Variance, exponential - geometric distribution problem

[mihalaras]

Problem Description:
The size X (in thousands of euros) of a compensation paid by an insurance company has the exponential distribution with parameter λ=1/4.
The number of claims paid by the insurance company over a week has a geometric distribution with parameter p=1/2.
How much is the variance of the (total) amount of claims paid by the insurance company over a week?

Solution: 64

Can someone help me with the methodology I should follow in order to reach a solution in this problem? Thank you in advance!

 

[BigBeachBanana]

Let [imath]X[/imath] be the claim amount, [imath]N[/imath] be the number of claims, and [imath]S[/imath] be the aggregate loss, then:
[math]Var(S) = E(N)Var(X)+E(X)^2Var(N)[/math]
[imath][/imath]

 

[mihalaras]

Let [imath]X[/imath] be the claim amount, [imath]N[/imath] be the number of claims, and [imath]S[/imath] be the aggregate loss, then:
[math]Var(S) = E(N)Var(X)+E(X)^2Var(N)[/math]
[imath][/imath]

Thanks for your reply! I believe I need a little bit more context here.. I'm still quite confused.

If I'm not mistaken the aggregate loss (S) is equal to N*X and is the amount of claims paid by the insurance company.
My thought was to take VAR(NX) but I'm not quite sure if it's the same as the one you provided me with.

Also, may I ask how do we create the formula "Var(S)=E(N) Var(X)+E(X)^2 Var(N)" ?

 

[BigBeachBanana]

Thanks for your reply! I believe I need a little bit more context here.. I'm still quite confused.

If I'm not mistaken the aggregate loss (S) is equal to N*X and is the amount of claims paid by the insurance company.
My thought was to take VAR(NX) but I'm not quite sure if it's the same as the one you provided me with.

Also, may I ask how do we create the formula "Var(S)=E(N) Var(X)+E(X)^2 Var(N)" ?

The aggregate loss is conditioned on having a claim count.

[math]\tag{\text{Conditional Variance}}Var(S) = E[Var(S|N)] + Var[E(S|N)][/math][math]\tag{I.I.D assumption}=E[N\cdot Var(X)] + Var[N\cdot E(X)][/math][math]= E(N)Var(X)+E(X)^2Var(N)[/math]

 

[mihalaras]

The aggregate loss is conditioned on having a claim count.

[math]\tag{\text{Conditional Variance}}Var(S) = E[Var(S|N)] + Var[E(S|N)][/math][math]\tag{I.I.D assumption}=E[N\cdot Var(X)] + Var[N\cdot E(X)][/math][math]= E(N)Var(X)+E(X)^2Var(N)[/math]

I'm afraid I'm still getting stuck somewhere. I'm going to attach my thoughts on trying to solve this bellow. I am currently studying for exams and I spent a lot of hours trying to understand your answers, but I'm getting confused on the methodology and on why we need to get conditional variance.
Is my thought of Var(S) = Var(X*Y) correct or am I completely mistaken?

Again, thanks a lot for your time!

 

[BigBeachBanana]

I'm afraid I'm still getting stuck somewhere. I'm going to attach my thoughts on trying to solve this bellow. I am currently studying for exams and I spent a lot of hours trying to understand your answers, but I'm getting confused on the methodology and on why we need to get conditional variance.
Is my thought of Var(S) = Var(X*Y) correct or am I completely mistaken?

Again, thanks a lot for your time!

As I said You can't do S=X*Y because S is conditioned on having Y>0.
Theoretically, you can say I have 10,000 claims but the claim count is 0. This makes no practical sense so it's more correct to say S=XY|Y>0
You must incorporate conditional probability where Y>0 as shown in post #4.