Question

Besides a certain wealth of 100, Joe owns one house, the value of which is 80. The probability of full loss (due to fire) for this house is 0.10 for a given time period and Joe has no access to an insurance market. In the absence of fire, the value of a house remains at 80. Mary has the same initial wealth but owns two houses each valued at 40. The probability of full loss for each house is 0.10 and the fires are assumed independent random variables.

a.   Draw the cumulative distribution functions of final wealth for both Joe and for Mary. Compute expected final wealth for both Joe and for Mary.

b.   Show that Mary’s more diversified house situation stochastically dominates that of Joe in the second degree.

Answer 1

Consider the given problem here there are 2 possibility of “net wealth” of “Joe”, the probability distribution is given below.

The above table is the probability distribution of “net wealth” of “Joe”, there are 2 possibility. The 1^st one is “Joe” will loss nothing, => net wealth is “180” with probability “1-0.1=0.9”. The 2^nd one is “Joe” will loss house due to fire with probability “0.1”.

The Cumulative distribution function of “Joe’s” Net Wealth is given below.

=> F(X) = 0 for X < 100, = 0.1 for 100 ? X < 180, = 1 X ? 180. Consider the following fig.

Now, consider the probability distribution of “Mary”.

The above table is the probability distribution of “net wealth” of “Mary”, there are 3 possibility. The 1^st one is “Joe” will loss nothing, => net wealth is “180” with probability “0.9*0.9=0.81”, since here there are 2 house each have the same probability of getting fire of “0.1”, => here the required probability is “0.9*0.9. The 2^nd one is “Mary” will loss one house due to fire with probability “0.18”, here probability of losing one house is “0.1*0.9” and there are 2 such house , => the required probability is “2*0.9*0.1”. The 3^rd one is “Mary” will loss both of the house with probability “0.01”.

The Cumulative distribution function of “Joe’s” Net Wealth is given below.

=> F(X) = 0 for X < 100,

= 0 + 0.01 = 0.01 for 100 ? X < 140,

= 0.01 + 0.18 = 0.19 for 100 ? X < 140,

= 0.19 + 0.81 = 1 X ? 180. Consider the following fig.

Now, the expected final wealth of “Joe” is “E(X) = 180*0.9 + 100*0.1 = 172.

The expected final wealth of “Mary” is “E(X) = 180*0.81 + 140*0.18 + 100*0.01 = 172. So, both of them have same expected wealth.

b).

For the Variance of the net wealth of “Joe” is given below.

=> V(X) = E(X^2) – [E(X)]^, where E(X^2) = 180^2*0.9 + 100^2*0.1 = 30160.

=> V(X) = 30160 – 172^2 = 576.

For the Variance of the net wealth of “Mary” is given below.

=> V(X) = E(X^2) – [E(X)]^, where E(X^2) = 180^2*0.81 + 140^2*0.18 + 100^2*0.01 = 29872.

=> V(X) = 29872 – 172^2 = 288.

So, we can see that “Mary” have lower “variance”, => “Mary’s” have more diversified house situation.