If $E(XY)=E(X) \cdot E(Y)$, why is not $E(X^2)=[E(X)^2]$

Question

My book states that the expected value of the product of two independent random variables equal the product of their expected values. Why cannot one use this to give the following?

$$E(X \cdot X) = E(X) \cdot E(X) = [E(X)]^2$$

Of course, this is incorrect, as if this were true the variance would always be 0.

"the product of two independent random variables equal the product of their expected values." You seem to have overlooked that word. — Clement C., Mar 07 '17 at 18:28

score 8 · Answer 1 · edited Apr 13 '17 at 12:20

8

Because $X$ and $X$ are not independent.

For more info, see this question that asks something very similar if not the same.

edited Apr 13 '17 at 12:20

Community

1

answered Mar 07 '17 at 18:21

score 2 · Accepted Answer · answered Mar 07 '17 at 21:02

To see that $X$ is not independent of itself, consider $X$ to be the value of a coin flip: $P(X=0)=P(X=1)=1/2$.

Recall that $X$ and $Y$ are independent means for any (nice enough) sets $A,B\in\mathbb{R}$, $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$.

But $P(X=1,X=0)=0\neq 1/4=P(X=1)P(X=0).$

If $E(XY)=E(X) \cdot E(Y)$, why is not $E(X^2)=[E(X)^2]$

2 Answers2