Gamma-Mixed Exponential Rates and Lomax Laws

Introduction

An exponential waiting time has a rapidly decreasing tail when its rate is fixed. But in many models the rate varies across people, locations, or experimental settings. Integrating over that variation changes the distribution of waiting times: a gamma-distributed rate leads to a Lomax distribution, whose tail decreases polynomially rather than exponentially. That change matters because the resulting distribution may have a finite mean but an infinite variance, or may not have a finite mean at all.

We will derive the mixed density and survival function directly, then find the exact condition for a moment to exist. Throughout, the gamma distribution is parameterized by shape $\alpha>0$ and rate $\beta>0$; thus its density contains $e^{-\beta\lambda}$, not $e^{-\lambda/\beta}$.

Definition: Let $\Lambda\sim\operatorname{Gamma}(\alpha,\beta)$, with density $$ g(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0, $$ where $\Gamma(a)=\int_0^\infty u^{a-1}e^{-u}\,du$ for $a>0$. Conditional on $\Lambda=\lambda$, let $X$ have an exponential distribution with rate $\lambda$, so $$ f_{X\mid\Lambda}(x\mid\lambda)=\lambda e^{-\lambda x},\qquad x\geq 0. $$ The unconditional law of $X$ is the gamma-mixed exponential-rate distribution.

Deriving the Lomax Density and Survival Function

To obtain the unconditional density, average the conditional density over the distribution of the rate. For $x\geq0$, the integrand is nonnegative, so the mixture density is the integral of the conditional density against the gamma density:

\begin{aligned} f_X(x) &=\int_0^\infty \lambda e^{-\lambda x} \frac{\beta^\alpha}{\Gamma(\alpha)} \lambda^{\alpha-1}e^{-\beta\lambda}\,d\lambda\\ &=\frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty \lambda^\alpha e^{-(\beta+x)\lambda}\,d\lambda\\ &=\frac{\beta^\alpha}{\Gamma(\alpha)} \frac{\Gamma(\alpha+1)}{(\beta+x)^{\alpha+1}}\\ &=\frac{\alpha\beta^\alpha}{(\beta+x)^{\alpha+1}}. \end{aligned}

The third line follows by substituting $u=(\beta+x)\lambda$ into the defining gamma integral. The last line uses $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$, which follows by integration by parts. The density is zero for $x<0$.

Definition: The Lomax distribution with shape $\alpha>0$ and scale $\beta>0$ has density $$ f_X(x)=\frac{\alpha}{\beta}\left(1+\frac{x}{\beta}\right)^{-(\alpha+1)},\qquad x\geq0, $$ and density zero for $x<0$. The gamma-mixed exponential-rate model above has this distribution, with the same $\alpha$ and $\beta$. Here $\beta$ is the scale of the Lomax law and the rate parameter of the gamma mixing distribution.

The survival function gives the probability that the waiting time exceeds $x$. Conditional on $\Lambda=\lambda$, this probability is $e^{-\lambda x}$. Averaging over the rate gives

\begin{aligned} S_X(x)=P(X>x) &=\int_0^\infty e^{-\lambda x} \frac{\beta^\alpha}{\Gamma(\alpha)} \lambda^{\alpha-1}e^{-\beta\lambda}\,d\lambda\\ &=\frac{\beta^\alpha}{\Gamma(\alpha)} \frac{\Gamma(\alpha)}{(\beta+x)^\alpha}\\ &=\left(\frac{\beta}{\beta+x}\right)^\alpha,\qquad x\geq0. \end{aligned}

In particular, $S_X(0)=1$, and differentiating $1-S_X(x)$ for $x>0$ gives $\alpha\beta^\alpha(\beta+x)^{-(\alpha+1)}$, the density just derived. This also verifies the connection between the density and survival function. The distribution function is $F_X(x)=1-S_X(x)$ for $x\geq0$, and $F_X(x)=0$ for $x<0$.

Worked Example: Shape 2 and Scale 3

Suppose $\Lambda\sim\operatorname{Gamma}(2,3)$ in the shape-rate convention and $X\mid\Lambda=\lambda\sim\operatorname{Exponential}(\lambda)$. Substituting $\alpha=2$ and $\beta=3$ into the derived expressions gives

f_X(x)=\frac{2\cdot3^2}{(3+x)^3}=\frac{18}{(3+x)^3}, \qquad S_X(x)=\left(\frac{3}{3+x}\right)^2,\qquad x\geq0.

