- #1
zr95
- 25
- 1
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.
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.