Re: Covariance of two proportions?

On Aug 31, 11:51 pm, Jack Tomsky
<jtom...@xxxxxxxxxxxxx> wrote:
On Aug 31, 9:00 pm, Jack Tomsky
<jtom...@xxxxxxxxxxxxx> wrote:
Since these are from two different samples, the
covariance is zero.
In a different case where x1 and x2 are
observations from the same multinomial
distribution,
M(N;p_1, ..., p_(k-1)), the covariance between
p1hat
= x1/N and p2hat = x2/N is -p1*p2/N.

Hi, thanks Paul and Jack for replying. But maybe
I
need to explain a
bit more: The two samples are not from the same
population but they
are also not independent, there are
intersections:
p1=x1/n1
p2=(x1+x2)/(n1+n2)

As you see p2 include parts from p1 (namely x1
and
n1) plus
additionally other values (x2 and n2). How do you
treat this
situation?

So you're assuming that x1 ~ Bin(n1,p) and x2~
Bin(n2,p).

Then with

p1=x1/n1
p2=(x1+x2)/(n1+n2)

Cov(p1,p2) = Cov(x1/n1,(x1+x2)/(n1+n2)) =
Var(x1)/[n1(n1+n2)] + Cov(x1,x2)/[n1(n1+n2)] =
n1*p(1-p)/[n1(n1+n2)] + 0 = p(1-p)/(n1+n2).

Incidentally, this makes the correlation between p1
and p2 as

Corr(p1,p2) = Cov(p1,p2)/sqrt(Var(p1)*Var(p2)) =
[p(1-p)/(n1+n2)]/[sqrt(p(1-p)/n1)*sqrt(p(1-p)/(n1+n2))
] = sqrt(n1/(n1+n2)).

Thanks, but I have no indications that x1/n1 is equal
to x2/n2 (and
equal to p). Actually x1/n1 is different from x2/n2



I'm not assuming that x1/n1 is equal to x2/n2. All I'm assuming is that x1 and x2 are respectively distributed as Bin(n1,p) and Bin(n2,p). p is an unknown parameter common to both distributions.

I adapted your notation of p1 and p2 for the estimates, which should properly be called p1hat and p2hat instead. That's what be causing you confusion. There should be a distinction in the notation between sample estimates and unknown parameters. In this more proper notation, we have

Cov(p1hat,p2hat) = p(1-p)/(n1+n2)

Jack
.