calculation of variance from cdf (no mathematical expression available)

Question

Is it possible to calculate the variance of a continous random variable from the Cummulative distributive function plot ? We dont have the mathematical expression for cdf, all we have is just a plot of cdf on a graph sheet.

Answer 1

Yes there is a formula available which was proven by Pyke in 1965. It is given in Kendall and Stuart (Advanced Theory of Statistic vol 1, 4th ed. pp 56): $$ \text{Var}(x) = \int_{a}^{b} \left( \int_{x}^{b} 1-F(y) \, \text{d} y \right)^2 (1-F(x))^{-2} \, \text{d}F(x). $$ Here $F(a)=0$ and $F(b)=1$, where $F$ is the CDF.

The proof is by integration by parts for Stieltjes integrals.

Answer 2

Something useful might be $E[X^2] = 2 \int_0^\infty t P(|X|>t) dt$, $E[X] = \int_0^\infty (1-F(t)) dt - \int_{-\infty}^0 F(t) dt$ (or equivalently, integrating over the vertical axis, $E[X] = \int_0^1 F_X^{-1}(t) dt$).

Note that this simplifies considerably for non-negative $X$: $E[X^2] = 2 \int_0^\infty t (1-F(t)) dt$, $E[X] = \int_0^\infty (1-F(t)) dt$.

Then, use the fact that $var(X) = E[X^2]-(E[X])^2$.

An old chemist's trick to do integrals (such as estimating fractions of a chemical from a chromatogram, back before GC's and what not had computers) was to take a carefully drawn copy of the curve, cut it out carefully and weigh it. Since the density of paper is so uniform, this gave a pretty good idea of the area under the curve by knowing the density of the paper. Maybe you can try this.