juniorychichoa70

2022-05-02

I am getting ${f}_{X,Y}(x,y)={f}_{X}(x){f}_{Y}(y)$ even if the correlation coefficient $\rho \ne 0$

vanvanvan4ie

Beginner2022-05-03Added 12 answers

Step 1

Firstly, the domain of $S=\{(x,y):-a+bx<y<a+bx\phantom{\rule{thinmathspace}{0ex}},-h<x<h\}$ means that

$\begin{array}{rl}{f}_{X}(x)& ={\int}_{-a+bx}^{a+bx}{\displaystyle \frac{{\mathbf{1}}_{-h<x<h}}{4ha}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}y\\ & ={\displaystyle \frac{1}{2h}}\text{}{\mathbf{1}}_{-hxh}\end{array}$

[Note: because we want the marginal of x we 'integrate out' y over where it is supported relative to x.]

Nextly, by rewriting the domain $S=\{(x,y):(y-a)/b<x<(y+a)/b\phantom{\rule{thinmathspace}{0ex}},-a-bh<y<a+bh\}$ , we likewise have:

$\begin{array}{rl}{f}_{Y}(y)& ={\int}_{(y-a)/b}^{(y+a)/b}{\displaystyle \frac{{\mathbf{1}}_{-(a+bh)<y<(a+bh)}}{4ha}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & ={\displaystyle \frac{1}{2bh}}{\mathbf{1}}_{-(a+bh)<y<(a+bh)}\end{array}$

So ${f}_{X}(x){f}_{Y}(y)={\displaystyle \frac{1}{4b{h}^{2}}}{\mathbf{1}}_{-h<x<h,-(a+bh)<y<(a+bh)}$ , which does not equal to the joint p.d. function.

Firstly, the domain of $S=\{(x,y):-a+bx<y<a+bx\phantom{\rule{thinmathspace}{0ex}},-h<x<h\}$ means that

$\begin{array}{rl}{f}_{X}(x)& ={\int}_{-a+bx}^{a+bx}{\displaystyle \frac{{\mathbf{1}}_{-h<x<h}}{4ha}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}y\\ & ={\displaystyle \frac{1}{2h}}\text{}{\mathbf{1}}_{-hxh}\end{array}$

[Note: because we want the marginal of x we 'integrate out' y over where it is supported relative to x.]

Nextly, by rewriting the domain $S=\{(x,y):(y-a)/b<x<(y+a)/b\phantom{\rule{thinmathspace}{0ex}},-a-bh<y<a+bh\}$ , we likewise have:

$\begin{array}{rl}{f}_{Y}(y)& ={\int}_{(y-a)/b}^{(y+a)/b}{\displaystyle \frac{{\mathbf{1}}_{-(a+bh)<y<(a+bh)}}{4ha}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & ={\displaystyle \frac{1}{2bh}}{\mathbf{1}}_{-(a+bh)<y<(a+bh)}\end{array}$

So ${f}_{X}(x){f}_{Y}(y)={\displaystyle \frac{1}{4b{h}^{2}}}{\mathbf{1}}_{-h<x<h,-(a+bh)<y<(a+bh)}$ , which does not equal to the joint p.d. function.

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