Due to quarantine, I had to take my probability course via online, so I'm not learning propperly about expectations and correlations.
I'm having trouble trying to prove this:
If $E(Y|X=x)=E(Y) $ for every $x$, then $X$ & $Y$ are not correlated.
I've tried putting the formulas and calculate the correlation coefficient but failed to obtain zero (that's what they told me about independence). If anyone can give a hint and a book recommendation it would be fantastic, thank you !
I shall first present the proof for the discrete case. Observe that by definition, $$P(Y=y|X=x)=\frac{P(Y=Y,X=x)}{P(X=x)}.$$ We have \begin{align*} E[XY] &= \sum_x \sum_y xy P(Y=y,X=x) \\ &= \sum_x \sum_y xy\cdot P(X=x)\cdot\frac{P(Y=y,X=x)}{P(X=x)} \\ &= \sum_x \sum_y xy\cdot P(X=x)\cdot P(Y=y|X=x) \\ &= \sum_x \left[xP(X=x)\sum_y yP(Y=y|X=x)\right] \\ &= \sum_x \left[x P(X=x) E[Y|X=x] \right] \\ &= \sum_x xP(X=x)E[Y]\\ &= E[Y]\sum_x xP(X=x) \\ &= E[Y]E[X]. \end{align*} Note carefully how I move the terms depending only on $x$ outside of the sum over $y$. Thus, $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]=0.$
For the continuous case, if $f_{X,Y}$ is the joint PDF of $X$ and $Y$, $f_X$ is the PDF of $X$, and $f_{Y|x}$ is the PDF of the random variable $Y|x$, then $\frac{f_{X,Y}(x,y)}{f_X(x)}=f_{Y|x}(y)$. Thus, we have
\begin{align*} E[XY] &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} xy f_{X,Y}(x,y) \mathrm{d}y\mathrm{d}x\\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} xy f_X(x)\cdot\frac{ f_{X,Y}(x,y)}{f_X(x)} \mathrm{d}y\mathrm{d}x\\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} xy f_X(x)\cdot f_{Y|x}(y) \mathrm{d}y\mathrm{d}x \\ &= \int_{-\infty}^{\infty}xf_X(x)\left[\int_{-\infty}^{\infty} y\cdot f_{Y|x}(y)\mathrm{d}y\right]\mathrm{d}x \\ &= \int_{-\infty}^{\infty}xf_X(x) E[Y|X=x]\mathrm{d}x \\ &= \int_{-\infty}^{\infty}xf_X(x) E[Y]\mathrm{d}x \\ &= E[Y]\int_{-\infty}^{\infty}xf_X(x) \mathrm{d}x \\ &= E[Y]E[X]. \end{align*}