what is the expectation of this random variable?

Question

Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).

And suppose Y = $X\choose2$

, what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?

Thanks.

Answer 1

First, $Y={ X \choose 2}=\frac{X!}{2!(X-2)!}=\frac{X(X-1)(X-2)!}{2!(X-2)!}=\frac{X(X-1)}{2}$.

Second, use the fact that for any random variable $X$ and $a, b \in \mathbb{R}$, $E(aX+b)=aE(X)+b$.

Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.

So, $E(Y)=E(\frac{X(X-1)}{2})=E(\frac{X^{2}-X}{2})=\frac{E(X^{2})}{2}-\frac{E(X)}{2}$

All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).

Answer 2

Guide:

• The mean and variace of Hypergeometric should be well known, that is we know the first two moments.

• Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.