If $r$-th population moment is finite, then $r$-th sample moment probability converge to $r$-th population moment?

Consider $$r=2$$ case

We suppose $$X_i$$ follows pdf $$f(x)$$ for $$i=1,2,...,n$$ with $$E(X^2)\lt \infty$$

Want to show

$$P\left(\left|\frac{\sum_{i=1}^{n}X_{i}^2}{n}-E(X^2)\right| \lt \epsilon\right) \to 1\qquad\text{ as }\quad n\to\infty$$

Let $$Y_i=X_{i}^{2}$$ and $$\mu_2^\prime=E(X^2)$$

Then, by Markov's inequality, we obtain

$$P(|\overline Y_n-\mu_2^\prime| \lt \epsilon) \le 1-\frac{E((\overline Y_n-\mu_2^\prime)^2) }{\epsilon}$$

Let us compute $$E((\overline Y_n-\mu_2^\prime)^2)=\operatorname{Var}(\overline Y_n)$$

$$\operatorname{Var}(\overline Y_n)=\sum_{i=1}^{n}\frac{\operatorname{Var}(X^2)}{n^2}$$

But I can't be certain existence of the variance. I want to ask we can induce the variance is also finite from our given finite value of expectation. Please give me a hint!

• For application of LLN it is not necessary that variance exists. There is a proof for it using characteristic functions. Have a look here. – drhab Oct 16 '18 at 8:20