Probability Mixtures and the Pareto Distribution

Suppose (\Omega, \mathcal{F}) and (\Theta,\mathcal{G}) are two measure spaces. Assume that \nu is a probability measure on \mathcal{G} and for each \theta \in \Theta, \mathbb{P}_{\theta} is a probability measure on \mathcal{F}. If the function

\displaystyle\Omega \rightarrow [0,1], \ \theta \mapsto \mathbb{P}_{\theta}(A), \indent \forall A \in \mathcal{F}

is \mathcal{G}-measurable for each A \in \mathcal{F}, then

\mu(A) := \displaystyle\int_{\Theta}\mathbb{P}_{\theta}(A)\nu(d\theta), \indent A \in \mathcal{F}

defines a probability measure \mu on (\Omega,\mathcal{F}).

Proof. Since 0 \leq \mathbb{P}_{\theta} \leq 1 for all \theta \in \Theta, it is immediate that 0 \leq \mu \leq 1. Moreover,

\displaystyle\mu(\emptyset)=\int_{\Theta}\mathbb{P}_{\theta}(\emptyset)\nu(d\theta)=\int_{\Theta}0\nu(d\theta) = 0


\displaystyle\mu(\Omega)=\int_{\Theta}\mathbb{P}_{\theta}(\Omega)\nu(d\theta)=\int_{\Theta}1\nu(d\theta)=\nu(\Theta) = 1

Let \left\{A_{k}\right\}_{k=1}^{\infty} be a countable collection of disjoint measurable subsets of \mathcal{F}. Set A := \bigcup_{k=1}^{\infty}. By the \sigma-additivity of \mathbb{P}_{\theta} for each \theta \in \Theta and the dominated convergence theorem,

\begin{array}{lcl}\mu(A) = \displaystyle\int_{\Theta}\mathbb{P}_{\theta}(A)\nu(d\theta)=\int_{\Theta}\sum_{k=1}^{\infty}\mathbb{P}_{\theta}(A_{k})\nu(d\theta)&=&\displaystyle\sum_{k=1}^{\infty}\int_{\Theta}\mathbb{P}_{\theta}(A_{k})\nu(d\theta)\\&=&\displaystyle\sum_{k=1}^{\infty}\mu(A_{k})\end{array}


We say that the probability measure \mu is a mixture of the \mathbb{P}_{\theta}‘s with mixing distribution \nu.

We now show how well-known distributions can be obtained from the mixture of other well-known distributions. Recall that a random variable X is said to have exponential distribution with parameter \lambda >0 if it has a probability density function (pdf)

\displaystyle f_{X}(x)=\begin{cases} \lambda e^{-\lambda x} & {x \geq 0}\\ 0 & {x < 0} \end{cases}

A random variable X is said to have a gamma distribution with parameters \alpha > 0, \beta > 0, if it has pdf

\displaystyle f_{X}(x)=\begin{cases}\frac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-\frac{x}{\beta}} & {x \geq 0} \\ 0 & {x < 0}\end{cases}

Lastly, a random variable X has a Pareto distribution with parameters \alpha > 0, \gamma > 0 if it has pdf

\displaystyle f_{X}(x)=\begin{cases}\frac{\alpha \gamma^{\alpha}}{(x+\gamma)^{\alpha+1}} & {x\geq 0} \\ 0 & {x < 0}\end{cases}

We can obtain the Pareto distribution as a mixture of exponential distributions with gamma mixing measure. Let X \sim \text{Exp}(\lambda) and \Lambda \sim \text{Gamma}(\alpha,\beta). Then for any x \geq 0, we have by Fubini’s theorem that

\begin{array}{lcl}\mu([0,x])= \displaystyle\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_{0}^{\infty}\left(\int_{0}^{x}\lambda e^{-\lambda t}dt\right)\lambda^{\alpha-1}e^{-\frac{\lambda}{\beta}}d\lambda &=&\displaystyle\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_{0}^{\infty}\left(1-e^{-\lambda x}\right)\lambda^{\alpha-1}e^{-\frac{\lambda}{\beta}}d\lambda\\&=&\displaystyle 1 \displaystyle-\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}e^{-x \beta y}y^{\alpha-1}e^{-y}dy\\&=&1-\displaystyle\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}y^{\alpha-1}e^{-(x\beta+1)y}dy\\&=&1 -\displaystyle\frac{1}{(x\beta+1)^{\alpha}\Gamma(\alpha)}\int_{0}^{\infty}z^{\alpha-1}e^{-z}dz\\&=&1-\displaystyle\frac{1}{(x\beta+1)^{\alpha}}\end{array}

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

\displaystyle f_{Y}(x)=\frac{\alpha\beta}{(x\beta+1)^{\alpha+1}}=\frac{\alpha\beta^{-\alpha}}{(x+\beta^{-1})^{\alpha+1}}

Hence, Y is Pareto with parameters \beta^{-1} and \alpha.

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