3,2)=1/(1+1)=0.5 v(3,4)=1+1=2 ; v(4,3)=1/(1+1)=0.5
The smallest payment at time 1 = 2v(0,1) + (-3) v(2,1) + 5v(3,2)v(2,1) + 10v(4,3)v(3,2)v(2,1)
= 2(1.25) + (-3)(0.8) + 5 (0.5)(0.8) + 10(0.5)(0.5)(0.8) = 2.5-2.4+2+2 = 4.1
The smallest payment you would accept is 4.1.

5. Let K be the initial payment.
a)9980.89
b)10117.4

6. E[X]=1, Var(X)=5
V(X) = E(X2) – [E(X)] 2
5=E(X2)-1
E(X2) = 6
c) E[(2+X) 2] = E(4+4X+ X2) = 4 + 4 E(X) + E(X2) = 4+4+6 = 14
d) Var (4+3X) = E [(4+3X) 2] – [E(4+3X)] 2 = E (16+6X+9 X2) – [4+3E(X) ] 2 = 16+6 E(X)+9 E(X2) – 16 - 24 E(X) - 9 [E(X) ] 2 = 16 + 6 +9(6) -16 -24 -9 = 27
7. C (2,0,-3,5,10)