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Show that the random variables $X$ and $Y$ are uncorrelated but not independent

The given joint density is

$f(x,y)=1\;\; \text{for } \; -y<x<y \; \text{and } 0<y<1$, otherwise $0$

My main concern here is how should we calculate $f_1(x)$

$f_1(x)=\int_y dy = \int_{-x}^{1}dy + \int_{x}^{1}dy = 1+x +1=2\; \; \forall -1 <x<1$

OR Should we do this?

$f_1(x)$=$$ \begin{cases} \int_{-x}^{1}dy = 1+x && -1<x<0 \\ \int_{x}^{1}dy = 1-x & & 0\leq x <1 \\ \end{cases} $$

In the second case, how do I show they are not independent.

I can directly say that the joint distribution does not have a product space but I want to show that $f(x,y)\neq f_1(x)f_2(y)$

Also, for anyone requiring further calculations,

$f_2(y) = \int dx = \int_{-y}^{y}dx = 2y$

$\mu_2= \int y f_2(y)dy = \int_{0}^{1}2y^2 = \frac23$

$\sigma_2 ^2 = \int y^2f_2(y)dy - (\frac23) ^2 = \frac12 - \frac49 = \frac1{18}$

$E(XY)= \int_{y=0}^{y=1}\int_{x=-y}^{x=y} xy f(x,y)dxdy =\int_{y=0}^{y=1}\int_{x=-y}^{x=y} xy dxdy$ which seems to be $0$? I am not sure about this also.

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$f_1(x)=1+x$ if $-1<x<0$ and $1-x$ if $0<x<1$. ( In other words $f_1(x)=1-|x|$ for $|x|<1$). As you have observed $f_2(y)=2y$ for $0<y<1$. Now it is basic fact that if the random variables are independent then we must have $f(x,y)=f_1(x)f_2(y)$ (almost everywhere). Since the equation $(1-|x|)(2y)=f(x,y)$ is not true we can conclude that $X$ and $Y$ are not independent.

$EXY=0$ is correct. Also $EX=\int_{-1}^{1}x(1-|x|)dx=0$ so $X$ and $Y$ are uncorrelated.

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A slightly different approach:

The joint density can be written as $$f(x,y)=\underbrace{\frac{1}{2y}\mathbf1_{-y<x<y}}_{f_{X\mid Y=y}(y)}\,\underbrace{2y\mathbf1_{0<y<1}}_{f_Y(y)}$$

Clearly, the (conditional) distribution of $X$ 'given $Y$' depends on $Y$, so that $X$ and $Y$ are not independent.

In fact, $X\mid Y\sim U(-Y,Y)$, which implies $\operatorname{E}(X\mid Y)=0$.

So,

\begin{align} \operatorname{E}\,(XY)&=\operatorname{E}\left[\operatorname{E}\left(XY\mid Y\right)\right] \\&=\operatorname{E}\left[Y\operatorname{E}\left(X\mid Y\right)\right] \\&=\operatorname{E}\left[Y\times 0\right]=0 \end{align}

Also, $$\operatorname{E}(X)=\operatorname{E}\left[\operatorname{E}(X\mid Y)\right]=0$$

This proves that $\operatorname{Cov}(X,Y)=\operatorname{E}(XY)-\operatorname{E}(X)\operatorname{E}(Y)=0$.

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  • 1
    $\begingroup$ Downvote for a different method than what OP is following? $\endgroup$ – StubbornAtom May 7 at 6:05

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