For example, the probability of exceeding 3 time units is $S_X(3)=(3/6)^2=1/4$. The density integrates to one: an antiderivative of $18(3+x)^{-3}$ is $-9(3+x)^{-2}$, so

\int_0^\infty \frac{18}{(3+x)^3}\,dx =\lim_{b\to\infty}\left[-\frac{9}{(3+x)^2}\right]_{0}^{b} =0-\left(-1\right)=1.

This parameter choice has a finite mean, but its second moment does not exist: the shape is 2, and the moment condition proved below requires the moment order to be strictly less than the shape.

Which Moments Exist?

For $r\geq0$, the $r$th raw moment means $E[X^r]$, allowing the value $+\infty$. The exact condition for finiteness is that the moment order be strictly smaller than the Lomax shape.

Theorem: If $X$ has the Lomax density with shape $\alpha>0$ and scale $\beta>0$, then for $r\geq0$, $$ E[X^r]<\infty\quad\text{if and only if}\quad r<\alpha. $$ When $0\leq r<\alpha$, $$ E[X^r] =\beta^r\frac{\Gamma(r+1)\Gamma(\alpha-r)}{\Gamma(\alpha)}. $$

Proof. Starting from the density, substitute $t=x/(\beta+x)$, so $x=\beta t/(1-t)$ and $dx=\beta(1-t)^{-2}\,dt$. As $x$ runs from 0 to infinity, $t$ runs from 0 to 1. Therefore

\begin{aligned} E[X^r] &=\int_0^\infty x^r\frac{\alpha\beta^\alpha}{(\beta+x)^{\alpha+1}}\,dx\\ &=\alpha\beta^r\int_0^1 t^r(1-t)^{\alpha-r-1}\,dt. \end{aligned}

For $r\geq0$, the factor $t^r$ is bounded near $t=1$ and tends to 1 there. Thus the integral near 1 is finite exactly when the exponent $\alpha-r-1$ is greater than $-1$, or equivalently $r<\alpha$. If $r\geq\alpha$, that exponent is at most $-1$, and the integral diverges at 1. Near $t=0$, the integrand is integrable because $r\geq0$.

When $r<\alpha$, the integral is the beta integral $\int_0^1 t^{a-1}(1-t)^{b-1}\,dt=\Gamma(a)\Gamma(b)/\Gamma(a+b)$, with $a=r+1$ and $b=\alpha-r$. This identity follows by writing $\Gamma(a)\Gamma(b)$ as a double integral over positive $u,v$, then changing variables to $s=u+v$ and $t=u/(u+v)$; the Jacobian is $s$, and the integral separates into $\Gamma(a+b)$ times the beta integral. Consequently,

E[X^r] =\alpha\beta^r\frac{\Gamma(r+1)\Gamma(\alpha-r)}{\Gamma(\alpha+1)} =\beta^r\frac{\Gamma(r+1)\Gamma(\alpha-r)}{\Gamma(\alpha)}.

The last equality uses $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$. This proves both the formula in the finite case and the claimed condition for existence. $\square$

For integer $k\geq0$, if $k<\alpha$, then $\Gamma(k+1)=k!$, so the formula simplifies to

E[X^k]=\beta^k\frac{k!\,\Gamma(\alpha-k)}{\Gamma(\alpha)}.

In particular, the mean exists when $\alpha>1$, and the variance exists when $\alpha>2$. Using the gamma recurrence gives

E[X]=\frac{\beta}{\alpha-1}\quad(\alpha>1), \qquad \operatorname{Var}(X)=\frac{\alpha\beta^2}{(\alpha-1)^2(\alpha-2)} \quad(\alpha>2).

For the variance formula, the second raw moment is $E[X^2]=2\beta^2/((\alpha-1)(\alpha-2))$. Subtracting the squared mean gives $2\beta^2/((\alpha-1)(\alpha-2))-\beta^2/(\alpha-1)^2 =\alpha\beta^2/((\alpha-1)^2(\alpha-2))$.

