Tag Archives: Probability

Mixed distributions as mixtures

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A random variable X is a mixture if its distribution function F_X is a weighted average of a family of conditional distribution functions. The random variable X is a mixed distribution if it is a distribution that has at least one probability mass (i.e. there is at least one point a in the support of X such that P[X=a]>0) and there is some interval (a,b) contained in the support such that P[X=c]=0 for every c \in (a,b). It turns out that a mixed distribution can be expressed as a mixture. Three examples of mixed distributions from insurance applications have been presented in this blog. We demonstrate that these three mixed distributions are mixtures. The links to some previous posts on mixtures can be found at the end of this post.

Example 1
Link: An insurance example of a mixed distribution – I The mixed distribution in this example is the “per loss” payout for an insurance contract that has a policy maximum.

Example 2
Link: An insurance example of a mixed distribution – II The mixed distribution in this example is the “per loss” payout for an insurance policy that has a deductible.

Example 3
Link: An insurance example of a mixed distribution – III The mixed distribution in this example is the “per loss” payout of an insurance contract where there are both a deductible and a policy maximum.

Throughout this post, let X be the unmodified random loss. We assume that X is a continuous random variable with support the nonnegative real numbers.

Discussion of Example 1
Let Y_1 be the “per loss” insurance payout for a policy where the payout is capped at m. The following are the payout rule, the distribution function and the density function:

\displaystyle Y_1=\left\{\begin{matrix}X&\thinspace X<m\\{m}&\thinspace X \ge m\end{matrix}\right.

\displaystyle F_{Y_1}(x)=\left\{\begin{matrix}0&\thinspace x<0\\{F_X(x)}&\thinspace 0 \le x<m\\{1}&\thinspace x \ge m\end{matrix}\right.

\displaystyle f_{Y_1}(x)=\left\{\begin{matrix}f_X(x)&\thinspace 0 < x<m\\{1-F_X(m)}&\thinspace x=m\end{matrix}\right.

We show that F_{Y_1} can be expressed as a weighted average of two distribution functions. One of the distributions is the random loss between 0 and m. This is a limited loss and call this loss U. The second distribution is the point mass at m. Call this point mass V. The following are the distribution functions:

\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{\displaystyle \frac{F_X(x)}{F_X(m)}}&\thinspace 0 \le x<m\\{1}&\thinspace x \ge m\end{matrix}\right.

\displaystyle F_V(x)=\left\{\begin{matrix}0&\thinspace x<m\\{1}&\thinspace m \le x\end{matrix}\right.

It follows that F_{Y_1}(x)= p \thinspace F_U(x) + (1-p) \thinspace F_V(x) where p=F_X(m). Note that the distribution of U only describes the loss within (0,m). Thus the distribution function F_U is obtained from F_X by a scaler adjustment.

Discussion of Example 2
Let Y_2 be the “per loss” insurance payout for a policy where there is a deductible d. For each loss, the insurer pays the insured in excess of the deductible d. The following are the payout rule, the distribution function and the density function:

\displaystyle Y_2=\left\{\begin{matrix}0&\thinspace X<d\\{X-d}&\thinspace X \ge d\end{matrix}\right.

\displaystyle F_{Y_2}(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace y \ge 0\end{matrix}\right.

\displaystyle f_{Y_2}(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace y > 0\end{matrix}\right.

We show that F_{Y_2} can be expressed as a weighted average of two distribution functions. One of the distributions is the random loss greater than d. Call this loss U. The second distribution is the point mass at 0. Call this point mass V. The following are the distribution functions:

\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{\displaystyle \frac{F_X(x+d)-F_X(d)}{1-F_X(d)}}&\thinspace 0 \le x\end{matrix}\right.

\displaystyle F_V(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.

It follows that F_{Y_2}(x)= p \thinspace F_U(x) + (1-p) \thinspace F_V(x) where p=1-F_X(d). The random variable V is a point mass at the origin reflecting the case where no claim is made by the insurer. This point mass has weight F_X(d). The random variable U is the distribution describing the random losses that are greater than d.

