The joint distribution for the claims amount X and Y is:
$$\begin{array}{c|c|c|}
& X=10 & X=50 & X=100\\ \hline
Y=25 & 0.08 & 0.14 & 0.13\\ \hline
Y=75 & 0.15 & 0.3 & 0.2 \\ \hline
\end{array}$$
Find the variance of $X – 3Y$.
My attempt:
Let $Z=X-3Y$. The probability distribution of $Z$ is given by
$$\begin{array}{c|c|c|}
Z & P(Z=z) & Z^2 \\ \hline
-65 & 0.08 & 4225\\ \hline
-215 & 0.15 & 46225\\ \hline
-25 & 0.14 & 625\\ \hline
-175 & 0.30 & 30625\\ \hline
25 & 0.13 & 625\\ \hline
-125 & 0.20 & 15625\\ \hline
\end{array}$$
Now I computed $Var(Z)$ using $E(Z^2)-[E(Z)]^2$.
$$Var(Z) = 19753-(121.7)^2$$
This did not lead to the correct answer. Is there something conceptually wrong with my approach (I've double-checked my computations many times so they should be correct)? Note: I do not want to use the formula $Var(Z) = Var(X)+9Var(Y)-6Cov(X, Y)$ for this question unless I have to.
Best Answer
Your table is correct, but your calculation of the expectation is not. To complete your table, we have:
$$\begin{array}{|c|c|c|c|c|} \hline Z & P(Z=z) & Z^2 & Z\Pr[Z = z] & Z^2 \Pr[Z = z] \\ \hline -65 & 0.08 & 4225 & -5.2 & 338. \\ \hline -215 & 0.15 & 46225 & -32.25 & 6933.75 \\ \hline -25 & 0.14 & 625 & -3.5 & 87.5 \\ \hline -175 & 0.3 & 30625 & -52.5 & 9187.5 \\ \hline 25 & 0.13 & 625 & 3.25 & 81.25 \\ \hline -125 & 0.2 & 15625 & -25. & 3125. \\ \hline & & & \color{red}{-115.2} & 19753 \\ \hline \end{array}$$