Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For the life of me, I cannot figure out how to solve this:

enter image description here

I know that the coefficient of variation is just the standard deviation over the mean, and in this case, I want the c.o.v. of $C = 165A + 305B$. The problem, however, is that the given means and variances are conditional.

To find the mean, I figured I would set up a partition: in $A$'s case, for example, the expected value of $A$ would be the sum of all the possible A values times their probabilities. Since the only $A$ values are $2$ for when a loss occurs and $0$ for when it doesn't, I would get $E(A) = 0.25(2) + 0.75(0) = 0.5$, right? [Eventually, this would all go into $E(C) = E(165A + 305B) = 165E(A) + 305E(B)$ ] I followed a similar pattern for $B$ and wanted to perhaps use these in finding the variances.

For variance, it feels like something is missing here. I'm given conditional variance, which, in $A$'s case, would mean (?):

$$2.3 = E(A^2|\mbox{ loss occurred }) - [E(A|\mbox{ loss occurred })]^2$$

Because of the part that cancelled when I was finding $E(A)$, I'll assume I can say $E(A) = E(A|\mbox{ loss occurred })$. Thus,

$$2.3 = E(A^2|\mbox{ loss occurred }) - [2]^2 , $$since $A$'s mean is $2$.

But I'm not sure what to do next (or if this is going to help me) in order to get from "the variance of loss, given loss occurs" to just, "the variance of loss," which I need for the overall problem. What should I do now?

share|improve this question
    
It really helps readability if you format your questions using MathJax (see FAQ at top right). Regards –  Amzoti Apr 1 '13 at 2:42

1 Answer 1

Let $L_A$ and $L_B$ be the losses of $A$ and $B$ and $X_A$ and $X_B$ indicate whether losses occured or not ($1$ if yes, $0$ if no). Using the law of total variance you get $$Var(L_A) = E(Var(L_A|X_A)) + Var(E(L_A|X_A)) $$ where

$$E(Var(L_A|X_A)) = P(X_A=1)Var(L_A|X_A=1) + P(X_A=0)Var(L_A|X_0) = 0.25\cdot2.3 + 0.75\cdot 0 $$ and $$Var(E(L_A|X_A)) = E((E(L_A|X_A)^2) - (E(E(L_A|X_A))^2 \\ P(L_A>0)(E(L_A|L_A>0))^2+P(L_A=0)(E(L_A|L_A=0))^2 - (E(P(L_A>0)E(L_A|L_A>0)+P(L_A=0)E(L_A|L_A=0))^2 \\ 0.25\cdot 2^2+0 - (0.25\cdot 2)^2=0.125\cdot 2^2 $$ Thus $$ Var(L_A) = 0.25\cdot 2.3 +0.125\cdot 2^2 $$ Calculate $Var(L_B)$ analogously and then you have by independence $$Var(C) = 165 \cdot Var(L_A) + 305 \cdot Var(L_B)$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.