# Examples of mixtures

The notion of mixtures is discussed in this previous post. Many probability distributions useful for actuarial modeling are mixture distributions. The previous post touches on some examples – negative binomial distribution (a Poisson-Gamma mixture), Pareto distribution (an exponential-gamma mixture) and the normal-normal mixture. In this post we present additional examples. We discuss the following examples.

1. Poisson-Gamma mixture = Negative Binomial.
2. Normal-Normal mixture = Normal.
3. Exponential-Gamma mixture = Pareto.
4. Exponential-Inverse Gamma mixture = Pareto.
5. Gamma-Gamma mixture = Generalized Pareto.
6. Weibull-Exponential mixture = Loglogistic.
7. Gamma-Geometric mixture = Exponential.
8. Normal-Gamma mixture = Student t.

The first three examples are discussed in the previous post. We discuss the remaining examples in this post.

The Pareto Family

Examples 3 and 4 show that Pareto distributions are mixtures of exponential distributions with either gamma or inverse gamma mixing weights. In Example 3, $X \lvert \Theta$ is an exponential distribution with $\Theta$ being a rate parameter. When $\Theta$ follows a gamma distribution, the resulting mixture is a (Type I Lomax) Pareto distribution. In Example 4, $X \lvert \Theta$ is an exponential distribution with $\Theta$ being a scale parameter. When $\Theta$ follows an inverse gamma distribution, the resulting mixture is also a (Type I Lomax) Pareto distribution.

As a mixture, Example 5 is like Example 3, except that it is a gamma-gamma mixture resulting in a generalized Pareto distribution. Example 3 has been discussed in the previous post. We now discuss Example 4 and Example 5.

Example 4. Suppose that $X \lvert \Theta$ has an exponential distribution where $\Theta$ is a scale parameter.
Further suppose that the random parameter $\Theta$ follows an inverse gamma distribution with parameters $\alpha$ and $\beta$. Then the unconditional distribution for $X$ is a (Type I Lomax) Pareto distribution with shape parameter $\alpha$ and scale parameter $\beta$.

The following gives the cumulative distribution function (CDF) and survival function of the conditional random variable $X \lvert \Theta$.

$F(x \lvert \Theta)=1-e^{- x/\Theta}$

$S(x \lvert \Theta)=e^{- x/\Theta}$

The random parameter $\Theta$ follows an inverse gamma distribution with parameters $\alpha$ and $\beta$. The following is the pdf of $\Theta$:

$\displaystyle g(\theta)=\frac{1}{\Gamma(\alpha)} \ \biggl[\frac{\beta}{\theta}\biggr]^\alpha \ \frac{1}{\theta} \ e^{-\frac{\beta}{ \theta}} \ \ \ \ \ \theta>0$

We show that the unconditional survival function for $X$ is the survival function for the Pareto distribution with parameters $\alpha$ (shape parameter) and $\beta$ (scale parameter).

\displaystyle \begin{aligned} S(x)&=\int_0^\infty S(x \lvert \theta) \ g(\theta) \ d \theta \\&=\int_0^\infty e^{- x/\theta} \ \frac{1}{\Gamma(\alpha)} \ \biggl[\frac{\beta}{\theta}\biggr]^\alpha \ \frac{1}{\theta} \ e^{-\beta / \theta} \ d \theta \\&=\int_0^\infty \frac{1}{\Gamma(\alpha)} \ \biggl[\frac{\beta}{\theta}\biggr]^\alpha \ \frac{1}{\theta} \ e^{-(x+\beta) / \theta} \ d \theta \\&=\frac{\beta^\alpha}{(x+\beta)^\alpha} \ \int_0^\infty \frac{1}{\Gamma(\alpha)} \ \biggl[\frac{x+\beta}{\theta}\biggr]^\alpha \ \frac{1}{\theta} \ e^{-(x+\beta) / \theta} \ d \theta \\&=\biggl(\frac{\beta}{x+\beta} \biggr)^\alpha \end{aligned}

Note that the the integrand in the last integral is a density function for an inverse gamma distribution. Thus the integral is 1 and can be eliminated. The result that remains is the survival function for a Pareto distribution with parameters $\alpha$ and $\beta$. The following gives the CDF and density function of this Pareto distribution.

