Exponential Mixtures And Decreasing Hazards

Why Exponential Mixtures Have Changing Hazards

An exponential severity has a constant hazard: its rate does not change as the loss threshold rises. A finite mixture of exponential severities behaves differently. The mixture can represent a population in which losses come from several latent severity groups, each with its own exponential rate. Among all losses, the groups have fixed mixture probabilities; among losses exceeding a high threshold, however, the slower-decaying groups become more prominent. That changing composition makes the mixture hazard decrease.

The distinction matters whenever a hazard is used to describe the risk of exceeding a threshold. The hazard of a mixture is not generally a fixed average of the component rates. It is an average using the component weights among survivors at the particular threshold.

Definition: Let $X$ be a nonnegative loss severity from a finite mixture of $m$ exponential distributions. Component $j$ has rate $\lambda_j>0$ and original mixture probability $p_j>0$, where $\sum_{j=1}^{m}p_j=1$. Its density is $f_j(x)=\lambda_j e^{-\lambda_j x}$ and its survival function is $S_j(x)=e^{-\lambda_j x}$, for $x\geq 0$. The mixture density and survival function are $f_X(x)=\sum_{j=1}^{m}p_j f_j(x)$ and $S_X(x)=\sum_{j=1}^{m}p_j S_j(x)$. Its hazard is $h_X(x)=f_X(x)/S_X(x)$.

Each component rate is measured per unit of loss. A larger exponential rate means faster decay and a shorter tail; a smaller rate means slower decay and a heavier tail. The original probabilities $p_j$ describe the composition before observing the loss. To describe the composition after learning that $X>x$, use the survivor weights

q_j(x)=P(\text{component }j\mid X>x) =\frac{p_j e^{-\lambda_j x}}{\sum_{k=1}^{m}p_k e^{-\lambda_k x}}.

These weights are positive and sum to one. The component density and survival functions give the hazard in a useful weighted-average form:

h_X(x) =\frac{\sum_{j=1}^{m}p_j\lambda_j e^{-\lambda_j x}} {\sum_{j=1}^{m}p_j e^{-\lambda_j x}} =\sum_{j=1}^{m}q_j(x)\lambda_j.

Thus, at each threshold, the hazard lies between the smallest and largest component rates. As the threshold rises, faster-decaying components lose survivor weight. The average rate therefore falls unless all components have the same rate.

Deriving the Decreasing-Hazard Result

The decrease can be proved by differentiating the survivor-weighted average. First, since $S_X'(x)=-f_X(x)$, the derivative of $\log S_X(x)$ is $-h_X(x)$. Differentiating the definition of $q_j(x)$ then gives

q_j'(x) =q_j(x)\left(-\lambda_j+h_X(x)\right).

Differentiate $h_X(x)=\sum_j q_j(x)\lambda_j$, substitute this expression for $q_j'(x)$, and use $\sum_j q_j(x)\lambda_j=h_X(x)$:

\begin{aligned} h_X'(x) &=\sum_{j=1}^{m}\lambda_j q_j'(x)\\ &=\sum_{j=1}^{m}\lambda_j q_j(x)\left(h_X(x)-\lambda_j\right)\\ &=h_X(x)\sum_{j=1}^{m}q_j(x)\lambda_j -\sum_{j=1}^{m}q_j(x)\lambda_j^2\\ &=\left(\sum_{j=1}^{m}q_j(x)\lambda_j\right)^2 -\sum_{j=1}^{m}q_j(x)\lambda_j^2\\ &=-\operatorname{Var}_{q(x)}(\lambda). \end{aligned}

Here $\operatorname{Var}_{q(x)}(\lambda)$ is the variance of the component rates when rate $\lambda_j$ has probability $q_j(x)$. It is nonnegative, so the mixture hazard is nonincreasing. If at least two component rates differ and their mixture probabilities are positive, all survivor weights remain positive at every finite threshold, the variance is positive, and the hazard is strictly decreasing.

Theorem: A finite mixture of exponential severities with positive rates has a nonincreasing hazard. The hazard decreases strictly at every finite threshold if the mixture contains at least two distinct rates with positive probabilities. At zero, $h_X(0)=\sum_j p_j\lambda_j$. As $x$ tends to infinity, $h_X(x)$ tends to the smallest component rate.

