Zoagliaj

2022-07-16

The process to generate the two RVs is as follows. We first draw T from $Uniform(0,1)$ . If $T\le 0.5$ we take $X=T$ and draw Y from $Uniform(0,1)$ . Otherwise if $T>0.5$ , we take $Y=T$ and draw X from $Uniform(0,1)$ . Running a simulation it seems like X and Y are positively correlated, though intuitively it seems like they should have no effect on each other. What is the explanation?

minotaurafe

Beginner2022-07-17Added 22 answers

Step 1

Let's compute $E[XY]-E[X]E[Y]$ . We have $E[Y]=\frac{1}{2}(E[Y\mid T\le 0.5]+E[Y\mid T>0.5])=\frac{0.5+0.75}{2}$ . Similarly, we have $E[X]=\frac{1}{2}(E[X\mid T\le 0.5]+E[X\mid T>0.5])=\frac{0.25+0.5}{2}$ .

Then, we have $E[XY]=\frac{1}{2}(E[XY\mid T\le 0.5]+E[XY\mid T>0.5])$ . If $T\le 0.5$ , Y is independent from X, so $E[XY\mid T\le 0.5]=0.25\cdot 0.5$ . Similarly, when $T>0.5$ , Y is indepdent from X, so $E[XY\mid T>0.5]=0.5\cdot 0.75$ . Thus, we have $\mathrm{cov}(X,Y)=\frac{1}{2}(0.125+0.375)-\frac{1.25\cdot 0.75}{4}\approx 0.0156$ , indicating a positive correlation.

Let's compute $E[XY]-E[X]E[Y]$ . We have $E[Y]=\frac{1}{2}(E[Y\mid T\le 0.5]+E[Y\mid T>0.5])=\frac{0.5+0.75}{2}$ . Similarly, we have $E[X]=\frac{1}{2}(E[X\mid T\le 0.5]+E[X\mid T>0.5])=\frac{0.25+0.5}{2}$ .

Then, we have $E[XY]=\frac{1}{2}(E[XY\mid T\le 0.5]+E[XY\mid T>0.5])$ . If $T\le 0.5$ , Y is independent from X, so $E[XY\mid T\le 0.5]=0.25\cdot 0.5$ . Similarly, when $T>0.5$ , Y is indepdent from X, so $E[XY\mid T>0.5]=0.5\cdot 0.75$ . Thus, we have $\mathrm{cov}(X,Y)=\frac{1}{2}(0.125+0.375)-\frac{1.25\cdot 0.75}{4}\approx 0.0156$ , indicating a positive correlation.

Darian Hubbard

Beginner2022-07-18Added 7 answers

Step 1

Of course they are correlated. For one thing they are not independent (as X can only be bigger than 0.5 if $Y>0.5$ and conversely Y can only be smaller than 0.5 if $X\le 0.5$ ).

But in fact they are positively correlated, as the cases with $X>0.5$ , $Y<0.5$ cannot happen.

Let’s calculate this:

$\mathbb{E}(X)={\int}_{0}^{0.5}x\phantom{\rule{thinmathspace}{0ex}}dx+0.5{\int}_{0}^{1}x\phantom{\rule{thinmathspace}{0ex}}dx=({0.5}^{2}+0.5)/2=0.75/2=0.375$

and

$\mathbb{E}(Y)=0.5{\int}_{0}^{1}x\phantom{\rule{thinmathspace}{0ex}}dx+{\int}_{0.5}^{1}x\phantom{\rule{thinmathspace}{0ex}}dx=(0.5+(1-{0.5}^{2}))/2=0.625$

Then

$\mathbb{E}((X-0.375)(Y-0.625))={\int}_{0}^{0.5}(x-0.375){\int}_{0}^{1}(y-0.625)\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{thinmathspace}{0ex}}dx+{\int}_{0}^{1}(x-0.375){\int}_{0.5}^{1}(y-0.625)\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{thinmathspace}{0ex}}dx$

which amounts to $1/64$ for the covariance. So the correlation is positive. We’d still need to calculate the variance of X, Y (which is the same) for the correlation:

$\mathbb{E}({X}^{2})={\int}_{0}^{0.5}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx+0.5{\int}_{0}^{1}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx=({0.5}^{3}+0.5)/3=(5/8)/3=5/24$

so

$\mathrm{V}\mathrm{a}\mathrm{r}X=5/24-(3/8{)}^{2}=13/192$

so we get a correlation of

$(1/64)/(13/192)=3/13\approx 0.23$

Of course they are correlated. For one thing they are not independent (as X can only be bigger than 0.5 if $Y>0.5$ and conversely Y can only be smaller than 0.5 if $X\le 0.5$ ).

But in fact they are positively correlated, as the cases with $X>0.5$ , $Y<0.5$ cannot happen.

Let’s calculate this:

$\mathbb{E}(X)={\int}_{0}^{0.5}x\phantom{\rule{thinmathspace}{0ex}}dx+0.5{\int}_{0}^{1}x\phantom{\rule{thinmathspace}{0ex}}dx=({0.5}^{2}+0.5)/2=0.75/2=0.375$

and

$\mathbb{E}(Y)=0.5{\int}_{0}^{1}x\phantom{\rule{thinmathspace}{0ex}}dx+{\int}_{0.5}^{1}x\phantom{\rule{thinmathspace}{0ex}}dx=(0.5+(1-{0.5}^{2}))/2=0.625$

Then

$\mathbb{E}((X-0.375)(Y-0.625))={\int}_{0}^{0.5}(x-0.375){\int}_{0}^{1}(y-0.625)\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{thinmathspace}{0ex}}dx+{\int}_{0}^{1}(x-0.375){\int}_{0.5}^{1}(y-0.625)\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{thinmathspace}{0ex}}dx$

which amounts to $1/64$ for the covariance. So the correlation is positive. We’d still need to calculate the variance of X, Y (which is the same) for the correlation:

$\mathbb{E}({X}^{2})={\int}_{0}^{0.5}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx+0.5{\int}_{0}^{1}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx=({0.5}^{3}+0.5)/3=(5/8)/3=5/24$

so

$\mathrm{V}\mathrm{a}\mathrm{r}X=5/24-(3/8{)}^{2}=13/192$

so we get a correlation of

$(1/64)/(13/192)=3/13\approx 0.23$

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