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I have a question related to the implications of a zero second moment conditions. Consider a real-valued random variable $X$ defined on the probability space $(\Omega, \mathcal{F}, P)$. As we can read here, $E(X^2)=0 \rightarrow E(X)=0$.

If we keep to write the implications, we have $E(X^2)=0 \rightarrow E(X)=0\rightarrow Var(X)=0\leftrightarrow X(\omega)=K \text{ }\forall \omega \in \Omega, K \in \mathbb{R} \leftrightarrow E(X)=X$

Can we conclude that $E(X^2)=0 \rightarrow X=0$?

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  • Depends a bit on the precise space - I think in general, the best we can say that $X=0$ almost surely. – πr8 Jan 08 '16 at 15:39
  • I also think that your last $\leftrightarrow$ ought to be a $\rightarrow$ - you can have $E(X)=X$ without $X$ being a.s. constant. – πr8 Jan 08 '16 at 15:44

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Since it is always the case that $$0 \leq \text{Var}(X) \leq \mathbb{E}(X^2)\ ,$$ knowing that the second moment is zero implies zero variance. If the variance is zero then (with probability 1) the random variable $X$ is constant (equal to $\mathbb{E}(X)$).

Anders Muszta
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