$\displaystyle F(x)=1-\biggl(\frac{\beta}{x+\beta} \biggr)^\alpha$

$\displaystyle f(x)=\frac{\alpha \ \beta^{\alpha}}{(x+\beta)^{\alpha+1}}$

See here for further information on Pareto Type I Lomax distribution.

Example 5. Suppose that $X \lvert \Theta$ has a gamma distribution with shape parameter $k$ (a known constant) and rate parameter $\Theta$. Further suppose that the random parameter $\Theta$ follows a gamma distribution with shape parameter $\alpha$ and rate parameter $\beta$. Then the unconditional distribution for $X$ is a generalized Pareto distribution with parameters $\alpha$, $\beta$ and $k$.

Conditional on $\Theta=\theta$, the following is the density function of $X$.

$\displaystyle f(x \lvert \theta)=\frac{1}{\Gamma(k)} \ \theta^k \ x^{k-1} \ e^{-\theta x} \ \ \ \ \ x>0$

The following is the density function of the random parameter $\Theta$.

$\displaystyle g(\theta)=\frac{1}{\Gamma(\alpha)} \ \beta^\alpha \ \theta^{\alpha-1} \ e^{-\beta \theta} \ \ \ \ \ \ \theta>0$

The following gives the unconditional density function for $X$.

\displaystyle \begin{aligned} f(x)&=\int_0^\infty f(x \lvert \theta) \ g(\theta) \ d \theta \\&=\int_0^\infty \frac{1}{\Gamma(k)} \ \theta^k \ x^{k-1} \ e^{-\theta x} \ \frac{1}{\Gamma(\alpha)} \ \beta^\alpha \ \theta^{\alpha-1} \ e^{-\beta \theta} \ d \theta \\&=\int_0^\infty \frac{1}{\Gamma(k)} \ \frac{1}{\Gamma(\alpha)} \ \beta^\alpha \ x^{k-1} \ \theta^{\alpha+k-1} \ e^{-(x+\beta) \theta} \ d \theta \\&= \frac{1}{\Gamma(k)} \ \frac{1}{\Gamma(\alpha)} \ \beta^\alpha \ x^{k-1} \frac{\Gamma(\alpha+k)}{(x+\beta)^{\alpha+k}} \int_0^\infty \frac{1}{\Gamma(\alpha+k)} \ (x+\beta)^{\alpha+k} \ \theta^{\alpha+k-1} \ e^{-(x+\beta) \theta} \ d \theta \\&=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha) \ \Gamma(k)} \ \frac{\beta^\alpha \ x^{k-1}}{(x+\beta)^{\alpha+k}} \end{aligned}

Any distribution that has a density function described above is said to be a generalized Pareto distribution with the parameters $\alpha$, $\beta$ and $k$. Its CDF cannot be written in closed form but can be expressed using the incomplete beta function.

\displaystyle \begin{aligned} F(x)&=\int_0^x \frac{\Gamma(\alpha+k)}{\Gamma(\alpha) \ \Gamma(k)} \ \frac{\beta^\alpha \ t^{k-1}}{(t+\beta)^{\alpha+k}} \ dt \\&=\int_0^x \frac{\Gamma(\alpha+k)}{\Gamma(\alpha) \ \Gamma(k)} \ \biggl(\frac{t}{t+\beta} \biggr)^{k-1} \ \biggl(\frac{\beta}{t+\beta} \biggr)^{\alpha-1} \ \frac{\beta}{(t+\beta)^2} \ dt \\&=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha) \ \Gamma(k)} \ \int_0^{\frac{x}{x+\beta}} u^{k-1} \ (1-u)^{\alpha-1} \ du, \ \ \ u=\frac{t}{t+\beta} \\&=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha) \ \Gamma(k)} \ \int_0^{w} t^{k-1} \ (1-t)^{\alpha-1} \ dt, \ \ \ w=\frac{x}{x+\beta} \end{aligned}

The moments can be easily derived for the generalized Pareto distribution but on a limited basis. Since it is a mixture distribution, the unconditional mean is the weighted average of the conditional means.

