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jmcarson
April 28th 2007, 12:41 PM
"Exam MLC Sample Questions and Solutions" (MLC-09-07): I don't see how to get the solution to problem #14, calculating the covariance of a joint life and last survivor remaining lifetimes. Can anyone help? Thanks for your time.

mrsevansc
April 29th 2007, 06:45 PM
ASM ch 35 page 547
If lives are independend that's the correct formula.
if they are not you need to add Cov(T(x),T(y))

.Godspeed.
April 29th 2007, 08:24 PM
"Exam MLC Sample Questions and Solutions" (MLC-09-07): I don't see how to get the solution to problem #14, calculating the covariance of a joint life and last survivor remaining lifetimes. Can anyone help? Thanks for your time.They are using a very powerful shortcut here.

Let's step back and figure out what the question is asking using first principles. It wants us to find the covariance of a joint-life status random variable and a last-survivor status random variable. More generally, it wants us to find the covariance of two random variables, Txy and Txybar. This sounds bad, but recall that the covariance of two random variables X and Y, Cov(X,Y), equals E(X*Y)-E(X)E(Y).

So, analagously, (*) Cov(Txy,Txybar)=E(Txy*Txybar)-E(Txy)E(Txybar). The E(Txy)E(Txybar) term is much easier to calculate than the E(Txy*Txybar) term, but note that Txy*Txybar = Tx*Ty.

Again using first principles, Cov(Tx,Ty)=E(Tx*Ty)-E(Tx)E(Ty) --> E(Tx*Ty)=Cov(Tx,Ty)+E(Tx)E(Ty).

Plugging this result into our original equation (*), we get (**) Cov(Txy,Txybar)=Cov(Tx,Ty)+E(Tx)E(Ty)-E(Txy)E(Txybar).

Using the relationship E(Txybar)=E(Tx)+E(Ty)-E(Txy) and some algebraic manipulation, (**) can be rewritten as Cov(Tx,Ty)+(E(Tx)-E(Txy))(E(Ty)-E(Txy)).

In the question, lives X and Y are independent, so (**) reduces to (E(Tx)-E(Txy))(E(Ty)-E(Txy)).

Hope I didn't make a mistake anywhere, and I hope that helps.

jmcarson
April 30th 2007, 12:30 PM
That was perfect - much appreciated.