$X\sim N(52,6)$, $Y\sim (40,8)$. What's the standard deviation of $Z=X+Y$?
I'm considering to transform the linear relationship to matrix form
$$Z=\begin{pmatrix}
1& 1\\
\end{pmatrix}\begin{pmatrix}
X\\
Y
\end{pmatrix}$$
and apply $$\Sigma_{z}=A \Sigma^{-1} A'$$
However, I don't know the next step. How to calculate $\Sigma$?
Thank you!
Best Answer
It's interesting that you are trying to use the matrix form when this is more of a elementary problem.
Remember $$ \mbox{Var}(X + Y) = \mbox{Var}(X) + \mbox{Var}(Y) + 2 \mbox{Cov}(X,Y). $$
You don't say anything about the relation between $X$ and $Y$, so either you assume a covariance of zero or the problem is not solvable due to missing informations.
The matrix form of this is $$ \mbox{Cov}(AX) = A \cdot \mbox{Cov}(X)\cdot A'. $$
This covariance matrix seems to be your $\Sigma$. I don't know why you inverted it in your formula. Perhaps I misunderstand your equation? You know that $\Sigma_{1,1} = 6$ and $\Sigma_{2,2} = 8$. These are just the variances of $X$ and $Y$ respectively. Again, since you do not have a covariance and since you cannot calculate it from the information given, the question is either unsolvable or you have have to assume independence, which implies $\Sigma_{1,2}=\Sigma_{2,1}=0$. This will then give the same result as the equation for $\mbox{Var}(X+Y)$. Remember to take the root to get the standard deviations.