Is $f(Y | X)$ in the same family as $f(Y)$?

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

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.
• +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? Nov 7, 2013 at 17:20
• 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? Nov 7, 2013 at 17:43