show that EX2 geq EX2 Hint Consider the random variable XEX

show that E(X^2) geq E(X))^2 . Hint: Consider the random variable (X-E(X))^2

Solution

E(X) is defined as Sum (X*P(X))

Var(X) is defined as Sum ((X- E(X))^2*P(X)

(X- E(X))^2= X^2- 2E(X)X+ E(X)^2 so that second sum is

Sum[X^2*P(X)- 2E(X) X*P(X)+ E(X)^2 *P(X)]

We can break that into the three separate sums. Also because E(X) is a number and does not depend on the specific values of X, we can take it out of the sum:

Sum[X^2*P(X)]- 2E(X) Sum[X*P(X)]+ E(X)^2 Sum[P(X)]

Of course, the second sum, Sum[X*P(X)], is just E(X) again and the third sum, Sum[P(X)] = 1. That is,
Var(X)= Sum[X^2*P(X)]- 2 E(X)*E(X)+ E(X)^2= Sum[X^2*P(x)]- E(X)^2

Sum[X^2* P(X)] is, by definition E(X^2) so that is precisely
Var(X)= E(X^2)- E(X)^2.