variance of mixed probability distribution

[jazzman]

Hey.
Let's say I have a random Variable \(\displaystyle X\) which has a mixed probability distribution (it has a discrete and continuous part).
Meaning, it's cumulative distribution function looks like this:
\(\displaystyle F_{X}(x)=\alpha{F_{X_d}}(x)+(1-\alpha){F_{X_c}}(x)\)
When, \(\displaystyle {F_{X}}^{d}(x)\) is the cdf of it's discrete part (\(\displaystyle X_d\)) and \(\displaystyle {F_{X}}^{c}(x)\) is the cdf of it's continuous part (\(\displaystyle X_c\)).

I now want to calculate the variance of \(\displaystyle X\) according to this formula:
\(\displaystyle Var(X)=E(X^2)-(E(X))^2\)

I know that \(\displaystyle E(X)=\alpha{E(X_d)}+(1-\alpha)E(X_c)\)
However, I'm not sure how to calculate \(\displaystyle E(X^2)\).

Please help!
Thanks in advance.

 

[galactus]

You can find \(\displaystyle E(X^{2})\) from the integral \(\displaystyle E(X^{2})=\int_{a}^{b}\frac{x^{2}}{b-a}dx=\frac{a^{2}+ab+b^{2}}{3}\)

If that helps.

Say, we have a probability distribution like so:

\(\displaystyle \begin{array}{ccccc}x=0&1&2&3&4\\P(x)=\frac{1}{16}&\frac{1}{4}&\frac{3}{8}&\frac{1}{4}&\frac{1}{16}\\x^{2}=0&1&4&9&16\end{array}\)

\(\displaystyle E(X^{2})=\frac{1}{16}\cdot 0+\frac{1}{4}\cdot 1+\frac{3}{8}\cdot 4+\frac{1}{4}\cdot 9+\frac{1}{16}\cdot 16\)

Does that help or are you needing something else?.