Minimum conditions for defining joint PDF

In summary, the conversation discusses the conditions for changing the order of integration in a double integral involving a joint probability density function. It is commonly assumed that integrals with reversed order of integration are equivalent, but this is not always the case. A sufficient condition for changing the order of integration is that the probability density function is measurable, and that the integral is finite.
  • #1
winterfors
71
0
Suppose one knows a probability density [itex]p(x)[/itex] over a space [itex]X[/itex] (where [itex]x\in X[/itex]) and a conditional probability density [itex]p(y|x)[/itex] over a space [itex]Y[/itex] (where [itex]y\in Y[/itex]).

This implies the integral [itex]\int{p(x)dx}[/itex] is well defined as well as [itex]\int{p(y|x)dy}[/itex].

Defining a joint probability density

[tex]p(x,y)\ =\ p(y|x)p(x)[/tex] ,​

[itex]p(x)[/itex] will clearly be its marginal density over [itex]X[/itex], and the double integral

[tex]P(C)\ = \ \iint\limits_{(x,y)\in C}{p(x,y)dydx}[/tex]​

is well defined for all measurable subsets [itex]C\ \subseteq \ X\times Y[/itex].One commonly assumes that integrals with reversed order of integration are equivalent

[tex]P(C)\ = \ \iint\limits_{(x,y)\in C}{p(x,y)dxdy}\qquad ,[/tex]​

which also implies that the marginal probability density over [itex]Y[/itex] exists and is uniquely defined [itex]p(y)\ = \ \int{p(x,y)dx}[/itex].This is not necessarily the case, since the change in order of integration poses restrictions on the integrand [itex]p(x,y)[/itex]. One sufficient condition, I believe, is that [itex]p(x,y)[/itex] is continuous, but that is clearly not the minimal condition required. For instance, change of order of integration can be done for a [itex]p(x,y)[/itex] that involves step functions, which are obviously not continuous.

Does anyone know what the minimal conditions are for changing the order of integration, in this case?
 
Last edited:
Physics news on Phys.org
  • #2
If p>=0 then you just need to know that p is measurable. More generally, if p is measurable then
[tex]
\int\int |p(x,y)|dxdy=\int\int |p(x,y)| dydx
[/tex]
and, as long as this is finite then you can commute the integrals in [itex]\int\int p(x,y)dxdy[/itex]. See http://en.wikipedia.org/wiki/Fubini%27s_theorem" .
 
Last edited by a moderator:

Related to Minimum conditions for defining joint PDF

1. What are minimum conditions for defining a joint PDF?

The minimum conditions for defining a joint PDF are that it must be non-negative, its integral over the entire sample space must equal to 1, and it must be defined for all possible combinations of the random variables.

2. Why is it important to define minimum conditions for a joint PDF?

Defining minimum conditions for a joint PDF ensures that it is a valid probability distribution and allows for the calculation of probabilities for different combinations of random variables.

3. Can the minimum conditions for a joint PDF be relaxed?

No, the minimum conditions for a joint PDF cannot be relaxed as they are essential for it to represent a valid probability distribution.

4. How do the minimum conditions for a joint PDF differ from those for a single variable PDF?

The minimum conditions for a joint PDF are similar to those for a single variable PDF, except that a joint PDF must be defined for all possible combinations of random variables, whereas a single variable PDF only needs to be defined for a single random variable.

5. Are the minimum conditions for a joint PDF the same for all types of random variables?

Yes, the minimum conditions for a joint PDF are the same for all types of random variables, whether they are discrete or continuous.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
2K
  • Set Theory, Logic, Probability, Statistics
2
Replies
43
Views
4K
  • Precalculus Mathematics Homework Help
Replies
4
Views
622
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
957
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Classical Physics
Replies
0
Views
362
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
1K
Replies
3
Views
647
Back
Top