How to find conditional distributions from joint I want to learn about how to do Gibbs sampling, starting with finding conditional distributions given a joint distribution.  While looking for examples, I found this blog post that I wanted to replicate on my own, but I am having trouble understanding how to algebraically find the conditional distributions given the joint distribution.
The joint distribution is 
$$ f(x, y) = kx^2 exp(-xy^2 - y^2 + 2y - 4x) $$ where $k$ is a normalization constant.  The marginal distributions are
$$x \mid y \sim Gamma(3, y^2 + 4) $$
and
$$y \mid x \sim N(\frac{1}{x + 1}, \frac{1}{2(x + 1)}) $$
What are the steps to derive the two marginals from the joint?  Is there a trick I missing?  I feel like since it was unexplained in the example it should be somewhat simple but I am just not seeing it...
Thanks!  Also, bonus points if anyone can point me to any (free to access) examples with solutions similar to this.  I want to practice deriving conditional distributions from joint ones.
 A: The "trick" is to observe that $f(x\mid y)=f(x,y)/f(y)$ is proportional to $f(x,y)$ up to terms that do not involve $x$. Hence, $f(x\mid y)\propto x^2\exp(-(y^2+4)x)$, and this is the "kernel" of a $\mathrm{Gamma}(3,y^2+4)$ density. The other full conditional $f(y\mid x)$ is obtained similarly after completing the square in the exponent.
A: Those distributions you call "marginal" are not marginal.  They are conditional distributions because you wrote $x \mid y$.  The marginal distribution of $X$, for example, is necessarily independent of the value of $Y$.
To see how the conditional distribution is gamma, all you have to do is write $$f_{X \mid Y}(x) = \frac{f_{X,Y}(x,y)}{f_Y(y)} \propto f_{X,Y}(x,y).$$  That is to say, the conditional distribution is proportional to the joint distribution, appropriately normalized.  So we have $$f_{X \mid Y}(x) \propto x^2 e^{-x(y^2+4)},$$ completely ignoring any factors that are not functions of $x$.  Then we recognize that the gamma distribution has density $$f_S(s) \propto s^{a-1} e^{-bs},$$ so the choice of shape $a = 3$ and rate $b = y^2+4$ demonstrates that the conditional distribution $X \mid Y \sim \operatorname{Gamma}(a = 3, b = y^2+4)$.
The conditional distribution of $Y \mid X$ is done in a similar fashion.  Just ignore constants of proportionality:  $$f_{Y \mid X}(y) \propto e^{-(x+1)y^2+2y},$$ but this one requires us to complete the square to get it to look like a normal density:  $$-(x+1)y^2 + 2y = (x+1)\left(-\left(y - \tfrac{1}{x+1}\right)^2 \right) + \tfrac{1}{x+1},$$ and after exponentiating and removing the $e^{1/(x+1)}$ factor, comparing this against $$f_W(w) \propto e^{-(w-\mu)^2/(2\sigma^2)},$$ we see that we have a normal density with mean $\mu = 1/(x+1)$ and variance $\sigma^2 = 1/(2(x+1))$.
Now, if you wanted the marginal distributions, you would need to integrate:  $$f_X(x) = \int_{y=-\infty}^\infty f_{X,Y}(x,y) \, dy,$$ for example.  And as you can see, this expression will not be a function of $Y$.  The difference is that if I simulated realizations of ordered pairs $(X_i, Y_i)$ from the joint distribution, the marginal density for $X$ would be what you would see if I only told you the values of $X_i$.  The conditional distribution of $X$ given $Y = y$ would be what you would see if I only gave you the $X_i$ for which the corresponding $Y_i$ was equal to $y$.
