Random variables $X$ and $Y$ follow a joint distribution
$$f(x, y) = \left\{ \begin{array}{ll} 2,& 0 < x \leq y < 1,\\ 0,&\text{otherwise}\end{array}\right.
$$
Determine the correlation coefficient between $X$ and
$Y$ .
[Math] How to find correlation coefficient given joint distribution
statistics
Related Solutions
I think I have found the discrepancy.The short answer is that the distribution for case (2) is a doubly noncentral beta distribution. The bulk correlation is the same under each case.
In the solution below, I've used slightly different notation. In particular, $\boldsymbol{s }$ $\rightarrow$ $\boldsymbol{w}$. I've also used $\hat{A}$ to describe the maximum liklehood estimate for the amplitude of the sample vector $\boldsymbol{w}$ that maximally correlates with $\boldsymbol{x}$. This notation and formulation is targeted at detection theory applications.
The relative square error in approximating approximating a data stream $ \boldsymbol{x} $ using a waveform template $\boldsymbol{w }$ is the ratio of the least-squares error $\vert\vert \boldsymbol{e} \vert\vert^{2}$ to measure signal energy $ \vert \vert \boldsymbol{x} \vert \vert^{2}$. This error may be re-written as a ratio of quadratic forms that has a doubly noncentral $\beta$ distribution:
\begin{equation} \begin{split} \cfrac{\vert \vert \boldsymbol{e} \vert \vert^{2}}{\vert \vert \boldsymbol{x} \vert \vert^{2}} &= \cfrac{\vert \vert \boldsymbol{x} - \hat{A} \boldsymbol{w } \vert \vert^{2} }{ \vert \vert \boldsymbol{x} \vert \vert^{2}} \\ &= \cfrac{\vert \vert \boldsymbol{x} - \cfrac{\langle \boldsymbol{x},\, \boldsymbol{w} \rangle}{ \vert \vert \boldsymbol{w } \vert \vert^{2}} \boldsymbol{w } \vert \vert^{2}} {\vert \vert \boldsymbol{x} \vert \vert^{2}} \\ &= \cfrac{ \vert \vert P_{\boldsymbol{w}}^{\perp} \left( \boldsymbol{x} \right) \vert \vert ^{2}} { \vert \vert P_{\boldsymbol{w}}^{\perp}\left( \boldsymbol{x} \right) \vert \vert ^{2} + \vert \vert P_{\boldsymbol{w}} \left( \boldsymbol{x} \right) \vert \vert ^{2}} \\ &\overset{d}{=} \cfrac{ \chi_{1}^{2}( \lambda^{\perp} )} { \chi_{1}^{2}( \lambda ) + \chi_{N_{E} - 1}^{2}( \lambda^{\perp} ) }, \end{split} \end{equation}
where the noncentrality parameters are defined by $\lambda$ $=$ $\cfrac{\vert \vert P_{\boldsymbol{w}} \left( \boldsymbol{x} \right) \vert \vert ^{2}}{\sigma^{2}}$ and $\lambda^{\perp}$ $=$ $\cfrac{\vert \vert P_{\boldsymbol{w}}^{\perp} \left( \boldsymbol{x} \right) \vert \vert ^{2}}{\sigma^{2}}$, and where $\overset{d}{=}$ indicates distributional equality. This ratio is also related to the sample correlation coefficient $r$:
\begin{equation} \begin{split} \cfrac{\vert \vert \boldsymbol{x} - \cfrac{\langle \boldsymbol{x},\, \boldsymbol{w} \rangle}{ \vert \vert \boldsymbol{w } \vert \vert^{2}} \boldsymbol{w } \vert \vert^{2}} {\vert \vert \boldsymbol{x} \vert \vert^{2}} &= 1- \cfrac{\langle \boldsymbol{x},\, \boldsymbol{w} \rangle^{2} }{ \vert \vert \boldsymbol{w } \vert \vert^{2} \vert \vert \boldsymbol{x } \vert \vert^{2}} \\ &= 1 - r^{2} \end{split} \end{equation}
Therefore:
\begin{equation} \begin{split} r^{2} &\overset{d}{=} \cfrac{ \chi_{1}^{2}( \lambda )} { \chi_{1}^{2}( \lambda ) + \chi_{N_{E} - 1}^{2}( \lambda^{\perp} ) } \\ &\sim \text{Beta} \left( \frac{1}{2}, \frac{1}{2}N_{E} ; \lambda, \lambda^{\perp} \right) \end{split} \end{equation}
Distinct hypotheses regarding the distribution for $\boldsymbol{x}$ simplify the form for this distribution. When the data stream contains only noise, the hypothesis $\mathcal{H}_{0}$ is satisfied and $r^{2}$ has a central Beta distribution, where $\lambda^{\perp}$ $=$ $\lambda$ $=$ $0$. In the presence of signal, a data stream $\boldsymbol{x}$ will generally have a non-zero projection $P_{\boldsymbol{w}}^{\perp}\left( \boldsymbol{x} \right)$ orthogonal to the noise-contaminated template data vector $\boldsymbol{w}$. In this case $\lambda$, $\lambda^{\perp}$ $\ne$ $0$, and $r^{2}$ has doubly noncentral Beta distribution. If the template signal has a very large SNR, then $ \boldsymbol{x}$ $\cong$ $A \boldsymbol{w}$ $+$ $\boldsymbol{n}$, $\lambda^{\perp}$ $=$ $0$, and $r^{2}$ is reasonably approximated by a noncentral Beta distribution. The noncentral Beta distribution therefore provides an absolute upper bound on the detection performance of a correlation detector.
It helps to instead write an equivalent definition,$$\rho_{X,\,Y}=\frac{\Bbb E((X-\Bbb EX)(Y-\Bbb EY))}{\sigma_X\sigma_Y}.$$This is a covariance divided by a product of standard deviations. I've explained before that covariance is an inner product (with some qualifying statements you'll find at that link). Then standard deviation is like a squared length, so the above formula is like$$\cos\theta=\frac{a\cdot b}{|a||b|}.$$In particular, perfectly correlated variables are "parallel" in a vector space of random variables, whereas uncorrelated ones are orthogonal.
Best Answer
We need to calculate cov(X,Y)=EXY - EX*EY, var(X) and var(Y). To do it, we have to know marginal distributions of both random variables X and Y. This can be done by "integrating the other variable out" of the joint density function. I will show you how to calculate the marginal density function of X:
$ f(x)= \int^{1}_{x} f(x,y) dy = \int^{1}_{x} 2 dy = 2(1-x)$ for $x \in (0,1)$.
Are you able to continue?