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In formulas: \begin{align} Cov(X,Y)=0 \quad \Leftrightarrow\quad \mathbb{E}[Y|X]=\mathbb{E}[Y] \end{align} is this true?

Felix Benning
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  • Not sure what the problem is - I mean if I actually would have expected an answer from someone else I would maybe have put more effort into it but it was just a question I couldn't find an answer to when looking so I thought I would post the answer I found since this seems like something that someone else might think about as well – Felix Benning Aug 12 '18 at 21:52

1 Answers1

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Only "$\Leftarrow$" is true:

\begin{align} Cov(X,Y)&=\mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] \\ &=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=\mathbb{E}[\mathbb{E}[XY|X]]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=\mathbb{E}[X\mathbb{E}[Y|X]]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=\mathbb{E}[X\mathbb{E}[Y]]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=0 \end{align}

While "$\Rightarrow$" is false in general:

Be $X\sim N(0,1)$ and $Y:=|X|$ then $\mathbb{E}[X|Y]=0$ which implies that $Cov(X,Y)=0$ as shown above. But $\mathbb{E}[Y|X]=Y$

Felix Benning
  • 2,889