0

Given two random variables, both continuous, $X$ and $Y$, we know that $P(X=x)=P(Y=y)=0$. Is it an abuse of notation or is it correct to have the following?

$$P(X<u | Y=y) > 0$$

In particular, can we claim the following?

$$P(X<u | Y=y) = \frac{P(X<u, Y=y)}{P(Y=y)}$$

We have \begin{align} F_X(u) = P(X< u) &= \int_{y=-\infty}^{\infty} \quad \int_{x=-\infty}^u f(x,y) dx \quad dy \\ &= \int_{y=-\infty}^{\infty} \quad \frac{ P( X < u, y < Y < y+dy)}{dy} \quad dy \\ &= \int_{y=-\infty}^{\infty} \quad { P( X < u| y < Y < y+dy)} f_Y(y)dy{} \\ \end{align}

In the expression above, it looks like we can claim the following

$$P(X<u | Y=y) = P(X<u | y < Y < y +dy)$$

however, in the left hand side we do not have $dy$ and in the right hand side we do?

What about the following?

$$P(X<u | Y=y) = \frac{P(X<u, y < Y < y+dy)}{P(y < Y < y+dy)} =\frac{\int_{x=-\infty}^u f_{X,Y}(x,y)dxdy}{f_Y(y) dy} $$

Is the best to say

$$P(X<u | Y=y) = \lim_{\Delta y \rightarrow 0} \frac{P(X<u, y < Y < y+ \Delta y)}{P(y < Y < y+ \Delta y)} $$

or

$$P(X<u | Y=y) = \lim_{\Delta y \rightarrow dy} \frac{P(X<u, y < Y < y+ \Delta y)}{P(y < Y < y+ \Delta y)} $$

?

Here is a related question:

How to formalize "conditional random variables"

Daniel S.
  • 823
  • Are they discrete or continuous random variables? It looks like you are taking them to be discrete so that $P({X=x})$ and $P({Y=y})$ are non-zero. – UserA Jul 30 '20 at 18:17
  • all continuous ... updating accordingly – Daniel S. Jul 30 '20 at 18:17
  • 1
    It has to do with a notion of conditional expectation. We define then conditional probability in a way $\mathbb P(X \in A | \mathcal G) = \mathbb E[1_A | \mathcal G]$ (it's a random variable), where $\mathcal G$ is some $\sigma-$field. Now it can be shown that for $\mathbb E[X|Y] := \mathbb E[X|\sigma(Y)]$ there exists $\phi$ borel such thath $\mathbb E[X|Y] = \phi(Y)$. Then, you would get some $\phi$ such that $\mathbb P(X \le u | Y) = \phi(Y)$. And one can then understand $" \mathbb P(X \le u | Y=y)"$ to be the value $\phi(y)$ – Presage Jul 30 '20 at 18:18
  • @DanielS. Are you familiar with basic notions from probability theory such as sample spaces and $\sigma$-algebras? – UserA Jul 30 '20 at 18:23
  • no, not really ... just the very basics – Daniel S. Jul 30 '20 at 18:25
  • 1
    In this case try the following https://www.statlect.com/fundamentals-of-probability/conditional-probability-as-a-random-variable . It is fairly simple and provides you with a remedy for defining $P(A|B)$ when $P(B)=0$. – UserA Jul 30 '20 at 18:28

0 Answers0