PDA

View Full Version : One more problem

Rup
December 25th 2005, 10:17 PM
Can anybody try to solve and explain this problem?

Define Yd= claim paid by the insurer if there is a deductible of d. For aggregate actual claims X, you are given 1. X takes only positive integer values 2. E(X) = E(Yo) =5/3 3. E(Y2) =1/6 4. E(Y3)=0. Determine p1=P(X=1) Ans:1/2

Thanks

Rup

Sam Broverman
December 25th 2005, 11:11 PM
Since E[Y3]=0 , it follows that the maximum possible value
of X is 3 (if there were any values of X that are larger than 3
then after a deductible of 3 there would still be payments
for some losses, so thte expected value after deductible of 3 would
be strictly larger than 0).

Since X is an integer, X = 1 , 2 or 3 (we are assuming that
when a loss X occurs it is strictly larger than 0).
Let us label the probabilities p1 , p2 and p3 .
It must be true that p1 + p2 + p3 = 1 (definition
of a probability distribution).

Also, Y0 = X (deductible of 0), so
E[Y0] = E[X] = p1 + 2p2 + 3p3 = 5/3 (this was given).

Y2 is the claim paid after a deductible of 2.
If X is 1 or 2 then Y2=0, and if X=3 then Y2=1.
Therefore, E[Y2] = p3 (this is 0xp1 + 0xp2 + 1xp3) .
We are given the E[Y2] = 1/6 , but this is p3 .

Now we have 3 equations in the unknown factors p1, p2 and p3.
We can solve the equations to find p1.

Rup
December 26th 2005, 12:30 AM
Thanks Sam!