The method of convolution is a great technique for finding the probability density function (pdf) of the sum of two independent random variables. We state the convolution formula in the continuous case as well as discussing the thought process. Some examples are provided to demonstrate the technique and are followed by an exercise.
The Convolution Formula (Continuous Case)
Let and be independent continuous random variables with pdfs and , respectively. Let . Then the following is the pdf of .
Let’s look at the thought process behind the formula. Since and are independent, the joint pdf of and is . The pdf of is simply the sum of the “joint density” at the points of the line . In Figure 1 below, every point at the line is of the form . The joint density at each such point is . Summing the values of these joint density produces the probability density function of .
Setting the limits of the integral depends on knowing the range of possible values of or for a given line . If and can only take on positive values, then for a given line , both or can range from to (see Example 1 below).
Let and be independent exponentially distributed variables with common density where . Then the following is the pdf of .
The above pdf indicates that the independent sum of two identically distributed exponential variables has a Gamma distribution with parameters and .
Let and be independent uniformly distributed variables, and , respectively. The pdf of is:
The convolution formula is applied three times. For the first case, the line ranges in . For each such line, we have . Figure 2 below is a representative diagram.
For the second case, the line ranges in . For each such line, ranges from to . Figure 3 below is a representative diagram.
For the third case, the line ranges in . Figure 4 below is a representative diagram.
The following is the graph of the pdf of .
Suppose that is an exponentially distributed variable with pdf and has the uniform distribution . Find the pdf of the independent sum .
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 and be independent such that each follows . Let . Then the joint pdf of and is over and . The following is the cumulative distribution function of :
For the case , is obtained by multiplying the shaded area in the following diagram (Figure 1) by the joint pdf . Note that the area of the shaded region in Figure 1 is .
For the case , is obtained by multiplying the complement of the shaded area in the following diagram (Figure 2) by the joint pdf . Note that the area of the shaded region in Figure 2 is . The complement of the shaded region is .
The following is the pdf of .
Problem B Answers
Suppose and are and , respectively. Suppose that is an independent sum. Then the cdf and pdf of are: