Losses follow an exponential distribution with mean 1. Two independent losses are observed.

Calculate the expected value of the smaller loss.

• 0.25
• 0.50
• 0.75
• 1.00
• 1.50

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Understand the Given Distributions

Let [math]X[/math] and [math]Y[/math] be the two independent losses observed. Each loss is given to follow an exponential distribution with a mean of 1. For an exponential distribution with mean [math]\theta[/math]:

• The probability density function (PDF) is [math]f(x) = \frac{1}{\theta}e^{-x/\theta}[/math] for [math]x \gt 0[/math].
• The survival function, [math]P(X \gt x)[/math], is [math]e^{-x/\theta}[/math].

Since the mean is given as 1, we have [math]\theta = 1[/math]. Therefore, the survival functions for the individual losses [math]X[/math] and [math]Y[/math] are:

[[math]]P(X \gt x) = e^{-x}[[/math]]

[[math]]P(Y \gt y) = e^{-y}[[/math]]

Step 2: Determine the Survival Function of the Smaller Loss

Let [math]Z[/math] represent the smaller of the two losses, so [math]Z = \min(X, Y)[/math]. To calculate the expected value of [math]Z[/math], we first need to identify its distribution. We can do this by finding its survival function [math]P(Z \gt z)[/math]. The event that the minimum of [math]X[/math] and [math]Y[/math] is greater than [math]z[/math] implies that both [math]X[/math] and [math]Y[/math] must be greater than [math]z[/math].

[[math]]P(Z \gt z) = P(\min(X,Y) \gt z)[[/math]]

[[math]]P(Z \gt z) = P(X \gt z \text{ and } Y \gt z)[[/math]]

Given that [math]X[/math] and [math]Y[/math] are independent losses, we can multiply their individual probabilities:

[[math]]P(Z \gt z) = P(X \gt z) \cdot P(Y \gt z)[[/math]]

Substituting the survival functions from Step 1:

[[math]]P(Z \gt z) = e^{-z} \cdot e^{-z}[[/math]]

[[math]]P(Z \gt z) = e^{-2z}[[/math]]

Step 3: Identify the Distribution of Z

The survival function of [math]Z[/math] is [math]P(Z \gt z) = e^{-2z}[/math]. This functional form is characteristic of an exponential distribution's survival function, which is generally expressed as [math]e^{-z/\theta_Z}[/math], where [math]\theta_Z[/math] is the mean of that exponential distribution. By comparing [math]e^{-2z}[/math] with [math]e^{-z/\theta_Z}[/math], we can equate the exponents:

[[math]]-2z = -\frac{z}{\theta_Z}[[/math]]

This implies:

[[math]]\frac{1}{\theta_Z} = 2[[/math]]

Solving for [math]\theta_Z[/math]:

[[math]]\theta_Z = \frac{1}{2}[[/math]]

Thus, [math]Z[/math] follows an exponential distribution with a mean parameter of [math]\frac{1}{2}[/math].

Step 4: Calculate the Expected Value of Z

For any exponential distribution, its expected value is equal to its mean parameter. Since [math]Z[/math] follows an exponential distribution with a mean of [math]\frac{1}{2}[/math]:

[[math]]E[Z] = \theta_Z = \frac{1}{2}[[/math]]

Therefore, the expected value of the smaller loss is [math]0.50[/math].

Key Insights

• The minimum of independent exponential random variables is also an exponential random variable.
• If [math]X_1, X_2, \dots, X_n[/math] are independent exponential random variables with respective rate parameters [math]\lambda_1, \lambda_2, \dots, \lambda_n[/math], then [math]\min(X_1, X_2, \dots, X_n)[/math] is an exponential random variable with a combined rate parameter of [math]\sum_{i=1}^n \lambda_i[/math].
• For an exponential distribution, the mean [math]\theta[/math] and the rate parameter [math]\lambda[/math] are reciprocals: [math]\lambda = 1/\theta[/math]. In this problem, [math]X[/math] and [math]Y[/math] each have a mean of 1, implying a rate parameter of [math]\lambda_X = \lambda_Y = 1[/math]. Thus, [math]\min(X,Y)[/math] is exponential with a rate of [math]1+1=2[/math], leading to a mean of [math]1/2[/math].
• The expected value of an exponential distribution is simply its mean parameter.

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Solution: B

Let X and Y be the two independent losses and Z = min(X,Y). Then,

[[math]] \operatorname{P}(Z \gt z) = \operatorname{P}(X \gt z \cap Y \gt z ) = \operatorname{P}(X \gt z ) \operatorname{P}(Y \gt z) = e^{-z}e^{-z} = e^{-2z} [[/math]]

[[math]] F_Z(z) = \operatorname{P}(Z \leq z) = 1-\operatorname{P}(Z \gt z) = 1-e^{-2z} [[/math]]

which can be recognized as an exponential distribution with mean 1/2.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.