Variance for linear combination of coefficients

Question

Why does:

$Var(aX)=a^2Var(X)$

and

$Var(aX\pm bY)=a^2var(X)+b^2var(Y)\pm 2ab Cov(X,Y)$?

I've looked at similar stackexchange links, but haven't been able to find a "simple" answer without extensive math proof.

Answer 1

Simply use the definition of Variance expanding

$$\mathbb{V}[aX+bY]=\mathbb{E}[(aX+bY)^2]-(\mathbb{E}[aX+bY])^2$$

$$\mathbb{E}[a^2X^2+2abXY+b^2Y^2]-a^2\mathbb{E}^2[X]-b^2\mathbb{E}^2[Y]-2ab\mathbb{E}[X]\mathbb{E}[Y]$$

Using linearity of the expectation and re-organizing the addends you get

$$a^2[\mathbb{E}(X^2)-\mathbb{E}^2(X)]+b^2[\mathbb{E}(Y^2)-\mathbb{E}^2(Y)]+2ab[\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)]$$

Now you are almost done

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$X$ and $Y$ are rv's thus Capital letters are needed

The second statement is valid also for $V(X-Y)$ thus I amended it.

Answer 2

Be aware that: $$\mathsf{Var}(Z)=\mathsf{Cov}(Z,Z)\tag1$$

where by definition: $$\mathsf{Cov}(X,Y)=\mathbb E(X-\mathbb EX)(Y-\mathbb EY)\tag2$$

Based on $(2)$ that we find that:

• $\mathsf{Cov}$ is symmetric in the sense that $\mathsf{Cov}(X,Y)=\mathsf{Cov}(Y,X)$
• $\mathsf{Cov}(aX,Y)=a\mathsf{Cov}(X,Y)$
• $\mathsf{Cov}(X+Z,Y)=\mathsf{Cov}(X,Y)+\mathsf{Cov}(Z,Y)$

This together makes it possible to prove that $\mathsf{Cov}$ is bilinear so that:

• $\mathsf{Cov}(aX,aX)=a^2\mathsf{Cov}(X,X)$
• $\mathsf{Cov}(aX+bY,aX+bY)=a^2\mathsf{Cov}(X,X)+2ab\mathsf{Cov}(X,Y)+b^2\mathsf{Cov}(Y,Y)$
• $\mathsf{Cov}(aX-bY,aX-bY)=a^2\mathsf{Cov}(X,X)-2ab\mathsf{Cov}(X,Y)+b^2\mathsf{Cov}(Y,Y)$

Now apply $(1)$ in those rules.