1.13 Properties of Variance

1.

\(\sigma ^2=\text {var}(X)=E(X-\overline {X})^2=E(X^2)-(E(X))^2\)

2.

If \(c\) is a constant, then var\((cX)=c^2\)var\((X)\).

3.

The variance is independent of the origin if \(c\) is a constant then var\((X+c)=\) var\((X)\).

4.

\(\operatorname {var}(X)\geq 0\), with \(\operatorname {var}(X)=0\) if and only if \(P(X=c)=1\) for some constant \(c\) – that is, exactly when \(X\) does not vary at all.

5.

If \(X\) and \(Y\) are independent, then \(\operatorname {var}(X\pm Y)=\operatorname {var}(X)+\operatorname {var}(Y)\).

Note both features of this statement. The independence condition is part of the property, not an afterthought: without it the result is false. And the variances add even when the variables are subtracted – subtracting a second quantity introduces its variability just as surely as adding it does.

Proof. Writing the deviation of the sum in terms of the separate deviations, \begin {align*} \operatorname {var}(X\pm Y) &=E\big [(X\pm Y)-(\overline {X}\pm \overline {Y})\big ]^2 =\frac {1}{N}\sum \big [(X-\overline {X})\pm (Y-\overline {Y})\big ]^2\\ &=\frac {1}{N}\sum \Big [(X-\overline {X})^2+(Y-\overline {Y})^2 \pm 2(X-\overline {X})(Y-\overline {Y})\Big ]\\ &=\underbrace {\frac {1}{N}\sum (X-\overline {X})^2}_{\operatorname {var}(X)} +\underbrace {\frac {1}{N}\sum (Y-\overline {Y})^2}_{\operatorname {var}(Y)} \pm \underbrace {\frac {2}{N}\sum (X-\overline {X})(Y-\overline {Y})}_{2\operatorname {cov}(X,Y)}. \end {align*}

The last term is twice the covariance of \(X\) and \(Y\). When they are independent the covariance is zero, and the cross term vanishes: \[\operatorname {var}(X\pm Y)=\operatorname {var}(X)+\operatorname {var}(Y). \qedhere \] □

In general, without independence, the cross term remains and \[\operatorname {var}(X\pm Y)=\operatorname {var}(X)+\operatorname {var}(Y)\pm 2\operatorname {cov}(X,Y).\]

6.

The expression \(E(X-c)^2\) is minimized over the constant \(c\) where \(c=E(X)\) so that \(E(X-c)^2\geq \) var\((X)\) for all \(c\) with equality where \(c=E(X)\)

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.