0
$\begingroup$

Suppose we have a discrete random variable $X$ and a continuous random variable $Y$. I am trying to understand how one defines/ find the joint PDF and joint CDF of $X$ and $Y$.

The joint CDF of $X$ and $Y$ is given by $$F_{X, Y}(x, y) = P(X \le x, Y \le y)$$

If both the random variables were discrete (continuous) then we could have found the joint PMF (joint PDF). But since $X$ and $Y$ are discrete and continuous, respectively, can we define "hybrid" joint PDF or "hybrid" joint PMF? For instance, can we find the "hybrid" joint PDF $f_{X, Y}(x, y)$ such that the marginal distributions of $X$ and $Y$ are given by

$$f_Y(y) = \sum_{x}f_{X, Y}(x, y)$$ and $$P(X = x) = \int_{- \infty} ^ {\infty}f_{X, Y}(x, y) \; \mathrm{d} y$$

Since I don't know measure theory yet, any answers which don't use references to measure theory will be really helpful.

$\endgroup$
  • 2
    $\begingroup$ This is what measure theory is for... $\endgroup$ – zoli May 10 '18 at 6:26
2
$\begingroup$

The answer is yes, and the PDF is exactly what you say it is. If you don't want to use measure theory, then you have to take what you say as the definition of the PDF in this setting. This is why everything in elementary probability has two versions, one for discrete and one for continuous. You can imagine many variants where discreteness and continuity mix themselves up in more complicated ways, and then you have to define the PDF for each.

The main reason that measure theory is interesting for elementary probability is that it turns all of these definitions into theorems. You just define one concept, of what it means for one measure to be "absolutely continuous" with respect to another measure, and then you get the correct definition of PDF in all cases.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.