4

Suppose we have $X \sim Be(p_1)$, $Y \sim Be(p_2)$ and $corr(X,Y) = r$.

How does the joint distribution table look like?

For example, if $r = 0$ it is $$ \begin{array}{c|c|c} \cdot & T & F \\ \hline T & p_1p_2 & p_1(1-p_2) \\ \hline F & (1-p_1)p_2 & (1-p_1)(1-p_2) \\ \end{array} $$

Veliko
  • 1,084

2 Answers2

4

Let $R=Cov(X,Y)= r*\sqrt{p_1(1-p_1)}*\sqrt{p_2(1-p_2)}$

Let $0\leqslant a,b,c,d \leqslant 1$ be the respective probabilities of TT, TF, FT and FF.

We need to solve the system:

$a+b=p_1$

$a+c=p_2$

$a+b+c+d=1$

$a=R+p_1p_2$

We get:

$a=R+p_1p_2$

$b=p_1(1-p_2)-R$

$c=p_2(1-p_1)-R$

$d=1+R-p_1-p_2+p_1p_2$

Thanks to @Henry for corrections.

Evargalo
  • 2,593
  • Your $a+d=r+p1p2$ looks unlikely if $r=0$, or if $r=1$ and $p1=p2$ – Henry Jun 20 '17 at 09:48
  • @Henry : thanks, corrected. – Evargalo Jun 20 '17 at 09:52
  • 1
    I suspect it may be more complicated than $a=r+p1p2$: consider $p1=p2=\frac12$ and $r=1$. Perhaps something like $a=p1p2 + r \sqrt{p1(1-p1)p2(1-p2)} $ unchecked – Henry Jun 20 '17 at 09:55
  • Oh yeah, I used covariance instead of correlation... – Evargalo Jun 20 '17 at 09:57
  • @Henry, do you think that there is some restriction on the values $p_1$ and $p_2$ depending on the $r$? For example is it possible that $r = 0.99$ and $p_1 = 0.5, p_2 = 0.9$? – Veliko Jun 20 '17 at 11:23
  • 1
    @Veliko: Yes - I think your example would give $a=0.5985$ and $b=-0.0985$ with the latter number being clearly impossible. I suspect with $p_1=0.5,p_2=0.9$ then you must have $-\frac13 \le r \le \frac13$ – Henry Jun 20 '17 at 12:37
  • @Henry, I think the s should go inside the min but the 1 must stay without $s$. – Veliko Jun 20 '17 at 19:07
  • @Evargalo $d = 1 +R -p_1-p_2-p_1 p_2$. – Veliko Jun 20 '17 at 19:24
  • @Veliko, Yes - I will delete my earlier comment. – Henry Jun 20 '17 at 20:34
  • 3
    @Veliko: given $p_1$ and $p_2$ and letting $s=\sqrt{p_1(1-p_1)p_2(1-p_2)}$ it may be that $r$ is constrained to $$\max\left(-1,-\dfrac{ p_1p_2}{s},-\dfrac{(1-p_1)(1-p_2)}{s}\right) \le r \le \min\left(1,\dfrac{p_1(1-p_2)}{s},\dfrac{(1-p_1)p_2}{s}\right)$$ – Henry Jun 20 '17 at 20:36
  • @Evargalo For sure $d = (1-p_{1})(1-p_{2}) + R$ in order for the margin probabilities to be correct. The sign of R is mistaken in your nice answer. – FraSchelle Dec 20 '19 at 16:01
3

Hint:

Note that here $\Pr(X=1\wedge Y=1)=\mathbb EXY$ and that the other $3$ probabilities can easily be deduced from this one.

So it is enough to find $\mathbb EXY$ and for this you can use the equality: $$\mathbb EXY-\mathbb EX\mathbb EY=\text{Cov}(X,Y)=\text{Corr}(X,Y)\sqrt{\text{Var}X\text{Var}Y}$$

drhab
  • 151,093
  • This is a good idea, thank you. – Veliko Jun 20 '17 at 11:00
  • Do you think that there is some restriction on the values $p_1$ and $p_2$ depending on the $r$? For example is it possible that $r = 0.99$ and $p_1 = 0.5, p_2 = 0.9$? – Veliko Jun 20 '17 at 11:21
  • 1
    Yes. With the mentioned values of $p_1,p_2$ we arrive at $p=0.15r+0.45$ where $p:=\mathbb EXY=P(X=1=Y)$. This however with $p\leq\min(p_1,p_2)=0.5$. Then $r$ cannot take value $0.99$ (since that leads to $p>0.5$). – drhab Jun 20 '17 at 12:50