I'm trying to understand how the rules of probability apply with probability density functions. If I denote the two continuous random variables as $X$ and $Y$ and their corresponding joint distribution as $f_{X,Y}(x,y)$ are the following derivations legit / true / correctly understood:
$$P(Y = y) = \lim_{\Delta y \rightarrow0} \int_{y}^{y+\Delta y}\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\,dy$$
$$P(X = x) = \lim_{\Delta x \rightarrow0} \int_{x}^{x+\Delta x}\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\,dx$$
$$P(Y=y, X=x) = \lim_{\Delta y, \Delta x \rightarrow 0}\int_{y}^{y+\Delta y}\int_{x}^{x+\Delta x}f_{X,Y}(x,y)\,dx\,dy$$
$$P(Y=y \;|\;X=x) = \frac{P(Y=y, X=x)}{P(X=x)} = \frac{\lim_{\Delta y, \Delta x \rightarrow 0}\int_{y}^{y+\Delta y}\int_{x}^{x+\Delta x}f_{X,Y}(x,y)\,dx\,dy}{\lim_{\Delta x \rightarrow0} \int_{x}^{x+\Delta x}\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\,dx}$$
The reason I ask this is because I try to understand why the rules of conditional probability (and therefore Bayes' theorem) hold also for density functions:
$$f_{X|Y}(x\mid y) = \frac{f_{Y|X}(y\mid x)f_X(x)}{f_Y(y)}$$
I asked a question related to this at this post: https://math.stackexchange.com/posts/592606/edit but I didn't get a satisfying answer yet..
Are my derivations correctly interpreted? Why does conditional probability rule also hold for density function? Or is it because:
is the same as
$$f_{X|Y}(x\mid y) = \frac{f_{Y|X}(y\mid x)f_X(x)}{f_Y(y)}$$
If we just cancel the $dx$:s and $dy$:s?? Hope everybody gets what I'm trying to ask x) In summary: If you look at what I've sketched and wrote, does it seem I've got the point? :) Or is there something I'm missing?
Thank you for any help :)
If $(X,Y)$ is a densitable couple of random variables, then the three displayed quantities are $0$ and the fourth is undefined, being $0/0$. The formula relating conditional and unconditional densities is correct though.
It sems obvious that when writing $P(Y=y)$, for example, what you mean is really the value at $y$ of the density function of $Y$, in which case one should replace the LHS by $f_Y(y)$ and at least divide the RHS by $Δy$ for the formula to make sense. What is written, when the density $f_Y$ exists, is just that $P(Y=y)=0$ is equal to the RHS, also equal to $0$. Note finally that you equate "being continuous" with "being densitable", although these conditions are not equivalent hence the limit of the RHS deivided by $Δy$ might not exist, even for continuous random variables $Y$.