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

Calculate the probability that either of the losses is more than twice the other.

• 1/6
• 1/4
• 1/3
• 1/2
• 2/3

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

Step 1: Understand the Problem and Define Variables

We are given two independent losses, let's call them [math]X_1[/math] and [math]X_2[/math]. Both losses follow an exponential distribution with a mean of 1. The probability density function (PDF) for an exponential distribution with mean [math]\theta[/math] is given by [math]f(x) = \frac{1}{\theta} e^{-x/\theta}[/math] for [math]x \gt 0[/math]. Since the mean is 1, we have [math]\theta = 1[/math]. Therefore, the PDF for each loss is:

[[math]] f(x) = e^{-x} \quad \text{for } x \gt 0 [[/math]]

The survival function (the probability that a loss exceeds a certain value) is [math]P(X \gt x) = e^{-x}[/math]. We need to calculate the probability that either of the losses is more than twice the other. This can be mathematically expressed as [math]P(X_1 \gt 2X_2 \text{ or } X_2 \gt 2X_1)[/math].

Step 2: Address Mutually Exclusive Events

Given that both losses [math]X_1[/math] and [math]X_2[/math] are positive, the events [math]X_1 \gt 2X_2[/math] and [math]X_2 \gt 2X_1[/math] cannot occur simultaneously. If [math]X_1 \gt 2X_2[/math], then [math]X_1[/math] is strictly greater than [math]X_2[/math], implying [math]X_2[/math] cannot be greater than [math]2X_1[/math]. Specifically, if [math]X_1 \gt 2X_2[/math], then [math]X_2 \lt X_1/2[/math], which clearly means [math]X_2[/math] cannot be larger than [math]2X_1[/math]. Therefore, these two events are mutually exclusive. This allows us to calculate the total probability by summing the probabilities of each event:

[[math]] P(X_1 \gt 2X_2 \text{ or } X_2 \gt 2X_1) = P(X_1 \gt 2X_2) + P(X_2 \gt 2X_1) [[/math]]

Step 3: Calculate the Probability of One Event ([math]X_2 \gt 2X_1[/math])

To find [math]P(X_2 \gt 2X_1)[/math], we integrate over the joint probability density function (PDF) of [math]X_1[/math] and [math]X_2[/math]. Since [math]X_1[/math] and [math]X_2[/math] are independent, their joint PDF is the product of their individual PDFs:

[[math]] f(x_1, x_2) = f(x_1)f(x_2) = e^{-x_1}e^{-x_2} [[/math]]

We set up the integral for the region where [math]x_2 \gt 2x_1[/math]:

[[math]] P(X_2 \gt 2X_1) = \int_{0}^{\infty} \int_{2x_1}^{\infty} f(x_1, x_2) \, dx_2 \, dx_1 [[/math]]

Substitute the joint PDF:

[[math]] P(X_2 \gt 2X_1) = \int_{0}^{\infty} e^{-x_1} \left( \int_{2x_1}^{\infty} e^{-x_2} \, dx_2 \right) \, dx_1 [[/math]]

First, evaluate the inner integral with respect to [math]x_2[/math]:

[[math]] \int_{2x_1}^{\infty} e^{-x_2} \, dx_2 = [-e^{-x_2}]_{2x_1}^{\infty} = (0 - (-e^{-2x_1})) = e^{-2x_1} [[/math]]

Now, substitute this result back into the outer integral:

[[math]] P(X_2 \gt 2X_1) = \int_{0}^{\infty} e^{-x_1} e^{-2x_1} \, dx_1 [[/math]]

Combine the exponential terms:

[[math]] P(X_2 \gt 2X_1) = \int_{0}^{\infty} e^{-3x_1} \, dx_1 [[/math]]

Evaluate this integral:

[[math]] = \left[ -\frac{1}{3} e^{-3x_1} \right]_{0}^{\infty} = 0 - \left(-\frac{1}{3} e^{0}\right) = \frac{1}{3} [[/math]]

So, the probability that the second loss is more than twice the first loss is [math]\frac{1}{3}[/math].

Step 4: Utilize Symmetry and Calculate Total Probability

Due to the identical distributions of [math]X_1[/math] and [math]X_2[/math] and their independence, the probability that the first loss is more than twice the second loss, [math]P(X_1 \gt 2X_2)[/math], is the same as [math]P(X_2 \gt 2X_1)[/math]. We can verify this through integration as well:

[[math]] P(X_1 \gt 2X_2) = \int_{0}^{\infty} \int_{2x_2}^{\infty} e^{-x_2} e^{-x_1} \, dx_1 \, dx_2 [[/math]]

The inner integral [math]\int_{2x_2}^{\infty} e^{-x_1} \, dx_1 = e^{-2x_2}[/math]. Substituting back:

[[math]] P(X_1 \gt 2X_2) = \int_{0}^{\infty} e^{-x_2} e^{-2x_2} \, dx_2 = \int_{0}^{\infty} e^{-3x_2} \, dx_2 = \frac{1}{3} [[/math]]

Both probabilities are [math]\frac{1}{3}[/math]. Finally, we sum these probabilities to get the total probability that either loss is more than twice the other:

[[math]] P(\text{either loss is more than twice the other}) = P(X_1 \gt 2X_2) + P(X_2 \gt 2X_1) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} [[/math]]

The final answer is [math]\frac{2}{3}[/math].

Key Insights

• Understanding the properties of the exponential distribution, specifically its PDF [math]f(x) = \lambda e^{-\lambda x}[/math] and survival function [math]P(X \gt x) = e^{-\lambda x}[/math] (where [math]\lambda = 1/\theta[/math]).
• Recognizing that for positive random variables, the conditions [math]X \gt cY[/math] and [math]Y \gt cX[/math] (for [math]c \gt 1[/math]) are mutually exclusive, allowing for direct summation of probabilities.
• Calculating probabilities involving relationships between two independent random variables requires setting up and evaluating a double integral over the appropriate region defined by the inequality.
• Leveraging symmetry can significantly simplify calculations when the random variables have identical distributions. If [math]P(X_1 \gt cX_2)[/math] is calculated, [math]P(X_2 \gt cX_1)[/math] is often the same due to symmetry.

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Solution: E

If the first loss is [math]X[/math], then [math]\operatorname{P}(X \gt x) = e^{-x} [/math]; the probability that the second loss is more than twice [math]X[/math], would be [math]\operatorname{P}(X \gt 2x) = e^{-2x} [/math]. Thus, the probability that the second loss is more than twice the first loss is

[[math]]\int_0^{\infty} e^{-2x} e^{-x} dx = \frac{1}{3}[[/math]]

due to independence. By symmetry, the probability that the first loss 0 is more than twice the second loss is also 1/3. Thus, the answer is 2/3.

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