prove : if E(X) doesn't exist $E(x^2)$ too doesn't exist.

Question

$E(X^2) $ exists implies $\int x^2 f_X(x) \ dx < \infty$

now from the property of Riemann Integral $\int|x| f_X(x) \ dx \le \int x^2 f_X(x) \ dx $ . hence, existence of $E(X^2)$ implies existence of E(X) .

Is it right a right way to prove this ?

(i don't want a formal proof. I only want to know whether my logic is mathematical correct. actually i doubt that the property of Riemann integral that i have used is valid only for bounded functions. is it so ?

Answer 1

Suppose the measure is the Lebesgue measure on $[0,1]$. Let $A$ be a non measurable set and let $X=1_A-1_{A^c}$. Then $X^2 = 1_{[0,1]}$. Hence $EX^2 = 1$ , but $EX$ does not even exist.

If you are using the Riemann integral, choose $A=[0,1]\cap \mathbb{Q}$ instead.

Answer 2

If $E(X)$ doesn't exist, $E(X_+)$ or $E(X_-)$ or both are infinite. An application of the Jensen Inequality with the function $x^2$ will show you that $E(X^2) \geq E(X_+^2) \geq E(X_+)^2 = \infty$ (or the same calculation with $X_-$ if thats infinite).

Your calculation has the Problem, that not every random variable has to have a density.