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

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

The given joint density is

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

OR Should we do this?

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

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.

$$f_1(x)=1+x$$ if $$-1 and $$1-x$$ if $$0. ( In other words $$f_1(x)=1-|x|$$ for $$|x|<1$$). As you have observed $$f_2(y)=2y$$ for $$0. Now it is basic fact that if the random variables are independent then we must have $$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.

A slightly different approach:

The joint density can be written as $$f(x,y)=\underbrace{\frac{1}{2y}\mathbf1_{-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 $$\operatorname{Cov}(X,Y)=\operatorname{E}(XY)-\operatorname{E}(X)\operatorname{E}(Y)=0$$.

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