Worked Example: Shape 3 and Scale 2

Let $\alpha=3$ and $\beta=2$. Since $3>2$, both the mean and variance exist. The density and survival function are

f_X(x)=\frac{3\cdot2^3}{(2+x)^4}=\frac{24}{(2+x)^4}, \qquad S_X(x)=\left(\frac{2}{2+x}\right)^3.

At $x=2$, the probability of exceeding 2 is $S_X(2)=(2/4)^3=1/8$. The mean and variance are

E[X]=\frac{2}{3-1}=1, \qquad \operatorname{Var}(X) =\frac{3\cdot2^2}{(3-1)^2(3-2)} =\frac{12}{4}=3.

As a check, the second raw moment is $E[X^2]=\operatorname{Var}(X)+(E[X])^2=3+1=4$. The integer-moment formula gives $2^2\cdot2!\,\Gamma(1)/\Gamma(3)=8/2=4$, in agreement.

Worked Example: Finite Mean but Infinite Variance

Now take $\alpha=3/2$ and $\beta=4$. Because $3/2>1$, the mean exists:

E[X]=\frac{4}{3/2-1}=\frac{4}{1/2}=8.

But $3/2$ is not greater than 2, so the second moment is infinite. The survival probability at 4 is $S_X(4)=(4/8)^{3/2}=1/(2\sqrt{2})$. For large $x$, the survival function is approximately $4^{3/2}x^{-3/2}=8x^{-3/2}$, illustrating the slow, polynomial tail. A finite mean does not guarantee a finite variance.

Why Mixing Changes the Tail

With a fixed exponential rate $\lambda$, the survival function is $e^{-\lambda x}$. In the mixture, some realizations of $\Lambda$ are small, and those rates correspond to long waiting times. Averaging over these low-rate cases produces the Lomax survival function $(\beta/(\beta+x))^\alpha$, which decreases as a power of $x$ for large $x$. The shape $\alpha$ controls how quickly that tail decreases and therefore which moments are finite; the scale $\beta$ sets the units and stretches the distribution.

One can also see the mean condition by averaging the conditional exponential mean $E[X\mid\Lambda]=1/\Lambda$. When $\alpha>1$,

E\left[\frac{1}{\Lambda}\right] =\frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty \lambda^{\alpha-2}e^{-\beta\lambda}\,d\lambda =\frac{\beta\Gamma(\alpha-1)}{\Gamma(\alpha)} =\frac{\beta}{\alpha-1}.

This integral diverges when $\alpha\leq1$, consistent with the moment condition. The calculation also highlights an important distinction: the mean waiting time is not generally the reciprocal of the mean rate.

Common Mistake

A tempting but incorrect calculation is to replace the random rate by its average and then use the fixed-rate exponential mean. Since $E[\Lambda]=\alpha/\beta$, this would give $1/E[\Lambda]=\beta/\alpha$. But the conditional mean is $1/\Lambda$, so averaging waiting times requires $E[1/\Lambda]$, not $1/E[\Lambda]$. For example, with $\alpha=3$ and $\beta=2$, the reciprocal of the mean rate is $2/3$, while the actual mean waiting time is $2/(3-1)=1$. In general, these quantities differ; moreover, $E[1/\Lambda]$ is infinite when $\alpha\leq1$, even though $E[\Lambda]$ is finite for every positive $\alpha$.

Key takeaway: Mixing an exponential distribution over a gamma-distributed rate gives a Lomax law with survival function $S_X(x)=(\beta/(\beta+x))^\alpha$. Its $r$th nonnegative moment is finite exactly when $r<\alpha$; in particular, the mean requires $\alpha>1$, and the variance requires $\alpha>2$.

Check Your Understanding

Use the shape-rate convention for the gamma mixing distribution.

If $\Lambda\sim\operatorname{Gamma}(\alpha,\beta)$ and $X\mid\Lambda=\lambda$ is exponential with rate $\lambda$, write the unconditional density and survival function.
For a Lomax distribution with shape $5/2$, which of its first three integer raw moments are finite?
For $\alpha=4$ and $\beta=3$, calculate $E[X]$ and $P(X>3)$.
Explain why $1/E[\Lambda]$ is not generally the mean waiting time in the mixed model.
What happens to the $r$th moment when $r=\alpha$, and which endpoint in the transformed integral causes the problem?