Discussion of Example 3
Let Y_3 be the “per loss” insurance payout for a policy where there are both a deductible d and a policy cap m with d<m. For each loss, the insurer pays the insured in excess of the deductible d up to the policy cap m. The following are the payout rule, the distribution function and the density function:

\displaystyle Y_3=\left\{\begin{matrix}0&\thinspace X<d\\{X-d}&\thinspace d \le X < d+m\\{m}&d+m \le X\end{matrix}\right.

\displaystyle F_{Y_3}(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace 0 \le y < m\\{1}&m \le y\end{matrix}\right.

\displaystyle f_{Y_3}(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace 0 < y < m\\{1-F_X(d+m)}&y=m\end{matrix}\right.

The distribution of Y_3 can be expressed as a mixture of three distributions – two point masses (one at the origin and one at m) and one continuous variable describing the random losses in between 0 and m. Consider the following distribution functions:

\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.

\displaystyle F_V(x)=\left\{\begin{matrix}0&\thinspace x<0\\{\displaystyle \frac{F_X(x+d)-F_X(d)}{F_X(d+m)-F_X(d)}}&\thinspace 0 \le x<m\\{1}&m \le x\end{matrix}\right.

\displaystyle F_W(x)=\left\{\begin{matrix}0&\thinspace x<m\\{1}&\thinspace m \le x\end{matrix}\right.

The random variables U and W represent the point masses at 0 and m, respectively. The variable V describes the random losses in between 0 and m. It follows that F_{Y_3} is the weighted average of these three distribution functions.

\displaystyle F_{Y_3}(x)=p_1 \thinspace F_U(x)+p_2 \thinspace F_V(x)+p_3 \thinspace F_W(x)

The weights are: p_1=F_X(d), p_2=F_X(d+m)-F_X(d), and p_3=1-F_X(d+m)

Here’s the links to examples of mixed distributions:
Example 1 An insurance example of a mixed distribution – I
Example 2 An insurance example of a mixed distribution – II
Example 3 An insurance example of a mixed distribution – III

Here’s the links to some previous posts on mixtures:
Examples of mixtures
Basic properties of mixtures

An insurance example of a mixed distribution – III

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In the previous two posts, we discuss mixed distributions that are derived from modifying coverage on insurance contracts. Let X be the dollar amount of an random loss covered by an insurance contract. Without any coverage modification, the insurer would be obligated to pay the entire amount of the loss X. With some type of coverage modification, we are interested in the amount Y paid out by the insurer. How do we model Y based on the distribution of X? In one previous post, we discussed the model of the insurance payout Y when the insurance contract has a policy maximum (An insurance example of a mixed distribution – I). In another post, the coverage modification is having a deductible (An insurance example of a mixed distribution – II). In this post, we consider an insurance contract that has a combination of a deductble and a policy maximum. We discuss the model for the insurance payout Y and illustrate the calculation with the exponential distribution.

Note that the model for Y in this post and in the previous two posts is to model the insurance per loss or per claim. In other words, we model the payment made by the insurer for each insured loss. In future posts, we will discuss models that describe the insurance payments per insurance policy during a policy period. Such models will have to take into account that there may be no loss (or claim) during a period or that there may be multiple losses or claims in a policy period.

Modifying insurance coverage (e.g. having a policy maximum and/or a deductible) is akin to censoring and/or truncating the random loss amounts. Each type of censoring creates a probability mass in the distribution in the “per loss” insurance payout. So the presence of a deductible and a policy maximum in the same contract creates two probability masses. Let d be the deductible and let m be the policy maximum where d<m. Specifically, the following is the payout rule:

\displaystyle Y=\left\{\begin{matrix}0&\thinspace X<d\\{X-d}&\thinspace d \le X < d+m\\{m}&d+m \le X\end{matrix}\right.

The two probability masses are at y=0 and y=m. Thus the distribution function F_Y of Y has two jumps:

\displaystyle F_Y(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace 0 \le y < m\\{1}&m \le y\end{matrix}\right.

