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I was wondering about this. I have a second moment that is zero. Can I conclude that if $E[x^2]=0$ then $Var[x]=E[x^2]-E[x]E[x]$ implies $E[x]=0$ since the variance must be non-negative? (If not, can I imply anything else from that, in particular, regarding if we also have mean square convergence?)

TooTone
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Majte
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1 Answers1

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The answer to your question is yes.

Alternatively, you can use Hölder's inequality and obtain the following inequality $$ 0\le|\operatorname EX|\le\operatorname E|X|\le(\operatorname E|X|^2)^{1/2}. $$

Cm7F7Bb
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  • Cool, thanks! I somehow find this by itself so counter-intuitive.. I mean if I take a constant, it has a zero variance right? But the expected value of a constant is a constant. I am dealing with some probability and mean square convergence theorems. I have some results that state $E[x^2]$ and then conclude $x\rightarrow_{ms} 0$ and I was wondering if it stems through that. – Majte Feb 17 '14 at 10:23
  • second moment $\not=$ variance! – kjetil b halvorsen Feb 17 '14 at 10:28
  • A constant has a non negative second moment that is equalled by its squared mean? – Majte Feb 17 '14 at 10:29
  • @Majte You're welcome! The variance in general is not equal to the second moment (we have that $\operatorname{Var}X=\operatorname E X^2-(\operatorname E X)^2$). Yes, if the random variable is constant almost surely, then the second moment $\operatorname EX^2=(\operatorname EX)^2$. – Cm7F7Bb Feb 17 '14 at 10:35
  • Then I trust you that's the solution :D Still, that a constant variable has a second moment is so odd. Can you give an intuition for that? – Majte Feb 17 '14 at 10:44
  • @Majte The $p$-th absolute moment of a random variable is defined as $\operatorname E|X|^p$. The expected value of a constant $c$ is equal to the constant $c$ itself, i.e. $\operatorname Ec=c$ for any $c\in\mathbb R$. Now if the random variable $X$ is a constant random variable, i.e. $X(\omega)=c$ for all $\omega\in\Omega$, then we have that $\operatorname E|X|^p=|c|^p$. So constant random variables have finite moments of all orders. – Cm7F7Bb Feb 17 '14 at 11:30
  • nice answer, just to check, Cauchy-Schwartz applies here as a special case of Holder's inequality? – TooTone Feb 17 '14 at 12:21
  • @TooTone Yes, you are precisely right. We actually don't need Hölder's inequality here, we can use the Cauchy-Schwartz inequality to conclude that $|\operatorname E X|\le(\operatorname E|X|^2)^{1/2}$, but Hölder's inequality came first to my mind. – Cm7F7Bb Feb 17 '14 at 12:35