\displaystyle \begin{aligned} E(X^w)&=\int_0^\infty E(X \lvert \theta) \ g(\theta) \ d \theta \\&=\int_0^\infty \frac{\Gamma(k+w)}{\theta^w \Gamma(k)} \ \frac{1}{\Gamma(\alpha)} \ \beta^\alpha \ \theta^{\alpha-1} \ e^{-\beta \theta} \ d \theta \\&=\frac{\beta^w \ \Gamma(k+w) \ \Gamma(\alpha-w)}{\Gamma(k) \ \Gamma(\alpha)} \int_0^\infty \frac{1}{\Gamma(\alpha-w)} \ \beta^{\alpha-w} \ \theta^{\alpha-w-1} \ e^{-\beta \theta} \ d \theta \\&=\frac{\beta^w \ \Gamma(k+w) \ \Gamma(\alpha-w)}{\Gamma(k) \ \Gamma(\alpha)} \ \ \ \ -k

Note that $E(X)$ has a simple expression $E(X)=\frac{k \beta}{\alpha-1}$ when $1<\alpha$.

When the parameter $k=1$, the conditional distribution for $X \lvert \Theta$ is an exponential distribution. Then the situation reverts back to Example 3, leading to a Pareto distribution. Thus the Pareto distribution is a special case of the generalized Pareto distribution. Both the Pareto distribution and the generalized Pareto distribution have thicker and longer tails than the original conditional gamma distribution.

It turns out that the F distribution is also a special case of the generalized Pareto distribution. The F distribution with $r_1$ and $r_2$ degrees of freedom is the generalized Pareto distribution with parameters $k=r_1/2$, $\alpha=r_2/2$ and $\beta=r_2/r_1$. As a result, the following is the density function.

\displaystyle \begin{aligned} h(x)&=\frac{\Gamma(r_1/2 + r_2/2)}{\Gamma(r_1/2) \ \Gamma(r_2/2)} \ \frac{(r_2/r_1)^{r_2/2} \ x^{r_1/2-1}}{(x+r_2/r_1)^{r_1/2+r_2/2}} \\&=\frac{\Gamma(r_1/2 + r_2/2)}{\Gamma(r_1/2) \ \Gamma(r_2/2)} \ \frac{(r_1/r_2)^{r_1/2} \ x^{r_1/2-1}}{(1+(r_1/r_2)x)^{r_1/2+r_2/2}} \ \ \ \ 0

Another way to generate the F distribution is from taking a ratio of two chi-squared distributions (see Theorem 9 in this previous post). Of course, there is no need to use the explicit form of the density function of the F distribution. In a statistical application, the F distribution is accessed using tables or software.

The Loglogistic Distribution

The loglogistic distribution can be derived as a mixture of Weillbull distribution with exponential mixing weights.

Example 6. Suppose that $X \lvert \Lambda$ has a Weibull distribution with shape parameter $\gamma$ (a known constant) and a parameter $\Lambda$ such that the CDF of $X \lvert \Lambda$ is $F(x \lvert \Lambda)=1-e^{-\Lambda \ x^\gamma}$. Further suppose that the random parameter $\Lambda$ follows an exponential distribution with rate parameter $\theta^{\gamma}$. Then the unconditional distribution for $X$ is a loglogistic distribution with shape parameter $\gamma$ and scale parameter $\theta$.

The following gives the conditional survival function for $X \lvert \Lambda$ and the exponential mixing weight.

$\displaystyle S(x \lvert \lambda)=e^{-\lambda \ x^\gamma}$

$\displaystyle g(\lambda)=\theta^\gamma \ e^{-\theta^\gamma \ \lambda}$

The following gives the unconditional survival function and CDF of $X$ as well as the PDF.