Note that the distribution function F_Y is obtained by shifting the graph of F_X between the points (d,F_X(d)) and (d+m,F_X(d+m)) leftward to the point (0,F_X(d)).

The point mass at y=0 has probability P[X<d] and the point mass at y=m has probability P[X \ge d+m]. Thus the following is the density function of the “per loss” insurance payout Y:

\displaystyle f_Y(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace 0 < y < m\\{1-F_X(d+m)}&y=m\end{matrix}\right.

Here’s the mean payout and the higher moments of the payout:

\displaystyle E[Y]=\int_0^m y \thinspace f_X(y+d) \thinspace dy + m \thinspace [1-F_X(d+m)]

\displaystyle E[Y^n]=\int_0^m y^n \thinspace f_X(y+d) \thinspace dy + m^n \thinspace [1-F_X(d+m)] for all integer n>1

Example
Suppose the random loss X follows an exponential distribution with parameter \lambda. Let Y be the “per loss” payout for an insurance contract that has a combination of a deductible d and a policy maximum m. Let Z be the payout for an insurance contract with the same policy maximum m but with no deductible. Interestingly, in this exponential example, we can express E[Y] and Var[Y] in terms of E[Z] and E[Z^2] (see An insurance example of a mixed distribution – I).

\displaystyle E[Y]=\int_0^m y \thinspace \lambda e^{-\lambda (y+d)} \thinspace dy + m \thinspace [e^{-\lambda (d+m)}]

\displaystyle =e^{-\lambda d} \thinspace \biggl(\int_0^m y \thinspace \lambda \thinspace e^{-\lambda y} \thinspace dy+m \thinspace e^{-\lambda m}\biggr)=e^{-\lambda d} E[Z]

\displaystyle E[Y^2]=\int_0^m y^2 \thinspace \lambda \thinspace e^{-\lambda (y+d)} \thinspace dy+m^2 \thinspace e^{-\lambda (d+m)}

\displaystyle =e^{-\lambda d} \thinspace \biggl(\int_0^m y^2 \thinspace \lambda \thinspace e^{-\lambda y} \thinspace dy+m^2 \thinspace e^{-\lambda m}\biggr)=e^{-\lambda d} E[Z^2]

\displaystyle Var[Y]=e^{-\lambda d} E[Z^2] - e^{-2 \lambda d} E[Z]^2

Comment about the exponential example
In the insurance contract with a policy maximum but no deductible, the expected insurance payout (per loss) is reduced from \displaystyle \frac{1}{\lambda} to \displaystyle \frac{1-e^{-\lambda m}}{\lambda}. With the addition of a deductible d, the expected payout is reduced from \displaystyle \frac{1}{\lambda} to \displaystyle e^{-\lambda d} \frac{1-e^{-\lambda m}}{\lambda}. The expected insurance payout is reduced by the amount \displaystyle \frac{1-e^{-\lambda d}(1-e^{-\lambda m})}{\lambda}. Then the fraction of the loss eliminated by the deductible and the policy cap is \displaystyle 1-e^{-\lambda d}(1-e^{-\lambda m}). Note that the fraction of the loss eliminated by the policy cap alone is \displaystyle 1-(1-e^{-\lambda m})=e^{-\lambda m}.

Intuitively, it is clear that a higher fraction of the random loss is eliminated if there is a deductible on top of the policy cap. In this example, this is also borne out by the following inequality:

\displaystyle e^{-\lambda m} < 1-e^{-\lambda d}(1-e^{-\lambda m})

It is also intuitively clear that the presence of a deductible and a policy maximum reduces the variance of the insurance payout. We show that this is the case for the exponential example. In fact, we show that Var[Y] is less than Var[Z], that is when there is a deductible on top of the policy maximum, the variance of the “per loss” payout is less than the variance when there is only a policy maximum. We have the following derivation and the following claim:

\displaystyle Var[Y]=e^{-\lambda d} E[Z^2] - e^{-2 \lambda d} E[Z]^2

\displaystyle =e^{-\lambda d} (Var[Z]+E[Z]^2) - e^{-2 \lambda d} E[Z]^2

\displaystyle =e^{-\lambda d} Var[Z] + (e^{-\lambda d}-e^{-2 \lambda d}) E[Z]^2

Claim.
\displaystyle Var[Z]>e^{-\lambda d} Var[Z] + (e^{-\lambda d}-e^{-2 \lambda d}) E[Z]^2

Suppose not. The following derives a contradiction:

\displaystyle Var[Z] \le e^{-\lambda d} Var[Z] + (e^{-\lambda d}-e^{-2 \lambda d}) E[Z]^2

\displaystyle (1-e^{-\lambda d}) Var[Z] \le e^{-\lambda d}(1-e^{-\lambda d})E[Z]^2

\displaystyle Var[Z] \le e^{-\lambda d}E[Z]^2

Note that the last inequality holds for all positive number d>0. Thus Var[Z] \le 0. This is a contradiction as Var[Z]>0. Thus Var[Y] is less than Var[Z].

An insurance example of a mixed distribution – II

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This is a continuation of the previous post, presenting another example of a mixed distribution involving an insurance contract. Suppose we have an insurance contract where the insurance payout per loss is subject to a deductible d. That is, if the random loss is less than d, the insurer pays nothing (the loss is assumed by the insured). If the loss exceeds d, the insurer pays X-d. This type of insurance contracts is also called the excess-of-loss contracts, since the insurer agrees to pay the insured the amount of the random loss X in excess of a fixed amount d. The following is another statement of the insurance payout rule:

\displaystyle Y=\left\{\begin{matrix}0&\thinspace X<d\\{X-d}&\thinspace X \ge d\end{matrix}\right.

We assume that the random loss X only takes on nonnegative numbers and is a continuous random variable. Let F_X be the distribution function of the random loss X. The insurance payout variable Y has a point mass at y=0. The following is the distribution function of Y.

\displaystyle F_Y(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace y \ge 0\end{matrix}\right.

The distribution function F_Y has a jump at y=0 and the size of the jump is F_X(d). The point mass at y=0 reflects the probability of having no insurance payout. As a result, the following is the density of Y.

\displaystyle f_Y(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace y > 0\end{matrix}\right.

The following is the calculation for the mean insurance payout and higher moments:

\displaystyle E[Y]=\int_o^{\infty} y \thinspace f_X(y+d) \thinspace dy

\displaystyle E[Y^n]=\int_o^{\infty} y^n \thinspace f_X(y+d) \thinspace dy for all integer n>1

Example of Calculation
Suppose we have an excess-of-loss policy with deductible d>0. Furthermore, the random loss X has an exponential distribution with parameter \lambda. Then the distribution function of X is F_X(x)=1-e^{-\lambda x}. The following is the distribution function of the insurance payout:

\displaystyle F_Y(y)=\left\{\begin{matrix}0&\thinspace y<0\\{1-e^{-\lambda (y+d)}}&\thinspace y \ge 0\end{matrix}\right.

The size of the jump at y=0 is 1-e^{-\lambda d}. As a result, the density of the insurance payout is:

\displaystyle f_Y(y)=\left\{\begin{matrix}1-e^{-\lambda d}&\thinspace y=0\\{\lambda e^{-\lambda (y+d)}}&\thinspace y > 0\end{matrix}\right.

The expected insurance payout is:

\displaystyle E[Y]=\int_o^{\infty} y \thinspace \lambda e^{-\lambda (y+d)} \thinspace dy=\frac{e^{-\lambda d}}{\lambda}

The following calculation produces the variance of the insurance payout:

\displaystyle E[Y^2]=\int_o^{\infty} y^2 \thinspace \lambda e^{-\lambda (y+d)} \thinspace dy=\frac{2 e^{- \lambda d}}{\lambda^2}

\displaystyle Var[Y]=\frac{2 e^{- \lambda d}}{\lambda^2}-\biggl(\frac{e^{-\lambda d}}{\lambda}\biggr)^2=\frac{1}{\lambda^2} \thinspace e^{-\lambda d} \thinspace (2-e^{-\lambda d})

Comment
The presence of a deductible reduces the “per loss” expected payout paid by the insurer and also has the effect of variance reduction. In the exponential example, the expected payout is reduced from \displaystyle \frac{1}{\lambda} to \displaystyle \frac{e^{-\lambda d}}{\lambda}. Due to the ductible, the “per loss” payout is reduced by the amount \displaystyle \frac{1-e^{-\lambda d}}{\lambda}. The fraction of the expected loss eliminated by the presence of the deductible is 1-e^{-\lambda d}.

