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

independenceprobabilityprobability distributionsprobability theory

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.

Best Answer

$f_1(x)=1+x$ 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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