For the limiting result, let $\lambda_{\min}$ be the smallest rate. Factor $e^{-\lambda_{\min}x}$ from both the numerator and denominator of $h_X(x)$. Terms with rate greater than $\lambda_{\min}$ then contain $e^{-(\lambda_j-\lambda_{\min})x}$, which tends to zero. The components with the smallest rate remain, and their weighted average rate is $\lambda_{\min}$. If all rates are equal, the hazard is constant at that common rate instead.

Worked Calculations

The following examples use hypothetical severity models. They involve no interest or mortality assumptions. The method is to calculate the component survival contributions first, convert them to survivor weights, and then average the rates using those weights.

Worked Example: Calculating a Mixture Hazard at a Threshold

A hypothetical loss $X$, measured in units of $10,000, is an exponential mixture with probability $0.70$ on rate $0.5$ per unit and probability $0.30$ on rate $2$ per unit. Calculate $S_X(1)$, the survivor weights at $x=1$, and the hazard there. State the limiting hazard. The method is survivor weighting; it avoids averaging the rates using the original weights at every threshold.

Step 1: Calculate the component survival contributions. At $x=1$, the two contributions are

0.70e^{-0.5}=0.4245715,\qquad 0.30e^{-2}=0.0406006.

Therefore, $S_X(1)=0.4245715+0.0406006=0.4651721$, or $0.4652$ to four decimal places.

Step 2: Find the weights among losses exceeding the threshold. The rate-$0.5$ weight is $0.4245715/0.4651721=0.9127$. The rate-$2$ weight is $0.0406006/0.4651721=0.0873$. These add to $1.0000$.

Step 3: Average the rates using those weights.

$$ h_X(1)=(0.9127)(0.5)+(0.0873)(2)=0.6309. $$

The hazard is $0.6309$ per loss unit. As the threshold tends to infinity, it tends to the smallest rate, $0.5$ per unit. A check using the density gives $f_X(1)=(0.70)(0.5)e^{-0.5}+(0.30)(2)e^{-2}=0.2934869$, and $f_X(1)/S_X(1)=0.2934869/0.4651721=0.6309$, as required.

Worked Example: Tracking Weights at an Exact Threshold

A hypothetical severity has probability $0.40$ on an exponential component with rate $1$ per $1,000 loss unit, and probability $0.60$ on a component with rate $2$ per unit. At threshold $x=\ln 2$, calculate each survivor weight, the hazard, and the hazard derivative. Compare the hazard with its value at zero. The method uses exact survival factors; it avoids treating the original probabilities as unchanged after conditioning on survival.

Step 1: Evaluate component survival contributions. Since $e^{-\ln 2}=1/2$ and $e^{-2\ln 2}=1/4$, the slow component contributes $0.40(1/2)=0.20$, and the fast component contributes $0.60(1/4)=0.15$. Thus $S_X(\ln 2)=0.35$.

Step 2: Normalize to obtain survivor weights. The rate-$1$ weight is $0.20/0.35=4/7$; the rate-$2$ weight is $0.15/0.35=3/7$. The weights sum to $1$.

Step 3: Calculate the hazard.

h_X(\ln 2)=\frac{4}{7}(1)+\frac{3}{7}(2)=\frac{10}{7}=1.4286.

Step 4: Calculate its derivative. The weighted mean rate is $10/7$, and the weighted second moment of the rates is $(4/7)(1^2)+(3/7)(2^2)=16/7$. Hence

h_X'(\ln 2) =\left(\frac{10}{7}\right)^2-\frac{16}{7} =\frac{100-112}{49} =-\frac{12}{49} \approx -0.2449.

At zero, the hazard is the original weighted average, $h_X(0)=0.40(1)+0.60(2)=1.6$. It has fallen to $1.4286$ at $x=\ln 2$, and its derivative there is negative. Both results are consistent with the decreasing-hazard theorem.

Worked Example: Three Components and the Limiting Rate

A hypothetical severity, measured in units of $5,000, has probabilities $0.20$, $0.50$, and $0.30$ on exponential rates $0.5$, $1$, and $2$ per unit, respectively. At $x=\ln 4$, calculate the survival probability, the survivor weights, and the hazard. Then identify the limiting hazard. The method factors the component survival contributions; it avoids assuming that the largest original group must dominate the far tail.

