Let $X$ and $Y$ be two (continuous) random variables with the joint p.d.f
$$f(x,y)=\dfrac{1}{4ah};\quad\text{for }-a+bx<y<a+bx,-h<x<h,\\ \text{and } f(x,y)=0;\quad\text{elsewhere. }$$
Here is the picture showing the distribution of $X$ and $Y$. Now the correlation coefficient of $X$ and $Y$ is defined by:
$$\rho=\dfrac{E[(X-\mu_{1})(Y-\mu_{2})]}{\sigma_{1}\sigma_{2}},$$ here $\mu$ denotes the expectation, and $\sigma$ for the deviatation. By this definition, we can show that for the above random variables, $$\rho=\dfrac{bh}{\sqrt{a^{2}+b^{2}h^{2}}}.$$
It is known that $\rho$ always satisfies: $-1\leq \rho\leq 1$. Also
if $\rho$ is close to $1$ then we would expect that $X$ and $Y$ have some strong correlation, that is the points $(X,Y)$ would almost concentrate near a line.
Look at the value of $\rho$ we see that when $h$ goes to $+\infty$ ($a,b$ are fixed) then $\rho\approx 1$. However, I am so confused that:
if we look at the picture, when $h$ is large then our parallelogram also becomes large. That is to say, we don't see any strong correlation between $X$ and $Y$ here. That's quite a contradiction to the above observation.
Could someone give me a hint?
Thanks a lot.
First consider two cases to illustrate the behavior:
When $a \to 0$, the parallelogram support turns to a straight line, i.e. $(X, Y)$ becomes perfectly linear and they degenerate to a one dimensional distribution. In such case $|\rho| \to 1$
When $h \to 0$, the parallelogram becomes more and more "rectangular" (at last become a vertical distribution, degenerate). So in such case $|\rho| \to 0$
Now when $h \to +\infty$, imagine in the above diagram the parallelogram is stretched to very very wide. In such case you need to zoom further and further away in order to view the entire parallelogram. As $a$ is unchanged and the zoom factor is high, it becomes very narrow and like a straight line as $h$ increase.