# An example of a mixture

We use an example to motivate the definition of a mixture distribution.

Example 1

Suppose that the loss arising from an insured randomly selected from a large group of insureds follow an exponential distribution with probability density function (pdf) $f_X(x)=\theta e^{-\theta x}$, $x>0$, where $\theta$ is a parameter that is a positive constant. The mean claim cost for this randomly selected insured is $\frac{1}{\theta}$. So the parameter $\theta$ reflects the risk characteristics of the insured. Since the population of insureds is large, there is uncertainty in the parameter $\theta$. It is more appropriate to regard $\theta$ as a random variable in order to capture the wide range of risk characteristics across the individuals in the population. As a result, the pdf indicated above is not an unconditional pdf, but, rather, a conditional pdf of $X$. The below pdf is conditional on a realized value of the random variable $\Theta$.

$\displaystyle f_{X \lvert \Theta}(x \lvert \theta)=\theta e^{-\theta x}, \ \ \ \ \ x>0$

What about the marginal (unconditional) pdf of $X$? Let’s assume that the pdf of $\Theta$ is given by $\displaystyle f_\Theta(\theta)=\frac{1}{2} \ \theta^2 \ e^{-\theta}$. Then the unconditional pdf of $X$ is the weighted average of the conditional pdf.

\displaystyle \begin{aligned}f_X(x)&=\int_0^{\infty} f_{X \lvert \Theta}(x \lvert \theta) \ f_\Theta(\theta) \ d \theta \\&=\int_0^{\infty} \biggl[\theta \ e^{-\theta x}\biggr] \ \biggl[\frac{1}{2} \ \theta^2 \ e^{-\theta}\biggr] \ d \theta \\&=\int_0^{\infty} \frac{1}{2} \ \theta^3 \ e^{-\theta(x+1)} \ d \theta \\&=\frac{1}{2} \frac{6}{(x+1)^4} \int_0^{\infty} \frac{(x+1)^4}{3!} \ \theta^{4-1} \ e^{-\theta(x+1)} \ d \theta \\&=\frac{3}{(x+1)^4} \end{aligned}

Several other distributional quantities are also weighted averages, which include the unconditional mean, and the second moment.

\displaystyle \begin{aligned}E(X)&=\int_0^{\infty} E(X \lvert \Theta=\theta) \ f_\Theta(\theta) \ d \theta \\&=\int_0^{\infty} \biggl[\frac{1}{\theta} \biggr] \ \biggl[\frac{1}{2} \ \theta^2 \ e^{-\theta}\biggr] \ d \theta \\&=\int_0^{\infty} \frac{1}{2} \ \theta \ e^{-\theta} \ d \theta \\&=\frac{1}{2} \end{aligned}

\displaystyle \begin{aligned}E(X^2)&=\int_0^{\infty} E(X^2 \lvert \Theta=\theta) \ f_\Theta(\theta) \ d \theta \\&=\int_0^{\infty} \biggl[\frac{2}{\theta^2} \biggr] \ \biggl[\frac{1}{2} \ \theta^2 \ e^{-\theta}\biggr] \ d \theta \\&=\int_0^{\infty} e^{-\theta} \ d \theta \\&=1 \end{aligned}

As a result, the unconditional variance is $Var(X)=1-\frac{1}{4}=\frac{3}{4}$. Note that the unconditional variance is not the weighted average of the conditional variance. The weighted average of the conditional variance only produces $\frac{1}{2}$.

\displaystyle \begin{aligned}E[Var(X \lvert \Theta)]&=\int_0^{\infty} Var(X \lvert \Theta=\theta) \ f_\Theta(\theta) \ d \theta \\&=\int_0^{\infty} \biggl[\frac{1}{\theta^2} \biggr] \ \biggl[\frac{1}{2} \ \theta^2 \ e^{-\theta}\biggr] \ d \theta \\&=\int_0^{\infty} \frac{1}{2} \ e^{-\theta} \ d \theta \\&=\frac{1}{2} \end{aligned}

It turns out that the unconditional variance has two components, the expected value of the conditional variances and the variance of the conditional means. In this example, the former is $\frac{1}{2}$ and the latter is $\frac{1}{4}$. The additional variance in the amount of $\frac{1}{4}$ is a reflection that there is uncertainty in the parameter $\theta$.

\displaystyle \begin{aligned}Var(X)&=E[Var(X \lvert \Theta)]+Var[E(X \lvert \Theta)] \\&=\frac{1}{2}+\frac{1}{4}\\&=\frac{3}{4} \end{aligned}

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The Definition of Mixture
The unconditional pdf $f_X(x)$ derived in Example 1 is that of a Pareto distribution. Thus the Pareto distribution is a continuous mixture of exponential distributions with Gamma mixing weights.

Mathematically speaking, a mixture arises when a probability density function $f(x \lvert \theta)$ depends on a parameter $\theta$ that is uncertain and is itself a random variable with density $g(\theta)$. Then taking the weighted average of $f(x \lvert \theta)$ with $g(\theta)$ as weight produces the mixture distribution.

A continuous random variable $X$ is said to be a mixture if its probability density function $f_X(x)$ is a weighted average of a family of probability density functions $f(x \lvert \theta)$. The random variable $\Theta$ is said to be the mixing random variable and its pdf $g(\theta)$ is said to be the mixing weight. An equivalent definition of mixture is that the distribution function $F_X(x)$ is a weighted average of a family of distribution functions indexed by a mixing variable. Thus $X$ is a mixture if one of the following holds.

$\displaystyle f_X(x)=\int_{-\infty}^{\infty} f(x \lvert \theta) \ g(\theta) \ d \theta$

$\displaystyle F_X(x)=\int_{-\infty}^{\infty} F(x \lvert \theta) \ g(\theta) \ d \theta$

Similarly, a discrete random variable is a mixture if its probability function (or distribution function) is a weighted sum of a family of probability functions (or distribution functions). Thus $X$ is a mixture if one of the following holds.

$\displaystyle P(X=x)=\sum \limits_{y} P(X=x \lvert Y=y) \ P(Y=y)$

$\displaystyle P(X \le x)=\sum \limits_{y} P(X \le x \lvert Y=y) \ P(Y=y)$