\displaystyle \begin{aligned} S(x)&=\int_0^\infty S(x \lvert \lambda) \ g(\lambda) \ d \lambda \\&=\int_0^\infty e^{-\lambda \ x^\gamma} \ \theta^\gamma \ e^{-\theta^\gamma \ \lambda} \ d \lambda \\&=\int_0^\infty \theta^\gamma \ e^{-(x^\gamma+\theta^\gamma) \ \lambda} \ d \lambda \\&=\frac{\theta^\gamma}{(x^\gamma+\theta^\gamma)} \int_0^\infty (x^\gamma+\theta^\gamma) \ e^{-(x^\gamma+\theta^\gamma) \ \lambda} \ d \lambda \\&=\frac{\theta^\gamma}{x^\gamma+\theta^\gamma} \end{aligned}

\displaystyle \begin{aligned} F(x)&=1-S(x)=1-\frac{\theta^\gamma}{x^\gamma+\theta^\gamma} =\frac{x^\gamma}{x^\gamma+\theta^\gamma} =\frac{(x/\theta)^\gamma}{1+(x/\theta)^\gamma} \end{aligned}

$\displaystyle f(x)=\frac{d}{dx} \biggl( \frac{x^\gamma}{x^\gamma+\theta^\gamma} \biggr)=\frac{\gamma \ (x/\theta)^\gamma}{x [1+(x/\theta)^\gamma]^2}$

Any distribution that has any one of the above three distributional quantities is said to be a loglogistic distribution with shape parameter $\gamma$ and scale parameter $\theta$.

One interesting point about loglogistic distribution that an inverse loglogistic distribution is another loglogistic distribution. Suppose that $X$ has a loglogistic distribution with shape parameter $\gamma$ and scale parameter $\theta$. Let $Y=\frac{1}{X}$. Then $Y$ has a loglogistic distribution with shape parameter $\gamma$ and scale parameter $\theta^{-1}$.

\displaystyle \begin{aligned} P[Y \le y]&=P[\frac{1}{X} \le y] =P[X \ge y^{-1}] =\frac{\theta^\gamma}{y^{-\gamma}+\theta^\gamma} \\&=\frac{\theta^\gamma \ y^\gamma}{1+\theta^\gamma \ y^\gamma} \\&=\frac{y^\gamma}{(\theta^{-1})^\gamma+y^\gamma} \end{aligned}

The above is a survival function for the loglogistic distribution with the desired parameters. Thus there is no need to specially call out the inverse loglogistic distribution.

In order to find the mean and higher moments of the loglogistic distribution, we take the approach of identifying the conditional Weibull means and the weight these means by the exponential mixing weights. Note that the parameter $\Lambda$ in the conditional CDF $F(x \lvert \Lambda)=1-e^{-\Lambda \ x^\gamma}$ is not a scale parameter. The Weibull distribution in this conditional CDF is equivalent to a Weibull distribution with shape parameter $\gamma$ and scale parameter $\Lambda^{-1/\gamma}$. According to formula (4) in this previous post, the $k$th moment of this Weillbull distribution is

$\displaystyle E[ (X \lvert \Lambda)^k]=\Gamma \biggl(1+\frac{k}{\gamma} \biggr) \Lambda^{-k/\gamma}$

The following gives the unconditional $k$th moment of the Weibull-exponential mixure.

\displaystyle \begin{aligned} E[X^k]&=\int_0^\infty E[ (X \lvert \Lambda)^k] \ g(\lambda) \ d \lambda \\&=\int_0^\infty \Gamma \biggl(1+\frac{k}{\gamma} \biggr) \lambda^{-k/\gamma} \ \theta^\gamma \ e^{-\theta^\gamma \ \lambda} \ d \lambda\\&=\Gamma \biggl(1+\frac{k}{\gamma} \biggr) \ \theta^\gamma \int_0^\infty \lambda^{-k/\gamma} \ e^{-\theta^\gamma \ \lambda} \ d \lambda \\&=\theta^k \ \Gamma \biggl(1+\frac{k}{\gamma} \biggr) \int_0^\infty t^{-k/\gamma} \ e^{-t} \ dt \ \ \text{ where } t=\theta^\gamma \lambda \\&=\theta^k \ \Gamma \biggl(1+\frac{k}{\gamma} \biggr) \int_0^\infty t^{[(\gamma-k)/\gamma]-1} \ e^{-t} \ dt \\&=\theta^k \ \Gamma \biggl(1+\frac{k}{\gamma} \biggr) \ \Gamma \biggl(1-\frac{k}{\gamma} \biggr) \ \ \ \ -\gamma

