Property of probabilities to add up to 1 : conditional case I was wondering in the case of conditional distributions, where let's say we have $P(X|Y)$, if we fix $Y = a$, would the probabilities of $P(X|Y=a)$ for all values of X necessarily all add up to 1? If so then why?
Additionally, our professor also told us that if we had the joint distribution of $X$ and $Y$, then when we are trying to derive $f_{X|Y}(x|y)$ we are essentially fixing y, and looking at a slice of our pdf, and scaling that so that the integral adds up to 1. I did not quite understand this, and if someone could give me some more geometric intuition that would be super helpful. Thank you!
 A: The conditional distribution of $X$ given some event $A$ is still a probability distribution, so yes, the "probabilities must sum to $1$." For instance, if $X$ is a discrete random variable, then
$$\sum_x P(X=x \mid A) = 1$$
where the sum is over all values $X$ can take. Here, $A$ can be any event, such as $\{Y=a\}$ for some other random variable $Y$.

In the case where $(X,Y)$ has a joint density $f_{X,Y}(x,y)$, what your professor is saying is that the conditional density of $X$ given $Y=y_0$ looks kind of like the function $g(x) := f_{X,Y}(x,y_0)$ (this is the "slice of the joint PDF" where the second component is fixed at $y_0$). However, we do not have $\int_{-\infty}^\infty g(x) \, dx = 1$ necessarily since $\int_{-\infty}^\infty f_{X,Y}(x,y_0) \, dx = f_Y(y_0)$. But if we divide $g$ by this quantity (this is the "rescaling") we can make the integral equal to one. The conditional density ends up being this resulting function: $f_{X \mid Y=y_0}(x) = \frac{f_{X,Y}(x,y_0)}{f_Y(y_0)}$ which does satisfy $\int_{-\infty}^\infty f_{X \mid Y=y_0}(x) \, dx = 1$.
[Proving why this actually is the conditional density is a bit technical, but I guess your professor is just trying to give you intuition. It resembles the discrete case $P(X=x \mid Y=y_0) = \frac{P(X=x, Y=y_0)}{P(Y=y_0)}$.]
