# Proving that central moment is finite

I'm having trouble showing that the 2nd central moment is finite. I have $X_1,\ldots,X_n \overset{iid}{\sim} f(x)$ with $E[X_1]=\mu$ and $E[X_1^k]$ exists and is finite for any integer $k \geq 1$.

I would like to use Law of Large Numbers, so I need to show that either $E[|X_1|]$ is finite or that $E[(X_1-\mu)^2]$ is finite. I tried proving the first one with Jensen's but got stuck since absolute value is convex, not concave.

So now I'm stuck trying to show second central moment is finite.

• It's good you're having trouble, because the conclusion is false unless $k$ has some special values. What do you know about $k$? – whuber Jan 25 '13 at 16:53
• If $E[X_1] = \mu$ where $\mu$ is finite, then $E[|X_1|]$ is also finite. – Dilip Sarwate Jan 25 '13 at 17:12
• You need to clarify exactly what the the $E[X^k]$ statement means. Is it all $k$?, all $k<M$?, some fixed $k$?... – Dave Jan 25 '13 at 22:31
• For $k \geq 1$ an integer. – bdeonovic Jan 26 '13 at 20:15
• You say you want to use the LLN. For what and on what quantity? Note that the statement $\mathbb E X_1 = \mu$ for $\mu \in \mathbb R$ automatically implied by definition that $\mathbb E |X_1|$ is finite! (@Dilip also alludes to this in a previous comment.) So, there is nothing to prove in that respect. That $\mathbb E X_1^k$ exists and is finite for any integer $k \geq 1$ is a very strong statement. – cardinal Jan 26 '13 at 22:37

Isn't the second central moment just the variance? So $$E[(X_{1}-\mu)^{2}] = E(X_{1}^{2})-(E(X_{1}))^{2}$$

Since we know that $E[X_1] = \mu$ and $E[X_{1}^{2}]$ exists and is finite it follows that the variance is finite.