Let X have density f and mean mu=E(X). When the second moment is finite, variance is the expected squared deviation from the mean. sigma2=Var(X)=E[(X-mu)2] = _all x(x-mu)2f(x),dx =E(X2)-mu2. Calculate E(X2)= _all xx2f(x),dx, then subtract the square of the mean. For a piecewise density, integrate each piece over its own interval. The standard deviation is sigma=sqrt(Var(X)). It has the same units as X; variance has squared units.

Let \(X\) have density \(\mathrm{f}\) and mean \(\mu=\mathrm{E}(X)\). When the second moment is finite, variance is the expected squared deviation from the mean.

\(\begin{aligned}\sigma^2=\operatorname{Var}(X)&=\mathrm{E}[(X-\mu)^2]\\&=\int_{\text{all }x}(x-\mu)^2\mathrm{f}(x)\,\mathrm{d}x\\&=\mathrm{E}(X^2)-\mu^2.\end{aligned}\)

Calculate \(\mathrm{E}(X^2)=\int_{\text{all }x}x^2\mathrm{f}(x)\,\mathrm{d}x\), then subtract the square of the mean. For a piecewise density, integrate each piece over its own interval.

The standard deviation is \(\sigma=\sqrt{\operatorname{Var}(X)}\). It has the same units as \(X\); variance has squared units.

Continuous Random Variables

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Continuous Variance and Standard Deviation

Continuous Variance and Standard Deviation