The range $\gamma follows from the fact that the arguments of the gamma function must be positive. Thus the $k$th moments of the loglogistic distribution are limited by its shape parameter $\gamma$. If $\gamma=1$, then $E(X)$ does not exist. For a larger $\gamma$, more moments exist but always a finite number of moments. This is an indication that the loglogistic distribution has a thick (right) tail. This is not surprising since mixture distributions (loglogistic in this case) tend to have thicker tails than the conditional distributions (Weibull in this case). The thicker tail is a result of the uncertainty in the random parameter in the conditional distribution (the Weibull $\Lambda$ in this case).

Another Way to Obtain Exponential Distribution

We now consider Example 7. The following is a precise statement of the gamma-geometric mixture.

Example 7. Suppose that $X \lvert \alpha$ has a gamma distribution with shape parameter $\alpha$ that is a positive integer and rate parameter $\beta$ (a known constant). Further suppose that the random parameter $\alpha$ follows a geometric distribution with probability function $P[Y=\alpha]=p (1-p)^{\alpha-1}$ where $\alpha=1,2,3,\cdots$. Then the unconditional distribution for $X$ is an exponential distribution with rate parameter $\beta p$.

The conditional gamma distribution has an uncertain shape parameter $\alpha$ that can take on positive integers. The parameter $\alpha$ follows a geometric distribution. Here’s the ingredients that go into the mixture.

$\displaystyle f(x \lvert \alpha)=\frac{1}{(\alpha-1)!} \ \beta^\alpha \ x^{\alpha-1} \ e^{-\beta x}$

$P[Y=\alpha]=p (1-p)^{\alpha-1}$

The following is the unconditional probability density function of $X$.

\displaystyle \begin{aligned} f(x)&=\sum \limits_{\alpha=1}^\infty f(x \lvert \alpha) \ P[Y=\alpha] \\&=\sum \limits_{\alpha=1}^\infty \frac{1}{(\alpha-1)!} \ \beta^\alpha \ x^{\alpha-1} \ e^{-\beta x} \ p (1-p)^{\alpha-1} \\&=\beta p \ e^{-\beta x} \sum \limits_{\alpha=1}^\infty \frac{[\beta(1-p) x]^{\alpha-1}}{(\alpha-1)!} \\&=\beta p \ e^{-\beta x} \sum \limits_{\alpha=0}^\infty \frac{[\beta(1-p) x]^{\alpha}}{(\alpha)!} \\&=\beta p \ e^{-\beta x} \ e^{\beta(1-p) x} \end{aligned}

The above density function is that of an exponential distribution with rate parameter $\beta p$.

Student t Distribution

Example 3 (discussed in the previous post) involves a normal distribution with a random mean. Example 8 involves a normal distribution with mean 0 and an uncertain variance, which follows a gamma distribution such that the two gamma parameters are related to a common parameter $r$, which will be the degrees of freedom of the student t distribution. The following is a precise description of the normal-gamma mixture.

Example 8. Suppose that $X \lvert \Lambda$ has a normal distribution with mean 0 and variance $1/\Lambda$. Further suppose that the random parameter $\Lambda$ follows a gamma distribution with shape parameter $\alpha$ and scale parameter $\theta$ such that $2 \alpha=\frac{2}{\theta}=r$ is a positive integer. Then the unconditional distribution for $X$ is a student t distribution with $r$ degrees of freedom.

The following gives the ingredients of the normal-gamma mixture. The first item is the conditional density function of $X$ given $\Lambda$. The second is the density function of the mixing weight $\Lambda$.

$\displaystyle f(x \lvert \lambda)=\frac{1}{\sqrt{1/\lambda} \ \sqrt{2 \pi}} \ e^{-(\lambda/2) \ x^2}=\sqrt{\frac{\lambda}{2 \pi}} \ e^{-(\lambda/2) \ x^2}$

$\displaystyle g(\lambda)=\frac{1}{\Gamma(\alpha)} \biggl( \frac{1}{\theta} \biggr)^\alpha \ \lambda^{\alpha-1} \ e^{-\lambda/\theta}$

The following calculation derives the unconditional density function of $X$.

