I know that $\operatorname{Var}[aX+bY]=\operatorname{Cov}[aX+bY,aX+bY]=a^2\operatorname{Var}[X]+2ab\operatorname{Cov}[X,Y]+b^2\operatorname{Var}[Y]$ (by expanding $(ax+by)(ax+by)$ and letting $\operatorname{Var}[X]=x^2$-terms, $\operatorname{Var}[Y]=y^2$-terms and $\operatorname{Cov}[X,Y]=xy$-terms).
But what happens when we've got an $x$-term or a $y-$term on its own?
e.g. say we want $\operatorname{Var}[X+Y+1]$.
Then we have $(x+y+1)^2=x^2+2xy+2x+y^2+2y+1$, so: $$\operatorname{Var}[X+Y+1]=\operatorname{Var}[X]+2\operatorname{Cov}[X,Y]+2?+\operatorname{Var}[Y]+2?+1. $$ Could someone fill in the blanks?
Thanks.
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$\begingroup$If all you want is the variance, getting it through the covariance formula the way you're doing is a lot more complicated than it needs to be. You just observe: $${\rm Var}[X + Y + 1] = {\rm Var}[X + Y],$$ because ${\rm Var}[X + c] = {\rm Var}[X]$ for any constant $c$. Then you can use the covariance formula.
$\endgroup$ 3 $\begingroup$You almost got it except that $$ \text{Cov}(aX, bY+c) = \text{Cov}(aX, bY)+\text{Cov}(aX,c) = ab \text{Cov}(X,Y) +0 $$ because covariance of a random variable with a constant is $0$.
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