Suppose and are two measure spaces. Assume that is a probability measure on and for each , is a probability measure on . If the function

is -measurable for each , then

defines a probability measure on .

*Proof.* Since for all , it is immediate that . Moreover,

and

Let be a countable collection of disjoint measurable subsets of . Set . By the -additivity of for each and the dominated convergence theorem,

We say that the probability measure is a **mixture** of the ‘s with **mixing distribution** .

We now show how well-known distributions can be obtained from the mixture of other well-known distributions. Recall that a random variable is said to have exponential distribution with parameter if it has a probability density function (pdf)

A random variable is said to have a gamma distribution with parameters , if it has pdf

Lastly, a random variable has a Pareto distribution with parameters if it has pdf

We can obtain the Pareto distribution as a mixture of exponential distributions with gamma mixing measure. Let and . Then for any , we have by Fubini’s theorem that

If is a random variable with distribution , then differentiating with respect to , we obtain that has pdf

Hence, is Pareto with parameters and .

### Like this:

Like Loading...

*Related*