\displaystyle \begin{aligned} f(x)&=\int_{0}^\infty f(x \lvert \lambda) \ g(\lambda) \ d \lambda \\&=\int_{0}^\infty \sqrt{\frac{\lambda}{2 \pi}} \ e^{-(\lambda/2) \ x^2} \ \frac{1}{\Gamma(\alpha)} \biggl( \frac{1}{\theta} \biggr)^\alpha \ \lambda^{\alpha-1} \ e^{-\lambda/\theta} \ d \lambda \\&=\frac{1}{\Gamma(\alpha)} \ \biggl( \frac{1}{\theta} \biggr)^\alpha \ \frac{1}{\sqrt{2 \pi}} \int_0^\infty \lambda^{\alpha+\frac{1}{2}-1} e^{-(\frac{x^2}{2}+\frac{1}{\theta} ) \lambda} \ d \lambda \\&=\frac{\Gamma(\alpha+\frac{1}{2})}{\Gamma(\alpha)} \ \biggl( \frac{1}{\theta} \biggr)^\alpha \ \frac{1}{\sqrt{2 \pi}} \ \biggl(\frac{2 \theta}{\theta x^2+2} \biggr)^{\alpha+\frac{1}{2}} \\& \times \int_0^\infty \frac{1}{\Gamma(\alpha+\frac{1}{2})} \ \biggl(\frac{\theta x^2+2}{2 \theta} \biggr)^{\alpha+\frac{1}{2}} \lambda^{\alpha+\frac{1}{2}-1} e^{-\frac{\theta x^2+2}{2 \theta} \lambda} \ d \lambda \\&=\frac{\Gamma(\alpha+\frac{1}{2})}{\Gamma(\alpha)} \ \biggl( \frac{1}{\theta} \biggr)^\alpha \ \frac{1}{\sqrt{2 \pi}} \ \biggl(\frac{2 \theta}{\theta x^2+2} \biggr)^{\alpha+\frac{1}{2}} \ \ \ \ \ -\infty

The above density function is in terms of the two parameters $\alpha$ and $\theta$. In the assumptions, the two parameters are related to a common parameter $r$ such that $\alpha=\frac{r}{2}$ and $\theta=\frac{2}{r}$. The following derivation converts to the common $r$.

\displaystyle \begin{aligned} f(x)&=\frac{\Gamma(\frac{r}{2}+\frac{1}{2})}{\Gamma(\frac{r}{2})} \ \biggl( \frac{r}{2} \biggr)^{\frac{r}{2}} \ \frac{1}{\sqrt{2 \pi}} \ \biggl(\frac{2 \frac{2}{r}}{\frac{2}{r} x^2+2} \biggr)^{\frac{r}{2}+\frac{1}{2}} \\&=\frac{\Gamma(\frac{r}{2}+\frac{1}{2})}{\Gamma(\frac{r}{2})} \ \frac{r^{r/2}}{2^{r/2}} \ \frac{1}{2^{1/2} \sqrt{\pi}} \ \biggl(\frac{2/r}{x^2/r+1} \biggr)^{(r+1)/2} \\&=\frac{\Gamma \biggl(\displaystyle \frac{r+1}{2} \biggr)}{\Gamma \biggl(\displaystyle \frac{r}{2} \biggr)} \ \frac{1}{\sqrt{\pi r}} \ \frac{1 \ \ \ \ \ }{\biggl(1+\displaystyle \frac{x^2}{r} \biggr)^{(r+1)/2}} \ \ \ \ \ -\infty

The above density function is that of a student t distribution with $r$ degrees of freedom. Of course, in performing test of significance, the t distribution is accessed by using tables or software. A usual textbook definition of the student t distribution is the ratio of a normal distribution and a chi-squared distribution (see Theorem 6 in this previous post.

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$\copyright$ 2017 – Dan Ma