I was reading through wikipedia about joint probability. I found out that for all random variables $$ f_{Y\mid X = x}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)} $$ However, nowhere was I able to find how to algrebraically evaluate $ f_{Y\mid X = x}(y) $ or $ f_{X,Y}(x,y) $ except in terms of eachothers.
I tried deducing $ f_{X,Y}(x,y) $ myself where $ Y $ is a linear combination of the randoms $X$ and $Z$. This is where I got.
$$ Y = X + Z $$
$$ f_{Y\mid X = x}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)} $$
$$ f_{X,Y}(x,y) = P(X = x \cap Y = y) $$ $$ f_{X,Y}(x,y) = P(X = x \cap X + Z = y) $$ $$ f_{X,Y}(x,y) = P(X = x \cap Z + x = y) $$ $$ f_{X,Y}(x,y) = P(X = x \cap Z = y - x) $$ $$ f_{X,Y}(x,y) = f_{X,Z}(x,y - x) $$ $$ f_{X,Y}(x,y) = f_X(x) f_Z(y - x) $$ $$ f_{Y\mid X = x}(y) = f_Z(y - x) $$
It seems to make sense to me, although I'm most uncertain about the substitution of $X$ for $x$. Is this allowed between the arguments of the union, or is there something that makes it different because $X = x$ and $Y = y$ are events rather than equations?
Thanks for your time.
Probability densities are not probability measures, so you cannot equate them. $$~f_{Y,X}(y,x)\not=P(Y=y\cap X=x)~$$
However, the intuition is reasonably correct and you can manipulate them in an analogous manner; which is not always quite straightforward:
We use a change of variable transformation; based on the chain rule of calculus.
When $z=g(y,x)$ this is:
$$\begin{align}f_{Y,X}(y,x) = f_{Z,X}(g(y,x),x)~{\Big\lvert\dfrac{\partial g(y,x)}{\partial y}\Big\rvert} \end{align}$$
In this case $g(y,x)=y-x$ so it is simply:
$$f_{Y,X}(y,x) ~=~ f_{Z,X}(y-x,x)$$
However when, for example, $Y=2Z+X$ , then we use $g(y,x)=\tfrac {y-x}2$ so
$$f_{Y,X}(y,x) ~=~ \tfrac 12 f_{Z,X}(\tfrac {y-x}2,x)$$
[This should make sense when you realise that if continuous random variable $Y$ is spread twice as far as $Z$ (for any $X$), then its density will be halved.]