The time, [math]T[/math], that a manufacturing system is out of operation has cumulative distribution function

[[math]] F(t) = \begin{cases} 1-(2/t)^2, \, t \gt 2 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

The resulting cost to the company is [math]Y = T^2[/math]. Let [math]g[/math] be the density function for [math]Y[/math]. Determine [math]g(y)[/math], for [math]y \gt 4 [/math].

• [math]\frac{4}{y^2}[/math]
• [math]\frac{8}{y^{3/2}}[/math]
• [math]\frac{8}{y^3}[/math]
• [math]\frac{16}{y}[/math]
• [math]\frac{1024}{y^5}[/math]

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

Step 1: Analyze the Given Information

We are provided with the cumulative distribution function (CDF) for [math]T[/math], the time a manufacturing system is out of operation:

[[math]] F(t) = \begin{cases} 1-\left(\frac{2}{t}\right)^2, & t \gt 2 \\ 0, & \text{Otherwise.} \end{cases} [[/math]]

The resulting cost to the company, [math]Y[/math], is defined as a transformation of [math]T[/math]:

[[math]] Y = T^2 [[/math]]

Our goal is to determine the probability density function (PDF) for [math]Y[/math], denoted as [math]g(y)[/math], specifically for the domain where [math]y \gt 4[/math].

Step 2: Derive the Cumulative Distribution Function (CDF) of [math]Y[/math]

To find the density function [math]g(y)[/math], we first need to determine the cumulative distribution function (CDF) of [math]Y[/math], which we denote as [math]G(y)[/math]. By definition, the CDF of [math]Y[/math] is:

[[math]] G(y) = P(Y \leq y) [[/math]]

Substitute the given relationship [math]Y = T^2[/math] into the expression:

[[math]] G(y) = P(T^2 \leq y) [[/math]]

Since [math]T[/math] represents time, it must be non-negative ([math]T \gt 0[/math]). For the inequality [math]T^2 \leq y[/math] to hold, [math]y[/math] must be positive. We can then take the square root of both sides:

[[math]] G(y) = P(T \leq \sqrt{y}) [[/math]]

By definition, [math]P(T \leq \sqrt{y})[/math] is the CDF of [math]T[/math] evaluated at [math]\sqrt{y}[/math], which is [math]F(\sqrt{y})[/math]. Using the given expression for [math]F(t)[/math], and substituting [math]t = \sqrt{y}[/math]:

[[math]] G(y) = F(\sqrt{y}) = 1 - \left(\frac{2}{\sqrt{y}}\right)^2 [[/math]]

Simplify the expression for [math]G(y)[/math]:

[[math]] G(y) = 1 - \frac{4}{y} [[/math]]

The condition for [math]F(t)[/math] to be non-zero is [math]t \gt 2[/math]. Since [math]t = \sqrt{y}[/math], we must have [math]\sqrt{y} \gt 2[/math], which implies [math]y \gt 4[/math]. Thus, the CDF of [math]Y[/math] for [math]y \gt 4[/math] is:

[[math]] G(y) = 1 - \frac{4}{y}, \quad \text{for } y \gt 4 [[/math]]

Step 3: Calculate the Probability Density Function (PDF) of [math]Y[/math]

The probability density function [math]g(y)[/math] is obtained by differentiating the cumulative distribution function [math]G(y)[/math] with respect to [math]y[/math]:

[[math]] g(y) = \frac{d}{dy} G(y) [[/math]]

Substitute the derived CDF of [math]Y[/math]:

[[math]] g(y) = \frac{d}{dy} \left(1 - \frac{4}{y}\right) [[/math]]

To differentiate, it's helpful to rewrite [math]\frac{4}{y}[/math] as [math]4y^{-1}[/math]:

[[math]] g(y) = \frac{d}{dy} (1 - 4y^{-1}) [[/math]]

Perform the differentiation:

[[math]] g(y) = 0 - 4(-1)y^{-1-1} [[/math]]

[[math]] g(y) = 4y^{-2} [[/math]]

This density function is valid for [math]y \gt 4[/math].

Key Insights

• To find the PDF of a transformed random variable [math]Y = h(T)[/math] from the CDF of [math]T[/math], the standard approach is to first find the CDF of [math]Y[/math], [math]G(y) = P(Y \leq y)[/math].
• The transformation [math]P(h(T) \leq y)[/math] is inverted to express it in terms of [math]T \leq h^{-1}(y)[/math], allowing the direct use of the original CDF [math]F(t)[/math].
• After obtaining the CDF [math]G(y)[/math], the PDF [math]g(y)[/math] is found by differentiating [math]G(y)[/math] with respect to [math]y[/math].
• It is essential to correctly determine the valid domain for [math]y[/math] based on the transformation and the original variable's domain. For [math]Y=T^2[/math], if [math]T \gt c[/math], then [math]Y \gt c^2[/math].

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

Solution: A

The distribution function of Y is given by

[[math]] G(y) = Pr( T^2 \leq y ) = Pr( T \leq \sqrt{y} ) = F( \sqrt{y} ) = 1-4/y [[/math]]

for [math] y \gt 4 [/math]. Differentiate to obtain the density function [math]g(y) = 4y^{-2}[/math].

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