ASM Q7.16

[Kabaka]

Help me understand this one. It sounds easy, but it's Friday.

Losses follow a single parameter Pareto distribution with alpha = 3 and theta = 500.

Determine the deductible d needed to achieve a loss elimination ratio of 20%.

Spoiler:

Here is my solution. Why is it incorrect?

Solve E[X^d] / E[X] = 0.2 for d with parameters for single Pareto. My result d = 322.75

Correct solution:

Spoiler:

"For single parameter Pareto, E[X] = (alpha) (theta) / (alpha - 1) = 750.

If the deductible, d, is less than 500, E[X^d]=d (since the probability that a loss is less than 500 is 0, so the random variable P(X<=0) = 0). But d = 150 eliminates 20% of the losses."

[Jski]

The LER is E(X^d) / E(X), right? So use the tables and the parameters given and plug / chug. Am I missing something?

[Kazodev]

He's saying that's what he did, but the solution seems to state that d = 150 is the correct solution since 150/750 = .2?

[Gandalf]

Are you recognizing that the formula for E(X^d) is different for d > theta and for d < theta in the single parameter Pareto.

[Kabaka]

Originally Posted by Gandalf

Are you recognizing that the formula for E(X^d) is different for d > theta and for d < theta in the single parameter Pareto.

Right, I see now that it says in the tables that the formula for E[X^d] is valid for d>= theta, or in this case for d>=500.

I still need some help here.

[Kabaka]

So I think what the solution is saying, is that since d<500, we can't use the formula in the tables and we have to use another method (that I can't find explained anywhere). How do we know that d<500? Is it just because after finding my solution of 322.75, I check it with the F(x) formula to find that F(x) < 0 (ie. invalid)?

[Abraham Weishaus]

Originally Posted by Kabaka

we have to use another method (that I can't find explained anywhere).

But I thought the explanation in the manual was quite clear. If the loss size is guaranteed to be at least 500, and I give you the minimum of the loss size and 300, what is the expected value of what I give you?

[Kabaka]

Originally Posted by Abraham Weishaus

But I thought the explanation in the manual was quite clear. If the loss size is guaranteed to be at least 500, and I give you the minimum of the loss size and 300, what is the expected value of what I give you?

I get 300.

.......

Ahhhhhhh, I get it. The solution in #17 helps to clarify for me. It says, "We can't use the trick of the last problem though since 0.8 (750) > theta..."

So we DO use the LER formula 0.2 = E[X^d] / E[X] to solve for E[X^d]. Then we have E[X^d] = 0.2 (750) = 150. But E[X^d] = d for the reason you state above. Ok, all good. Thanks for the prodding.

[DudeMan]

Originally Posted by Kabaka

So I think what the solution is saying, is that since d<500, we can't use the formula in the tables and we have to use another method (that I can't find explained anywhere). How do we know that d<500? Is it just because after finding my solution of 322.75, I check it with the F(x) formula to find that F(x) < 0 (ie. invalid)?

Kabaka.....I believe in a single parameter pareto, the value theta (500 in this case) is the minimum possible value of a claim. A deductible of 150 effects EVERY claim since S(500)=1.....E(X)=750......E(X^150)=150*S(150)=150 .....150/750=0.2

[beck]

Originally Posted by Kabaka

I get 300.

.......

Ahhhhhhh, I get it. The solution in #17 helps to clarify for me. It says, "We can't use the trick of the last problem though since 0.8 (750) > theta..."

So we DO use the LER formula 0.2 = E[X^d] / E[X] to solve for E[X^d]. Then we have E[X^d] = 0.2 (750) = 150. But E[X^d] = d for the reason you state above. Ok, all good. Thanks for the prodding.

you were definately "tricked"