How to Model the Distribution of the Sum of Two Exponential Random Variables
The topic of modeling the distribution of the sum of two exponential random variables is crucial in many fields, including statistics, engineering, and economics. This article will guide you through understanding, modeling, and deriving the distribution of such a sum using both theoretical and practical methods.
Exponential Random Variables
To begin, let us define what exponential random variables are. An exponential random variable X1 is a continuous random variable that models the time between events in a Poisson process. The probability density function (PDF) of an exponential random variable X1 with rate parameter λ1 is given by:
fX1(x1) λ1 e-λ1x1 for x1 ≥ 0
Similarly, another exponential random variable X2 with rate parameter λ2 has the PDF:
fX2(x2) λ2 e-λ2x2 for x2 ≥ 0
Sum of Random Variables
When we are interested in the sum of these two random variables, i.e., Y X1 X2, the problem becomes more complex. The nature of the distribution of the sum depends on whether the rate parameters are the same or different.
Same Rate Parameters
If the two exponential random variables have the same rate parameter, i.e., λ1 λ2 λ, the sum Y follows a gamma distribution with shape parameter k 2 and scale parameter θ 1/λ. The PDF of Y in this case is:
fY(y) λ2 y e-λy
The name of the gamma distribution in this context is Gamma(2, λ).
Different Rate Parameters
When the rate parameters are different, i.e., λ1 ≠ λ2, the distribution of the sum can be derived using the convolution of the two PDFs. The convolution integral is given by:
fY(y) ∫0y fX1(x) fX2(y - x) dx
Substituting the PDFs of X1 and X2, we get:
fY(y) ∫0y λ1 e-λ1x λ2 e-λ2(y - x) dx
Simplifying this expression, we obtain:
fY(y) λ1 λ2 e-λ2y ∫0y e-λ1 - λ2x dx
The integral evaluates to:
∫0y e-λ1 - λ2x dx 1 - e-λ1 - λ2y
Thus, the PDF of Y can be expressed as:
fY(y) (λ1 λ2 / (λ1 - λ2)) (e-λ1 y - e-λ2 y) for y ≥ 0
Summary
In summary, when X1 and X2 are independent exponential random variables with the same rate λ, their sum Y follows a gamma distribution:
Y ~ Gamma(2, λ)
If the rate parameters are different, the sum can be computed using the convolution method, resulting in:
fY(y) (λ1 λ2 / (λ1 - λ2)) (e-λ1 y - e-λ2 y)
These results provide a comprehensive framework for understanding the distribution of the sum of two independent exponential random variables in various applications.