Variance 11/10

E[X] does not say anything about the the spread of the values.

This is measured by

$$Variance(X)=E[(X-\mu)^2],\mbox{ where }\mu=E[X]$$

which can also be written in the computational formula:

$$Var(X)=E[X^2]-(E[X])^2 \mbox{often noted } \sigma^2$$

The unit in which this is measured is not coherent with that of X, we very often use the standard deviation

$$SD(X)=\sqrt{Var(X)}=\sigma$$

$$Var(aX+b)=a^2Var(X), \qquad SD(aX+b)=\vert a\vert SD(X)$$

For the Bernouill(p): X²=X

var(X)=E(X²)-(E(X))²=p-p²=p(1-p)=pq

Property 1:
For two independent random variables, X and Y:

Var(X+Y)=Var(X)+Var(Y)

This essential property allowed us to compute the variance of a binomial S_n, because we can write a binomial as the sum of n independent Bernouilli(p) random variables X_i so that:

$$var(S_n)=var(\sum_i X_i)= n var(X_i)=npq$$

Example(which I `did' in class!):
What is the variance of the geometric? Using the computational formula, we compute first E(X²):

$$\begin{eqnarray*}E(X^2)&=&\sum_{j=1}^{+\infty} j^2 P(X=j) =\sum_{j=1}^{+\infty} ... ...rac{1}{p}+ \frac{2q}{p^2}- \frac{1}{p^2}\\ &=& \frac{q}{p^2}\\ \end{eqnarray*}$$

From the sums of independent variables theorem, and the fact that a Negative Binomial Y_r can be written as the sum of r independent Geometrics, we have:

$$var(Y_r)=\frac{rq}{p^2}$$

I also showed in class, that the variance of the Poisson($\lambda$) random variable is $\lambda$, a fact that helps recognize a Poisson random variable.