Given Random Variable X ~ N(0,1), Y ~ N(1,2)
How would I go about finding the new mean and standard deviation in Z = 4X + 5Y?
I know the means get added linearly, and
VAR(X+Y) = VAR(X) + VAR(Y) + 2Cov(X,Y).
The constants are throwing me off of this equation..
Best Answer
Linearity also implies $$\mathsf{E}(aU)=a\mathsf{E}(U)$$ Therefore,
$$\mathsf{E} (Z)=\mathsf{E}(4X+5Y)=4\mathsf{E}(X)+5\mathsf{E}(Y)$$ Due to the linearity of $\mathsf{E}$. Also Note that $$\mathsf{Var}(aU)=a^2\mathsf{Var}(U)$$ Hence, $\mathsf{Var}$ is not linear.
$$\begin{align} \mathsf{Var}(aX + bY) &= \mathsf{E}((aX+bY)^2) -[\mathsf{E}(aX+bY)]^2\\ & = \mathsf{E}(a^2X^2+2abXY +b^2Y^2) -[a\mathsf{E}(X)+b\mathsf{E}(Y)]^2\\ & = a^2 \mathsf{E}(X^2) + 2ab\mathsf{E}(XY) +b^2\mathsf{E}(Y^2) - [a^2 \mathsf{E}^2(X) + 2ab \mathsf{E}(X)\mathsf{E}(Y) + b^2 \mathsf{E}^2(Y)]\\ & = a^2 [\mathsf{E}(X^2) - \mathsf{E}^2(X)] + b^2[\mathsf{E}(Y^2)-\mathsf{E}^2(Y)] + 2ab [\mathsf{E}(XY) - \mathsf{E}(X)\mathsf{E}(Y)] \\ & = a^2 \mathsf{Var}(X) + b^2 \mathsf{Var}(Y) + 2ab \mathsf{Cov}(X,Y) \end{align}$$
Hence, $$\mathsf{Var}(Z)=(4)^2\mathsf{Var}(X)+(5)^2\mathsf{Var}(Y)+2(4)(5)\mathsf{Cov}(X,Y)$$