The presence of the deductible is clearly variance reducing. Note that Var[Y] < \frac{1}{\lambda^2}.

Comment
Note that the model for Y in this post and in the previous post is to model the insurance per loss or per claim. In other words, we model the payment made by the insurer for each insured loss. In future posts, we will discuss models that describe the insurance payments per insurance policy during a policy period. Such “per policy” models will have to take into account that there may be no loss (or claim) during a period or that there may be multiple losses or claims in a policy period.

An insurance example of a mixed distribution – I

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Mixed distributions arise naturally in many applications. In this post and in the next several posts we discuss several examples of mixed distributions based on insurance concepts. We then illustrate the calculation using the exponential distribution as the model for the random loss.

The support of a random variable X is the subset S of the real numbers on which the probability mass function (if X is discrete) or the probability density function (if X is continuous) is positive. If X is any continuous random variable, P[X=a]=0 for any a \in S. On the other hand, if X is any discrete random variable, P[X=a] > 0 for any a \in S. In other words, a discrete distribution consists of a finite or countably infinite number of probability masses (or point masses) while continuous distributions have no point masses. This is one distinguishing characteristic between continuous random variables and discrete random variables. Mixed distributions (or mixed random variables) are nether continuous nor discrete. This means if X has a mixed distribution, X has at least one probability mass (i.e. P[X=a]>0 for at least one a \in S) and it is also true that there is some interval (a,b) \subset S such that P[X=c]=0 for all c \in (a,b).

Let’s consider one insurance example of a mixed distribution. Let X be the size (in dollar amount) of a random loss for an insurance contract. Suppose that the insurance contract has a policy maximum of M. For the purpose of this discussion, we assume that X is a continuous random variable. We also assume that P[X>M]>0. If the random loss is less than M, the insurer pays the entire random loss amount. If the random loss exceeds M, the insurance payout is capped at M. Let Y be the “per loss” amount payable by the insurer. Then Y is determined by the following rule:

\displaystyle Y=\left\{\begin{matrix}X&\thinspace X<M\\{M}&\thinspace X \ge M\end{matrix}\right.

Since the random loss X has a continuous distribution, F_X, the distribution function of X, is a continuous function. Then the insurance payout Y has a mixed distribution. The distribution of Y has a probability mass (point mass) at x=M and the distribution is continuous on the interval (0,M). The cumulative distribution function F_Y of Y is a step function with a jump at x=M. The following is F_Y(x):

\displaystyle F_Y(x)=\left\{\begin{matrix}0&\thinspace x<0\\{F_X(x)}&\thinspace 0 \le x<M\\{1}&\thinspace x \ge M\end{matrix}\right.

Since the right tail \displaystyle \int_M^{\infty} f_X(x) \thinspace dx is positive, F_Y(x) has a jump at x=M. The size of the jump is 1-F_X(M), which is the size of the point mass at x =M. The density function of Y is a hybrid one:

\displaystyle f_Y(x)=\left\{\begin{matrix}f_X(x)&\thinspace X<M\\{1-F_X(M)}&\thinspace X=M\end{matrix}\right.

The following derives the mean and the higher moments of the insurance payout.

