# Minimum conditions for defining joint PDF

1. Feb 23, 2008

### winterfors

Suppose one knows a probability density $p(x)$ over a space $X$ (where $x\in X$) and a conditional probability density $p(y|x)$ over a space $Y$ (where $y\in Y$).

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

Defining a joint probability density

$$p(x,y)\ =\ p(y|x)p(x)$$ ,​

$p(x)$ will clearly be its marginal density over $X$, and the double integral

$$P(C)\ = \ \iint\limits_{(x,y)\in C}{p(x,y)dydx}$$​

is well defined for all measurable subsets $C\ \subseteq \ X\times Y$.

One commonly assumes that integrals with reversed order of integration are equivalent

$$P(C)\ = \ \iint\limits_{(x,y)\in C}{p(x,y)dxdy}\qquad ,$$ ​

which also implies that the marginal probability density over $Y$ exists and is uniquely defined $p(y)\ = \ \int{p(x,y)dx}$.

This is not necessarily the case, since the change in order of integration poses restrictions on the integrand $p(x,y)$. One sufficient condition, I believe, is that $p(x,y)$ is continuous, but that is clearly not the minimal condition required. For instance, change of order of integration can be done for a $p(x,y)$ 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: Feb 23, 2008
2. Feb 24, 2008

### gel

If p>=0 then you just need to know that p is measurable. More generally, if p is measurable then
$$\int\int |p(x,y)|dxdy=\int\int |p(x,y)| dydx$$
and, as long as this is finite then you can commute the integrals in $\int\int p(x,y)dxdy$. See http://en.wikipedia.org/wiki/Fubini%27s_theorem" [Broken].

Last edited by a moderator: May 3, 2017