Build mixture distributions and describe how they arise and what features they have.
The idea
A mixture distribution is built by first drawing which component applies, with probabilities (weights) that sum to one, then drawing from that component. Its density and its CDF are the weighted averages of the component densities and CDFs. Do not confuse this with a weighted sum of random variables. The mean is the weighted average of the component means, with weights wi and component means μi: Mean = Σ wi × μi. The variance uses each component's second moment, with σi² the component variance: Variance = Σ wi × (σi² + μi²) − μ². Here wi are the weights and μ is the mixture mean. So the variance is not the weighted average of the variances unless the means are equal. Mixing normals with the same mean but different volatilities gives a symmetric distribution with excess kurtosis, which is how regime-switching between calm and stressed markets produces fat tails. Mixing components with well separated means gives bimodality, and unequal means can give skewness. A mixture of normals is not normal. Exam questions usually ask for a tail probability or a standard deviation, so weight probabilities, not standard deviations.
Formulas
Variance of a mixture distribution
σ² = Σ wᵢ × (σᵢ² + μᵢ²) − μ²
- wᵢ
- the weight of component i
- σᵢ
- the standard deviation of component i
- μᵢ
- the mean of component i
- μ
- the mixture mean, Σ wᵢ × μᵢ
The weights sum to one. It equals the weighted average of variances only when the component means are equal; never weight the standard deviations.
A practice question
Which statement correctly describes a mixture of two component distributions with weights w and 1 - w?
- It is the distribution of the sum X1 + X2 of two independent random variables, one drawn from each component.Why not: Describes a convolution (sum of random variables), whereas a mixture selects one component at random for each draw.
- Its density is w x f1(x) + (1 - w) x f2(x): the weighted average of the two component densities. - correct
- It is the distribution of the weighted sum w x X1 + (1 - w) x X2 of two random variables.Why not: Confuses a mixture with a portfolio-style weighted sum of variables; weights in a mixture are probabilities of selecting a component.
- Its density is the product f1(x) x f2(x) of the two component densities.Why not: Multiplies the densities, which is a joint density of independent variables, not a mixture, and does not integrate to one.
A mixture first draws the component (with probability w for the first) and then draws from it, so the density and CDF are weighted averages of the component densities and CDFs. A weighted sum of variables is a different object.
A harder one
Daily returns are a mixture: with probability 90 percent a normal with mean 0 and standard deviation 1 percent, and with probability 10 percent a normal with mean 0 and standard deviation 4 percent. The standard deviation of the mixture is closest to:
- 1.58% - correct
- 1.30%Why not: Averages the standard deviations (0.9 x 1 + 0.1 x 4); standard deviations do not mix linearly, variances do.
- 3.81%Why not: Swaps the weights, using 10 percent on the calm component and 90 percent on the stress component.
- 2.50%Why not: Computes the weighted variance 0.9 x 1 + 0.1 x 16 = 2.5 but reports it without taking the square root.
Means are equal at zero, so variance = 0.9 x 1² + 0.1 x 4² = 0.9 + 1.6 = 2.5 (in percent squared). Standard deviation = sqrt(2.5) = 1.58 percent.
The hardest tier
Project profit (USD million) is modelled as a mixture: with probability 50 percent a normal with mean 5 and standard deviation 4, and with probability 50 percent a normal with mean 15 and standard deviation 10. The standard deviation of the mixture is closest to:
- 7.6Why not: Takes the square root of the weighted average of the variances only, ignoring the spread caused by the different component means.
- 13.5Why not: Adds the weighted squared means to the weighted variances but forgets to subtract the squared mixture mean.
- 7.0Why not: Averages the standard deviations (0.5 x 4 + 0.5 x 10), which is not how mixture dispersion works.
- 9.1 - correct
Mixture mean = 0.5 x 5 + 0.5 x 15 = 10. E[X²] = 0.5 x (16 + 25) + 0.5 x (100 + 225) = 20.5 + 162.5 = 183. Variance = 183 - 10² = 83, so the standard deviation is 9.1.
In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.
Start free →