Mixture Distribution

Study guide: Exam MAS-I

A Mixture Distribution describes a variable X drawn from a population made of subpopulations with different distributions: first a subpopulation is chosen at random, then X is drawn from it. Its density (and its CDF) is the weighted average of the component densities, with the weights given by a mixing distribution — discrete for a finite mixture, continuous for a continuous mixture.

f_X(x) = \sum_{i=1}^{k} a_i\, f_{X_i}(x), \quad a_i \ge 0,\ \sum_i a_i = 1

f_X(x) = \int f_{X \mid \Theta}(x \mid \theta)\, g_\Theta(\theta)\, d\theta

E[X^m] = \sum_{i=1}^{k} a_i\, E[X_i^m] = E\big[E[X^m \mid \Theta]\big]

• a_i is the proportion of the population in subpopulation i; in the continuous case \Theta indexes the subpopulations and g_\Theta is its density. The CDF, survival function, raw moments and moment generating function mix with the same weights.
• The mixing method. Any quantity that is linear in the density — a probability, a mean, a second moment — can be found by computing it within each subpopulation and averaging. The variance is not linear: compute E[X^2] - (E[X])^2 from the mixed moments, or use the law of total variance (Variance for Conditional and Marginal Distributions):

\text{Var}(X) = E\big[\text{Var}(X \mid \Theta)\big] + \text{Var}\big(E[X \mid \Theta]\big)

• Averaging the component variances misses the second term, the spread between subpopulation means, and so understates the variance.
• Standard results. A Poisson whose mean is gamma with parameters \alpha and \theta is negative binomial with r = \alpha, \beta = \theta — the basis of the Mixed Poisson Process. An exponential whose rate is gamma is Pareto.
• Mixed (discrete–continuous) type. A mixture of a point mass and a continuous distribution gives a CDF with a jump, such as an annual loss that is 0 with positive probability and continuous otherwise. Its moments combine a sum over the mass points with an integral over the continuous part.

Small and Large Claims in One Book 70% of a portfolio's claims are exponential with mean 1,000 and 30% are Pareto with \alpha = 3, \theta = 4{,}000. Find P(X > 5{,}000), E[X] and \text{Var}(X) for a randomly chosen claim.

Answer Mix the survival probabilities and the first two moments. For the Pareto, E[X] = \theta/(\alpha - 1) = 2{,}000 and E[X^2] = 2\theta^2/[(\alpha-1)(\alpha-2)] = 16{,}000{,}000; for the exponential, E[X] = 1{,}000 and E[X^2] = 2{,}000{,}000:

\begin{align*}
P(X > 5{,}000) &= 0.7\,e^{-5} + 0.3\left(\tfrac{4{,}000}{9{,}000}\right)^{3} \\
  &= 0.00472 + 0.02634 \\
  &= 0.0311 \\
E[X] &= 0.7(1{,}000) + 0.3(2{,}000) \\
  &= 1{,}300 \\
E[X^2] &= 0.7(2{,}000{,}000) + 0.3(16{,}000{,}000) \\
  &= 6{,}200{,}000 \\
\text{Var}(X) &= 6{,}200{,}000 - 1{,}300^2 \\
  &= 4{,}510{,}000
\end{align*}

Averaging the component variances, 0.7(1{,}000{,}000) + 0.3(12{,}000{,}000) = 4{,}300{,}000, falls short by exactly the variance of the means, 0.7 \times 0.3 \times (2{,}000 - 1{,}000)^2 = 210{,}000.

Gamma-Mixed Exponential Severity Each insured's claim size is exponential with rate \Lambda (mean 1/\Lambda), and \Lambda varies across insureds as a gamma distribution with \alpha = 3 and \theta = 0.001. Find the unconditional survival function and P(X > 1{,}000).

Answer Mix the conditional survival function e^{-\Lambda x} over \Lambda; the expectation is the gamma moment generating function (1 - \theta t)^{-\alpha} at t = -x:

\begin{align*}
S(x) &= E\big[e^{-\Lambda x}\big] \\
  &= (1 + 0.001x)^{-3} \\
  &= \left(\frac{1{,}000}{x + 1{,}000}\right)^{3}
\end{align*}

This is a Pareto with \alpha = 3 and \theta = 1{,}000, so P(X > 1{,}000) = (1/2)^3 = 0.125 and E[X] = 1{,}000/2 = 500. This Pareto has moments only of order below 3, so its tail is heavier than any single exponential's, which has every moment. A finite mixture of exponentials, by contrast, still has every moment and stays light-tailed.

A Loss With a Mass at Zero A policy has no loss with probability 0.6; otherwise the loss is exponential with mean 500. Find F(0), F(200) and E[X].

Answer X mixes a point mass at 0 (weight 0.6) with an exponential (weight 0.4):

\begin{align*}
F(0) &= 0.6 \\
F(200) &= 0.6 + 0.4\left(1 - e^{-200/500}\right) \\
  &= 0.6 + 0.4(0.3297) \\
  &= 0.7319 \\
E[X] &= 0.6(0) + 0.4(500) \\
  &= 200
\end{align*}

The CDF jumps by 0.6 at zero and rises continuously after it — neither a discrete nor a continuous distribution.