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Is it the case that for random variables $Y$ and $X$, $f(Y | X)$ in the same family as $f(Y)$?

If so how can I prove it? If not are there any situations (some families of distributions) where they can be the same? Or is it related to the link function(s)?

Thanks

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There are some specific cases where it is true, such as a bivariate normal, then $f(Y)$ and $f(Y|X)$ are both normal.

But consider the case where $f(Y|X)$ is normal (with mean depending on $X$) and $X$ follows a uniform distribution. Then $f(Y)$ is not normally distributed.

There are also distributions where the marginals ($f(X)$ and $f(Y)$) are both $\text{uniform}(0,1)$, but there is a hole in the square where the probability is $0$, so the conditional $f(Y|X)$ would not be uniform and for some values of $X$ would be disjoint.

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    $\begingroup$ +1, this is the right answer. For an illustration (eg, pictures) of how $f(Y)$ & $f(Y|X)$ can differ, there are some at my answer here: What if residuals are normally distributed but Y is not? $\endgroup$ Nov 7, 2013 at 17:20
  • $\begingroup$ What about members of exponential family? Or distributions with -inf->inf support? Can we relate $f(Y), f(X), f(Y|X)$ in some way? $\endgroup$ Nov 7, 2013 at 17:43
  • $\begingroup$ @CagdasOzgenc, I have seen a bivariate distribution where both marginals are normal, but there is a hole of 0 probability in the center (where the bivariate normal would have the highest probability), so for that distribution f(Y|X) is certainly not normal (may be approximate for some values of X, but clearly not when passing through the hole). $\endgroup$
    – Greg Snow
    Nov 11, 2013 at 15:20
  • $\begingroup$ @Greg, what if the distributions are nice? Meaning they are smooth, no discontinuities, support from -inf to +inf. $\endgroup$ Nov 11, 2013 at 19:17
  • $\begingroup$ Being 'nice' doesn't solve the problem; You can have both margins and the conditionals all 'nice' ... but still not of the same form, even with normal margins. $\endgroup$
    – Glen_b
    Nov 14, 2013 at 11:23

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