Once a fire is reported to a fire insurance company, the company makes an initial estimate, [math]X[/math], of the amount it will pay to the claimant for the fire loss. When the claim is finally settled, the company pays an amount, [math]Y[/math], to the claimant. The company has determined that [math]X[/math] and [math]Y[/math] have the joint density function

[[math]] f(x,y) = \begin{cases} \frac{2}{x^2(x-1)}y^{-(2x-1)/(x-1)}, \,\, x \gt1, y \gt 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Given that the initial claim estimated by the company is 2, calculate the probability that the final settlement amount is between 1 and 3.

• 1/9
• 2/9
• 1/3
• 2/3
• 8/9

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

Step 1: Understand the Problem and Conditional Probability Formula

The problem asks us to calculate the probability that the final settlement amount [math]Y[/math] is between 1 and 3, given that the initial claim estimate [math]X[/math] is 2. This is a conditional probability problem for continuous random variables, which can be expressed as:

[[math]]P[1 \lt Y \lt 3 | X = 2] = \int_1^3 f_{Y|X}(y|X=2) dy[[/math]]

where [math]f_{Y|X}(y|X=2)[/math] is the conditional probability density function of [math]Y[/math] given [math]X=2[/math]. The conditional PDF is defined as:

[[math]]f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)}[[/math]]

Here, [math]f(x,y)[/math] is the given joint density function, and [math]f_X(x)[/math] is the marginal density function of [math]X[/math]. Therefore, we need to calculate:

[[math]]P[1 \lt Y \lt 3 | X = 2] = \int_1^3 \frac{f(2,y)}{f_X(2)} dy[[/math]]

Step 2: Calculate the Joint Density at X = 2, f(2,y)

The joint density function is given by:

[[math]]f(x,y) = \begin{cases} \frac{2}{x^2(x-1)}y^{-(2x-1)/(x-1)}, \,\, x \gt1, y \gt 1 \\ 0, \, \textrm{Otherwise.} \end{cases}[[/math]]

We need to evaluate this function at [math]x=2[/math]. Substituting [math]x=2[/math] into the expression:

[[math]]f(2,y) = \frac{2}{2^2(2-1)}y^{-(2(2)-1)/(2-1)}[[/math]]

Simplifying the expression:

[[math]]f(2,y) = \frac{2}{4(1)}y^{-(4-1)/(1)} = \frac{1}{2}y^{-3}[[/math]]

This is valid for [math]y \gt 1[/math].

Step 3: Calculate the Marginal Density of X at X = 2, f_X(2)

To find the marginal density of [math]X[/math] at [math]x=2[/math], denoted as [math]f_X(2)[/math], we integrate [math]f(2,y)[/math] over all possible values of [math]y[/math]. Given the domain [math]y \gt 1[/math]:

[[math]]f_X(2) = \int_1^{\infty} f(2,y) dy = \int_1^{\infty} \frac{1}{2}y^{-3} dy[[/math]]

Now, we evaluate the integral:

[[math]]f_X(2) = \frac{1}{2} \left[ \frac{y^{-2}}{-2} \right]_1^{\infty}[[/math]]

[[math]]f_X(2) = -\frac{1}{4} [y^{-2}]_1^{\infty}[[/math]]

Evaluating the limits:

[[math]]f_X(2) = -\frac{1}{4} \left( \lim_{y \to \infty} y^{-2} - 1^{-2} \right)[[/math]]

[[math]]f_X(2) = -\frac{1}{4} (0 - 1) = \frac{1}{4}[[/math]]

Step 4: Calculate the Conditional Probability P[1 < Y < 3 | X = 2]

Now we have both [math]f(2,y)[/math] and [math]f_X(2)[/math]. We can substitute these into the conditional probability integral:

[[math]]P[1 \lt Y \lt 3 | X = 2] = \int_1^3 \frac{f(2,y)}{f_X(2)} dy = \int_1^3 \frac{\frac{1}{2}y^{-3}}{\frac{1}{4}} dy[[/math]]

Simplify the integrand:

[[math]]P[1 \lt Y \lt 3 | X = 2] = \int_1^3 \left( \frac{1}{2}y^{-3} \times 4 \right) dy = \int_1^3 2y^{-3} dy[[/math]]

Now, integrate the expression:

[[math]]P[1 \lt Y \lt 3 | X = 2] = 2 \left[ \frac{y^{-2}}{-2} \right]_1^3[[/math]]

[[math]]P[1 \lt Y \lt 3 | X = 2] = - [y^{-2}]_1^3[[/math]]

Evaluate the definite integral using the limits:

[[math]]P[1 \lt Y \lt 3 | X = 2] = - \left( 3^{-2} - 1^{-2} \right)[[/math]]

[[math]]P[1 \lt Y \lt 3 | X = 2] = - \left( \frac{1}{9} - 1 \right)[[/math]]

[[math]]P[1 \lt Y \lt 3 | X = 2] = - \left( -\frac{8}{9} \right) = \frac{8}{9}[[/math]]

Thus, the probability that the final settlement amount is between 1 and 3, given an initial claim estimate of 2, is [math]\frac{8}{9}[/math].

Key Insights

• Conditional probability for continuous random variables requires computing the ratio of the joint density to the marginal density.
• To find the marginal density [math]f_X(x)[/math] from a joint density [math]f(x,y)[/math], integrate [math]f(x,y)[/math] with respect to [math]y[/math] over its entire domain.
• Careful substitution of parameters (like [math]X=2[/math] in this case) into the density function is crucial for simplifying the expressions before integration.
• Accurate evaluation of definite integrals, including proper handling of limits (especially for improper integrals with infinity), is essential.
• The domain constraints for [math]X[/math] and [math]Y[/math] ([math]x\gt1, y\gt1[/math]) must be considered when setting up integral limits.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: E

[[math]] \begin{align*} \operatorname{P}[1 \lt Y \lt 3 | X = 2] = \int_1^3 \frac{f(2,y)}{f_x(2)} dy \\ f(2,y) = \frac{2}{4(2-1)} y^{-(4-1)/2 -1} = \frac{1}{2}y^{-3} \\ f_x(2) = \int_1^{\infty} \frac{1}{2}y^{-3} dy = - \frac{1}{4}y^{-2} \Big |_0^{\infty} = \frac{1}{4} \end{align*} [[/math]]

Finally,

[[math]] \operatorname{P}[1 \lt Y \lt 3 | X = 2] = \frac{\int_1^3 \frac{1}{2} y^{-3} dy}{\frac{1}{4}} = -y^2 \Big |_1^3 = 1- \frac{1}{9} = \frac{8}{9}. [[/math]]