Life Table Question

[DADDY_MUMMY]

The following table gives the conditional survival probabilities of a populatiton.
X P_x
0 0.9
1 0.7
2 0.5
3 0.3
4 0.1
5 0.0

a)Calculate S_x(x) for x = 0,1,2,...,6
b)Construct a life table for the population with l_0=1000000, giving values 1_x and d_x.
c)Calculate 4d0,3q2,and 2p3.

[Actuarialsuck]

So what's the problem? Also I think you mean l_x not 1_x

For starters express S(x) in terms of p_{x} or vice versa.

[DADDY_MUMMY]

This is my answer,hope that got any people can check for my answer,thank.

1)


2)


Is it correct??

[DADDY_MUMMY]

Got anyone can help to check my answer??

[Actuarialsuck]

For some reason I can't view your answer at work, can you write it out using TeX?

[DADDY_MUMMY]

S(0)=1
P(0)=S(1)/S(0) => 0.9 = S(1)/1 => S(1) = 0.9
P(1)=S(2)/S(1) => 0.7 = S(2)/0.9 => S(2) = 0.63
P(2)=S(3)/S(2) => 0.5 = S(3)/0.63 => S(3) = 0.315
P(3)=S(4)/S(3) => 0.3 = S(4)/0.315 => S(4) = 0.0945
P(4)=S(5)/S(4) => 0.1 = S(5)/0.0945 => S(5) = 0.00945
S(6)=0

l_0 = 1000000
p_1 = l_1 / l_0 =>0.9 = l_1 / 1000000 => l_1 = 900000
d_0 = l_0 - l_1 = 100000

p_2 = l_2 / l_1 =>0.7 = l_2 / 900000 => l_2 = 630000
d_1 = l_1 - l_2 = 270000

p_3 = l_3 / l_2 =>0.5 = l_3 / 630000 => l_3 = 315000
d_2 = l_2 - l_3 = 315000

p_4 = l_4 / l_3 =>0.3 = l_4 / 315000 => l_4 = 94500
d_3 = l_3 - l_4 = 220500

p_5 = l_5 / l_4 =>0.1 = l_5 / 94500 => l_5 = 9450
d_4 = l_4 - l_5 = 85050

I'm sorry that i don't know use TeX