\displaystyle E[Y]=\int_0^{M} x \thinspace f_X(x) \thinspace dx + M \thinspace [1-F_X(M)]

\displaystyle E[Y^n]=\int_0^{M} x^n \thinspace f_X(x) \thinspace dx + M^n \thinspace [1-F_X(M)] for all integers n > 1

Consequently, the following computes the variance of the insurance payout Y:

\displaystyle Var[Y]=E[Y^2]-\biggl(E[Y]\biggr)^2

Example of Calculation
Suppose we have an insurance contract with a policy maximum M. Furthermore, the random loss amount in the insurance contract has an exponential distribution with parameter \lambda. Then F_X(x)=1-e^{-\lambda x}. The distribution function of the insurance payout Y is:

\displaystyle F_Y(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1-e^{- \lambda x}}&\thinspace 0 \le x<M\\{1}&\thinspace x \ge M\end{matrix}\right.

In the distribution function, the size of the jump at x=M is e^{- \lambda M}. As a result, the density function of the insurance per loss payout Y is:

\displaystyle f_Y(x)=\left\{\begin{matrix}\lambda e^{- \lambda x}&\thinspace X<M\\{e^{- \lambda M}}&\thinspace X=M\end{matrix}\right.

The mean expected insurance payout is:

\displaystyle E[Y]=\int_0^{M} x \thinspace \lambda \thinspace e^{-\lambda x} \thinspace dx + M \thinspace [e^{-\lambda M}]= \frac{1}{\lambda}(1-e^{-\lambda M})

The following calculation derives The variance of the insurance payout:

\displaystyle E[Y^2]=\int_0^{M} x^2 \thinspace \lambda \thinspace e^{-\lambda x} \thinspace dx + M^2 \thinspace [e^{-\lambda M}]

\displaystyle = \frac{2}{\lambda^2} - \frac{2M}{\lambda}e^{-\lambda M}-\frac{2}{\lambda^2}e^{-\lambda M}

\displaystyle Var[Y]=\frac{2}{\lambda^2} - \frac{2M}{\lambda}e^{-\lambda M}-\frac{2}{\lambda^2}e^{-\lambda M} - \frac{1}{\lambda^2} \biggl(1-e^{-\lambda M}\biggr)^2

\displaystyle = \frac{1}{\lambda^2}-\frac{2M}{\lambda}e^{-\lambda M}-\frac{1}{\lambda^2}e^{-2 \lambda M}

Comment
Not surprisingly the policy cap reduces risk for the insurer. The cap for the insurance benefit has the effect of reducing the amount paid out by the insurer. In the above caculation involving the exponential distribution, the expected insurance payout is reduced from \displaystyle \frac{1}{\lambda} to \displaystyle \frac{1}{\lambda} (1-e^{-\lambda M}). In other words, due to the policy cap, the expected insurance per loss payout is reduced by the amount \displaystyle \frac{e^{-\lambda M}}{\lambda}. The fraction of the loss eliminated by the policy cap is \displaystyle e^{-\lambda M}.

On the other hand, it is clear that the policy cap is variance reducing. For the exponential example, we show that Var[Y] is less than Var[X], that is, the variance of the “per loss” insurance payout when there is a policy maximum is less than the variance of the unmodified random loss. We have the following claim.

Claim
\displaystyle Var[X]=\frac{1}{\lambda^2}> \frac{1}{\lambda^2}-\frac{2M}{\lambda}e^{-\lambda M}-\frac{1}{\lambda^2}e^{-2 \lambda M}=Var[Y]

Suppose the claim is not true. Then the following derivation produces a contradiction.

\displaystyle \frac{1}{\lambda^2}\le \frac{1}{\lambda^2}-\frac{2M}{\lambda}e^{-\lambda M}-\frac{1}{\lambda^2}e^{-2 \lambda M}

\displaystyle 1 \le 1-2M \lambda e^{-\lambda M}-e^{-2 \lambda M}

\displaystyle 2M \lambda e^{-\lambda M}+e^{-2 \lambda M} \le 0

Note that the last inequality is false as the left hand side of the inequality is positive.

Comment
Note that the model for Y in this post is to model the insurance per loss or per claim. In other words, we model the payment made by the insurer for each insured loss. In future posts, we will discuss models that describe the insurance payments per insurance policy during a policy period. Such “per policy” models will have to take into account that there may be no loss (or claim) during a period or that there may be multiple losses or claims in a policy period.