Solved – Same Joint Distribution, different conditional and marginal distribution

The joint density uniquely determines the marginal densities:

$$p(x) = sum_y p(x,y) ~~text{or}~~ p(x) = int_{-infty}^{infty}p(x,y),mathrm dy$$ (similarly for $p(y)$) and so the conditional densities are also determined uniquely by the joint density. So the answer is

No, you cannot construct another group of samples as you desire.

If the joint density of a population is being estimated from the samples instead of being known a priori, then different samples from the population will typically give slightly different estimates, and so the estimated $p(x,y)$ will be slightly different for the two sets of samples, as will marginals etc. Again, your desire to have the same joint density and different marginal densities will not be satisfied.

What you can get is the same marginal densities but different joint densities. A simple example is two Bernoulli random variables $X$ and $Y$ each with parameter $frac{1}{2}$.

• If they are independent, $p(0,0)=p(0,1)=p(1,0)=p(1,1) = frac{1}{4}$.

• But if $X = 1-Y$, $~p(0,0)=p(1,1)= 0, ~p(1,0)=p(0,1) = frac{1}{2}$.

Consider also two standard normal random variables. If they are jointly normal, then $p(x,y)$ is the bivariate normal density. As a special case, if they are independent, then $p(x,y) = p(x)p(y)$. But they could be marginally normal random variables that are not jointly normal with joint density $$p(x,y) = begin{cases} 2p(x)p(y), & text{if}~ x geq 0, y geq 0, text{or}~ x < 0, y < 0,\ 0,& text{otherwise}.end{cases}$$ Note that the random variables are positively correlated in this case since the probability mass lies entirely in the first and third quadrants. This again illustrates that the same marginal densities can give rise to different joint densities.

What you want, the same joint density but different marginal densities, is, alas, not possible.