I Discrete Random Vectors vs. Continuous Random Vectors

[zr95]

Given a continuous random vector (X,Y) with a joint density function
In order to check whether it is indeed a joint density ƒ(x,y) the method is to check if ∫∫ƒ(x,y)dxdy=1 where the integrals limits follow the bounds of x and y.

However, is it the case that if given an arbitrary discrete random vector (X,Y) with a joint density function:
In order to check whether it is indeed a joint density ƒ(x,y) the method is to check if ∑∑ƒ(x,y)=1 where the summations are, given bounds, for all those x and y.

Or are they both the ∫∫ f(x,y)dxdy=1? I know further on for finding the expectation and such that discrete uses the summation and continuous takes the integral but not sure if it differs in the first step or why it does.

 

[mathman]

The sum can be looked at as an integral using delta functions. Is that what you are aiming for?

 

[chiro]

Technically the measure changes and the requirements for axiomatic probability are quite precise - and often difficult to follow.

If you are interested in this then you can find it in a graduate (or "pure") probability course that involves measure theory, sigma algebras, and the use of set theory to make probability rigorous over a variety of spaces (which as it turns out - is quite difficult).

A course on financial calculus (as a graduate offering) will cover this in one form or another.