RMI 300 Pooling Assignment Sandy and Sammy both have the same loss distribution

Question

RMI 300 Pooling Assignment Sandy and Sammy both have the same loss distribution as follows. Assume that their loss distributions are also independent of one another. Loss Amount $O $8000 Observations 60 20 a. What is Sandy's expected loss? b. What is Sandy's standard deviation? c. Assume that Sandy and Sammy decide to pool their losses together. What is the pooled (total) distribution of outcomes and the pooled expected loss? d. What is the pooled standard deviation of their losses (total)? Per person? e. If Sandy and Sammy invite 198 friends to join their pool, what will be the expected loss per person? What will happen to the standard deviation per person in the pool of 200 people? Describe in words Assume all 200 individuals are positively correlated. What would the effect on pooling? f. g.

Explanation / Answer

(a) Expected value=value* probability

here loss values are 0 and 8000

here probabilities are 60/ (60+20)=60/80=0.75 and 20/ (60+20)=20/80=0.25

0 x .75 + 8000 x .25 = 2000

(b)  mean=8000*20/ (60+20) = 160000/80 = 2000

variance=summation(p*(x-mean)^2)

Sqrt (.75 x (0 – 2000)^2) + .25 x (8000 – 2000)^2)

= Sqrt ( 3,000,000 + 9,000,000) = 3464.10

(c) No accident:Total Cost = 0 Probability = .5625

Sandy has accident: Total Cost = 8000 Probability = .1875

Sammy has accident: Total Cost = 8000 Probability = .1875

Both have accident: Total Cost = 16,000 Probability = .0625

pool expected loss = .5625 x 0 + .1875 x 8000 + .1875 x 8000 + 16,000 x .0625

= 0 + 1500 + 1500 + 1000

= 4000

(d) mean=8000*20/ (60+20) = 160000/80 = 2000

Total loss = Sqrt( .5625 x (0- 2000)^2 + .1875 x (8000 – 2000) ^2 + .1875 x (8000 – 2000) ^2 + .0625 x (16,000 –2000)^2)

=sqrt(2250000+6750000+6750000+12250000)

=5291.5026

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