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I was trying to show that if two variables have diagonal covariance, this does not necessarily guarantee their independence. For this, I was using an example where $x \sim U(-1,1)$ and $y=X^{2}$ to show this.

Here is my try:

$$ p\left(y | x\right)=\delta\left(y-x^{2}\right) $$

Now we would like to show that although the two variables are dependent, the covariance matrix between them is diagonal. $$ \operatorname{cov}[X, Y]=\mathrm{E}[X Y]-\mathrm{E}[X] \mathrm{E}[Y] $$

$$ p\left(x, y\right)=p\left(x\right) p\left(y | x\right) $$

\begin{align*} E[X Y]=& \iint x y p(x, y) d y d x \\ &= \iint x y p\left(x\right) p\left(y | x\right) d y d x \\ &= \int x p\left(x\right) \int y \delta\left(y-x^{2}\right) d y d x \\ % &= \int y p(y) d y \cdot \int x p(x) d x \\ % &= E[X] E[Y] \end{align*} Here is the deal: I was trying to show that all off-diagonal elements are in fact zero, but I'm stuck at that integral. Any suggestion on how to proceed? Thanks

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$E(XY)=E(X^3)=\frac{1}{2}\int_{-1}^1x^3dx=0$. Therefore the covariance $=0$, since $E(X)=0$.

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  • $\begingroup$ What about the probability term then? Also, we are explicitly asked to use $p(x, y)=p(x) p(y | x)$ $\endgroup$ – Blade Apr 7 at 22:00
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    $\begingroup$ Your definition of $p(y|x)$ is unfamiliar to me. Therefore I just went back to basics. $\endgroup$ – herb steinberg Apr 7 at 22:14
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    $\begingroup$ @Blade Your integral looks straightforward. $\int y\delta (y-x^2)dy=x^2$, so your final integral is $\int x^3p(x)dx$, which is identical to that I wrote. $\endgroup$ – herb steinberg Apr 8 at 3:32

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