Step 1: Calculate the survival contributions. For rates $0.5$, $1$, and $2$, the survival factors at $\ln 4$ are $1/2$, $1/4$, and $1/16$. Multiplying by the original probabilities gives $0.10$, $0.125$, and $0.01875$, respectively. Therefore,

S_X(\ln 4)=0.10+0.125+0.01875=0.24375=\frac{39}{160}.

Step 2: Find the survivor weights. Dividing each contribution by $39/160$ gives $16/39$, $20/39$, and $3/39$. They sum to $39/39=1$. In particular, the rate-$2$ component's weight has fallen from $0.30$ to $3/39=0.0769$, while the slowest component's weight has risen from $0.20$ to $16/39=0.4103$.

Step 3: Calculate the hazard.

h_X(\ln 4) =\frac{16}{39}(0.5)+\frac{20}{39}(1)+\frac{3}{39}(2) =\frac{34}{39} =0.8718.

For a check, the density is $0.10(0.5)+0.125(1)+0.01875(2)=0.2125$, and $0.2125/0.24375=34/39$. The smallest rate is $0.5$, so $h_X(x)$ approaches $0.5$ as $x$ increases without bound. Although the rate-$1$ component had the largest original probability, it does not determine the limiting hazard.

Interpretation and Exam Trap

The survivor weights explain why the hazard falls: an observation must survive every threshold up to $x$ to contribute to $S_X(x)$. A component with a larger rate loses its survival contribution more quickly, because $e^{-\lambda_jx}$ decays faster. This is a selection effect among exceedances, not a change in the component rates themselves.

A common mistake is to calculate the hazard at every threshold as the original weighted average $\sum_j p_j\lambda_j$. For the first example, that incorrect calculation gives $0.70(0.5)+0.30(2)=0.95$ at $x=1$. It is wrong because $0.70$ and $0.30$ are the probabilities before conditioning on $X>1$. The correct weights are $0.9127$ and $0.0873$, giving $h_X(1)=0.6309$. The original weighted average is valid at $x=0$, where $q_j(0)=p_j$, but not generally at a positive threshold.

Another useful check is the range of the result: a weighted average of the component rates must lie between the smallest and largest rates. A calculated hazard outside that range signals an error in the weights or arithmetic. For a mixture with distinct rates, a hazard that increases with the threshold also conflicts with the derivative result.

Key takeaway: At threshold $x$, calculate the survivor weights $q_j(x)$, then use $h_X(x)=\sum_j q_j(x)\lambda_j$. Its derivative is the negative survivor-weighted variance of the component rates, so faster-decaying components lose weight and the hazard approaches the smallest rate.

Check Your Understanding

For each hypothetical mixture, use the component survival contributions to check your work. Rates are per stated loss unit.

A severity has probability $0.25$ on rate $0.5$ and probability $0.75$ on rate $1.5$. Find $h_X(0)$ and $h_X'(0)$. (Answer: $h_X(0)=1.25$, $h_X'(0)=-0.1875$.)
A severity has probability $0.20$ on rate $0.5$ and probability $0.80$ on rate $1.5$. At $x=\ln 2$, find the survivor weights and hazard. (Answer: weights $1/3$ and $2/3$, respectively; $h_X(\ln 2)=1.1667$.)
A severity has equal probabilities on rates $1$ and $3$. At $x=\ln 2$, find the survival probability, survivor weights, and hazard. (Answer: $S_X(\ln 2)=0.3125$; weights $0.8000$ and $0.2000$; hazard $1.4000$.)
A severity has probabilities $0.20$, $0.30$, and $0.50$ on rates $0.25$, $0.75$, and $1.25$. Find its initial hazard and limiting hazard. (Answer: initial hazard $0.9000$; limiting hazard $0.25$.)
A severity has probability $0.60$ on rate $0.5$ and probability $0.40$ on rate $1.5$. At $x=\ln 4$, find the survivor weights and hazard. (Answer: weights $6/7$ and $1/7$, respectively; $h_X(\ln 4)=9/14=0.6429$.)

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