1. ## probability

an insurance company has sold a policy to a customer, who could incur in a loss X(w) with probability density:
f(x)= 10^-6 x e^(-x/1000) for x>0
= 0 elsewhere

1)assume that the probability of arrival of the loss is 2%. using such info compute the expected value of the loss.

") the policy covering this risk is of the type stop-loss: in the case the claim exceeds 3000\$ this is the amount paid by the insurer. Taking into account also that the claim addressed to the insurer, in the case of loss, is of amount Y(w)=0.9X(w) compute the expected claim.

how do I do these? I can't figure out the reasoning. thanks in advance for your great help.

2. Originally Posted by 0123
an insurance company has sold a policy to a customer, who could incur in a loss X(w) with probability density:
f(x)= 10^-6 x e^(-x/1000) for x>0
= 0 elsewhere

1)assume that the probability of arrival of the loss is 2%. using such info compute the expected value of the loss.
E(loss) = integral p(loss) loss dloss = integral(x=0, infty) 0.02 x^2 10^-6 e^(-x/1000) dx = 40

RonL

3. why are you multiplying probability of the loss x(=times) loss x density of the loss ? what is the logical process behind? thank you very much

4. Originally Posted by 0123
why are you multiplying probability of the loss x(=times) loss x density of the loss ? what is the logical process behind? thank you very much
what you need to integrate is the loss times the probability of that loss.

The probability of that loss is the probability of any loss (0.02) times the
probability that the loss is x given that there is a loss (10^-6 x e^(-x/1000) ).

RonL

5. Originally Posted by CaptainBlack
what you need to integrate is the loss times the probability of that loss.

The probability of that loss is the probability of any loss (0.02) times the
probability that the loss is x given that there is a loss (10^-6 x e^(-x/1000) ).

RonL
I do not understand. I mean, if the probability of arrival of the loss is 2%
always , shoudn't we simply multiply x times 0.02 and integarte it? sorry for insisting

6. Originally Posted by 0123
I do not understand. I mean, if the probability of arrival of the loss is 2%
always , shoudn't we simply multiply x times 0.02 and integarte it? sorry for insisting
integral(x=0, infty) 0.02 x dx = infty

RonL

7. Might you please rephrase this

Originally Posted by CaptainBlack
The probability of that loss is the probability of any loss (0.02) times the
probability that the loss is x given that there is a loss (10^-6 x e^(-x/1000) ).

RonL
?
thanks

8. Originally Posted by CaptainBlack
what you need to integrate is the loss times the probability of that loss.

The probability of that loss is the probability of any loss (0.02) times the
probability that the loss is x given that there is a loss (10^-6 x e^(-x/1000) ).

RonL
f(x)=10^-6 x e^(-x/1000)

is the probability that the loss is x, given that there is some loss.

0.02 is the probability that there is some loss.

Hence the probabability of loss x is p(x)=0.02 f(x).

Therefore the expected loss is:

integral(x=0, infty) x (0.02 f(x)) dx

RonL