relationship between correlation and average

#21
ok, i see what you mean. well i guess most statisticians aren't big on contract law so you have to let me catch up. In the original question discussing the combined population, there is the country itself which could be considered a "sources of happiness", so there's at least one. When talking about the two populations separately, this probably applies. Namely that we are talking about correlation in the marginal distribution already taken over any 'other factors'.

I was thinking one assumption that is implicit here is most likely that those with 0 wealth have 0 expected happiness. Well its not that insane is it? If you throw some multi-variate normality assumptions, or maybe something weaker, on the fire, and your basically there id think. Normality assumptions can't be insane, because then all statisticians are insane. Or maybe....
It ain't implicit where I live, in the land of the positive Y intercept. Sometimes known as the "b", as in y = mx + b. Mother Teresa.
 

Dason

Ambassador to the humans
#22
Also multivariate normality within the populations wouldn't get you there. It would need to be an even stronger assumption.
 
#24
Also multivariate normality within the populations wouldn't get you there
Yes staring at formulas looks like you'd need some equal variance type assumption as well i think, and then on top of that correlation is at least as large in one population or some combination thereof. Anyway we already agreed that mother Theresa forbids this. Thats too many assumptions so looks like Joeb3350 loses.

Probably.
 

spunky

Doesn't actually exist
#25
Also multivariate normality within the populations wouldn't get you there. It would need to be an even stronger assumption.
Ok, I'll bite now because this thread is starting to get interesting.

For a bivariate distribution (let's keep it simple), under which assumptions might one be able to claim that a positive correlation implies a larger mean in one marginal than the other?

The bivariate normal is obviously out because of the independence between the mean vector and the correlation matrix. I can't find any source of this but memory from back in the day is nudging me towards claiming that this property generalizes to elliptical distributions as well (i.e., the mean vector and corr matrix are independent). So elliptical distributions are also...out?

We need a bivariate distribution where the marginal means and the corr matrix are not independent. I'll need to think about this more carefully, but I'm willing to propose the bivariate Bernoulli as a candidate where one of the "mean vectors" (the p parameter) cannot equal p=0, 1 or 0.5.
 
#26
Yes staring at formulas looks like you'd need some equal variance type assumption as well i think, and then on top of that correlation is at least as large in one population or some combination thereof. Anyway we already agreed that mother Theresa forbids this. Thats too many assumptions so looks like Joeb3350 loses.
Probably.
God has spoken, no need for further discussion.
 
#27
We need a bivariate distribution where the marginal means and the corr matrix are not independent.
Thanks for joining the party.


Could wejust take bivariate normals, subject to the restriction that marginal means are proportional to marginal SD? To me it seems easiear than bivariate-bernoulli, ie 2x2 table, I considered that but just lead to a bunch of p's and q's on a napkin. I copy and paste a probably familiar wikipedia formula for reference:

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I think staring at this that taking sigmas proportional to their respective means is good, but seems like you still need info related to the LHS, E(X1 | X2 ), and of course we take rho > 0. For some reason I suspect this will be the case with 2 x 2 table as well, ie you will need info about the Probability that one is rich given not happy.

God has spoken, no need for further discussion.
Not unless god speaks through wikipedia.
 

spunky

Doesn't actually exist
#28
Could wejust take bivariate normals, subject to the restriction that marginal means are proportional to marginal SD? To me it seems easiear than bivariate-bernoulli, ie 2x2 table, I considered that but just lead to a bunch of p's and q's on a napkin. I copy and paste a probably familiar wikipedia formula for reference:
After reading your post it made realize that the problem, as I posted it, is ill-conceived and allows for a variety of trivial solutions. For instance, take two standard normal random variables and add a positive constant (say 2 for argument's sake). They are both positively correlated and one will have a population mean of 2, which satisfies the question. So a condition could simply be "take two independent normals and add a constant to one of them", which sort of makes this a lot less interesting than I initially thought it would be.

Meh, I guess one even needs to throw assumptions at the problem itself to make it worth spending time on it :(