Moment generating functions to compute variance of sum of independent random variables

Question

Can anyone guide me on this? How to use mgfs to show that if $X$ and $Y$ are independent, then

$$\operatorname{Var}(aX+bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$$

Appreciate your help!

Answer 1

Assume that $\exists \mathbb{E}\left(e^{Xt}\right),\mathbb{E}\left(e^{Ys}\right)$ in a neighbourhood of $0$. Then from independence: $$ \mathbb{E}\left(e^{t(aX+bY)}\right)=\mathbb{E}\left(e^{atX}\right)\mathbb{E}\left(e^{btY}\right) $$ Now differentiating and using the properties of mgf: $$ \mathbb{E}(aX+bY)=\left.\frac{\partial}{\partial t}\mathbb{E}\left(e^{t(aX+bY)}\right)\right|_{t=0}=\left.\frac{\partial}{\partial t}\mathbb{E}\left(e^{atX}\right)\mathbb{E}\left(e^{btY}\right)\right|_{t=0}= $$ $$ =\left.\frac{\partial}{\partial t}\mathbb{E}\left(e^{atX}\right)\right|_{t=0}\left.\mathbb{E}\left(e^{btY}\right)\right|_{t=0}+\left.\mathbb{E}\left(e^{atX}\right)\right|_{t=0}\left.\frac{\partial}{\partial t}\mathbb{E}\left(e^{btY}\right)\right|_{t=0}=a\mathbb{E}X+b\mathbb{E}Y $$ $$ \mathbb{E}(aX+bY)^2=\left.\frac{\partial^2}{\partial t^2}\mathbb{E}\left(e^{t(aX+bY)}\right)\right|_{t=0}= \left.\frac{\partial^2}{\partial t^2}\mathbb{E}\left(e^{atX}\right)\mathbb{E}\left(e^{btY}\right)\right|_{t=0}+ $$ $$ +2\left.\frac{\partial}{\partial t}\mathbb{E}\left(e^{atX}\right)\right|_{t=0}\left.\frac{\partial}{\partial t}\mathbb{E}\left(e^{btY}\right)\right|_{t=0}+\left.\mathbb{E}\left(e^{atX}\right)\right|_{t=0}\left.\frac{\partial^2}{\partial t^2}\mathbb{E}\left(e^{btY}\right)\right|_{t=0}= $$ $$ =a^2\mathbb{E}X^2+2ab\mathbb{E}X\mathbb{E}Y+b^2\mathbb{E}Y^2 $$ Thus: $$ \text{Var}(aX+bY)=\mathbb{E}(aX+bY)^2-(\mathbb{E}(aX+bY))^2= $$ $$ =a^2\mathbb{E}X^2+2ab\mathbb{E}X\mathbb{E}Y+b^2\mathbb{E}Y^2-a^2(\mathbb{E}X)^2-2ab\mathbb{E}X\mathbb{E}Y-b^2(\mathbb{E}Y)^2= $$ $$ =a^2(\mathbb{E}X^2-(\mathbb{E}X)^2)+b^2(\mathbb{E}Y^2-(\mathbb{E}Y)^2)=a^2\text{Var}X+b^2\text{Var}Y $$