Independent sum of uniform

Consider the folowing problems.

  • Problem A. Suppose that the losses under two insurance policies are independent such that the loss from each policy follows a continuous uniform distribution on the interval from 0 to 2. Find the probability density function and the cumulative distribution function of the sum of the losses.
  • Problem B. Suppose that the losses under two insurance policies are independent such that the loss from one policy follows a continuous uniform distribution on the interval from 0 to 10 and the loss from the other policy is continuously uniformly distributed on the interval from 0 to 20. Find the probability density function and the cumulative distribution function of the sum of the losses.

Discussion of Problem A
Finding information about the probability distribution of the sum of several independent random variables is a basic topic in probability theory and is an important one in the SOA Exam P. One elementary method in finding the probability distribution of an independent sum is through the moment generating functions (e.g. independent sum of normal distributions). Another technique is convolution. For the independent sum of two uniform distributions, we can actually obtain the distribution by a graphical approach. The method of convolution can be applied to the examples discussed here.

Let X and Y be independent such that each follows U(0,2). Let Z=X+Y. Then the joint pdf of X and Y is \displaystyle f(x,y)=\frac{1}{4} over 0 \le x \le 2 and 0 \le y \le 2. The following is the cumulative distribution function of Z=X+Y:

\displaystyle F_Z(z)=\left\{\begin{matrix}0& \ \ \ \ \ \ z<0 \\{\text{ }}& \\{\displaystyle \frac{z^2}{8}}&\ \ \ \ \ \ 0 \le z <2 \\{\text{ }}& \\{\displaystyle 1-\frac{(4-z)^2}{8}}&\ \ \ \ \ \ 2 \le z <4 \\{\text{ }}& \\{1}&\ \ \ \ \ \ 4 \le z \end{matrix}\right.

For the case 0 \le z <2, F_Z(z) is obtained by multiplying the shaded area in the following diagram (Figure 1) by the joint pdf f(x,y)=\frac{1}{4}. Note that the area of the shaded region in Figure 1 is \displaystyle \frac{z^2}{2}.

For the case 2 \le z <4, F_Z(z) is obtained by multiplying the complement of the shaded area in the following diagram (Figure 2) by the joint pdf f(x,y)=\frac{1}{4}. Note that the area of the shaded region in Figure 2 is \displaystyle \frac{(4-z)^2}{2}. The complement of the shaded region is \displaystyle 4-\frac{(4-z)^2}{2}.

The following is the pdf of Z=X+Y.

\displaystyle f_Z(z)=\left\{\begin{matrix}\displaystyle \frac{z}{4}&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \le z <2 \\{\text{ }}& \\{\displaystyle \frac{4-z}{4}}&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \le z <4 \end{matrix}\right.

Problem B Answers
Suppose X and Y are U(0,10) and U(0,20), respectively. Suppose that Z=X+Y is an independent sum. Then the cdf and pdf of Z are:

\displaystyle F_Z(z)=\left\{\begin{matrix}0& \ \ \ \ \ \ z<0 \\{\text{ }}& \\{\displaystyle \frac{z^2}{400}}&\ \ \ \ \ \ 0 \le z <10 \\{\text{ }}& \\{\displaystyle \frac{z}{20}-\frac{1}{4}}&\ \ \ \ \ \ 10 \le z <20 \\{\text{ }}& \\{\displaystyle 1-\frac{(30-z)^2}{400}}&\ \ \ \ \ \ 20 \le z <30 \\{\text{ }}& \\{1}&\ \ \ \ \ \ 30 \le z \end{matrix}\right.

\displaystyle f_Z(z)=\left\{\begin{matrix} \displaystyle \frac{z}{200}&\ \ \ \ \ \ 0 \le z <10 \\{\text{ }}& \\{\displaystyle \frac{1}{20}}&\ \ \ \ \ \ 10 \le z <20 \\{\text{ }}& \\{\displaystyle \frac{30-z}{200}}&\ \ \ \ \ \ 20 \le z <30 \end{matrix}\right.

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