An actuary models the lifetime of a device using the random variable [math]Y = 10X^{0.8}[/math], where [math]X[/math] is an exponential random variable with mean 1. Let [math]f(y)[/math] be the density function for [math]Y[/math].

Determine [math]f(y)[/math] for [math]y\gt0[/math].

• [math]10 y^{0.8} \exp(−8 y^{−0.2} )[/math]
• [math]8 y^{−0.2} \exp(−10 y^{0.8} )[/math]
• [math]8 y^{−0.2} \exp[−(0.1y)^{1.25} ][/math]
• [math](0.1y )^{1.25} \exp[−0.125(0.1y)^{0.25} ][/math]
• [math]0.125(0.1y )^{0.25} \exp[−(0.1y )^{1.25} ][/math]

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

Step 1: Understand the Given Information

The problem describes a random variable [math]Y[/math] derived from another random variable [math]X[/math].

• Random Variable [math]X[/math]: [math]X[/math] is an exponential random variable with a mean of 1.
• The probability density function (PDF) of [math]X[/math] is given by [math]f_X(x) = e^{-x}[/math] for [math]x \gt 0[/math].
• The cumulative distribution function (CDF) of [math]X[/math] is [math]F_X(x) = P(X \leq x) = 1 - e^{-x}[/math] for [math]x \gt 0[/math].
• Relationship between [math]Y[/math] and [math]X[/math]: The lifetime of the device [math]Y[/math] is related to [math]X[/math] by the equation [math]Y = 10X^{0.8}[/math].
• Objective: Determine the density function for [math]Y[/math], denoted as [math]f_Y(y)[/math], for [math]y \gt 0[/math]. We will use the method of transformation via the CDF.

Step 2: Express [math]X[/math] in Terms of [math]Y[/math]

To use the CDF method, we first need to express [math]X[/math] as a function of [math]Y[/math]. We are given:

[[math]]Y = 10X^{0.8}[[/math]]

Divide by 10:

[[math]]\frac{Y}{10} = X^{0.8}[[/math]]

To isolate [math]X[/math], we raise both sides to the power of [math]\frac{1}{0.8}[/math]:

[[math]]X = \left(\frac{Y}{10}\right)^{\frac{1}{0.8}}[[/math]]

Since [math]\frac{1}{0.8} = \frac{10}{8} = \frac{5}{4} = 1.25[/math], we have:

[[math]]X = \left(\frac{Y}{10}\right)^{5/4}[[/math]]

This expression for [math]X[/math] is valid for [math]y \gt 0[/math], which ensures [math]x \gt 0[/math].

Step 3: Derive the Cumulative Distribution Function (CDF) of [math]Y[/math], [math]F_Y(y)[/math]

The CDF of [math]Y[/math] is defined as [math]F_Y(y) = P(Y \leq y)[/math]. We use the relationship between [math]Y[/math] and [math]X[/math] derived in the previous step:

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

Substitute [math]Y = 10X^{0.8}[/math]:

[[math]]F_Y(y) = P(10X^{0.8} \leq y)[[/math]]

Isolate [math]X[/math]:

[[math]]F_Y(y) = P\left(X^{0.8} \leq \frac{y}{10}\right)[[/math]]

[[math]]F_Y(y) = P\left(X \leq \left(\frac{y}{10}\right)^{\frac{1}{0.8}}\right)[[/math]]

Using [math]\frac{1}{0.8} = \frac{5}{4}[/math]:

[[math]]F_Y(y) = P\left(X \leq \left(\frac{y}{10}\right)^{5/4}\right)[[/math]]

This expression is equivalent to the CDF of [math]X[/math] evaluated at [math]\left(\frac{y}{10}\right)^{5/4}[/math], i.e., [math]F_Y(y) = F_X\left(\left(\frac{y}{10}\right)^{5/4}\right)[/math]. Since [math]F_X(x) = 1 - e^{-x}[/math] for an exponential distribution with mean 1, we substitute the argument:

[[math]]F_Y(y) = 1 - \exp\left(-\left(\frac{y}{10}\right)^{5/4}\right)[[/math]]

This is the CDF of [math]Y[/math] for [math]y \gt 0[/math].

Step 4: Differentiate [math]F_Y(y)[/math] to Find the PDF [math]f_Y(y)[/math]

To find the probability density function [math]f_Y(y)[/math], we differentiate the CDF [math]F_Y(y)[/math] with respect to [math]y[/math]:

[[math]]f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} \left[1 - \exp\left(-\left(\frac{y}{10}\right)^{5/4}\right)\right][[/math]]

Let [math]u = \left(\frac{y}{10}\right)^{5/4}[/math]. Then [math]F_Y(y) = 1 - e^{-u}[/math]. Using the chain rule, [math]\frac{d}{dy}(1 - e^{-u}) = -(-e^{-u}) \frac{du}{dy} = e^{-u} \frac{du}{dy}[/math]. First, calculate [math]\frac{du}{dy}[/math]:

[[math]]\frac{du}{dy} = \frac{d}{dy} \left[\left(\frac{y}{10}\right)^{5/4}\right] = \frac{d}{dy} \left[\frac{1}{10^{5/4}} y^{5/4}\right][[/math]]

[[math]]\frac{du}{dy} = \frac{1}{10^{5/4}} \cdot \frac{5}{4} y^{\frac{5}{4}-1} = \frac{5}{4 \cdot 10^{5/4}} y^{1/4}[[/math]]

Now substitute this back into the expression for [math]f_Y(y)[/math]:

[[math]]f_Y(y) = \exp\left(-\left(\frac{y}{10}\right)^{5/4}\right) \cdot \frac{5}{4 \cdot 10^{5/4}} y^{1/4}[[/math]]

Simplify the constant term and [math]y[/math] terms:

[[math]]\frac{5}{4 \cdot 10^{5/4}} y^{1/4} = \frac{5}{4 \cdot 10 \cdot 10^{1/4}} y^{1/4} = \frac{1}{8 \cdot 10^{1/4}} y^{1/4} = \frac{1}{8} \left(\frac{y}{10}\right)^{1/4}[[/math]]

So, the PDF [math]f_Y(y)[/math] is:

[[math]]f_Y(y) = \frac{1}{8} \left(\frac{y}{10}\right)^{1/4} \exp\left(-\left(\frac{y}{10}\right)^{5/4}\right)[[/math]]

To match the given options, we convert fractions to decimals and use the [math](0.1y)[/math] form:

• [math]\frac{1}{8} = 0.125[/math]
• [math]\left(\frac{y}{10}\right)^{1/4} = (0.1y)^{0.25}[/math]
• [math]\left(\frac{y}{10}\right)^{5/4} = (0.1y)^{1.25}[/math]

Substituting these into the expression for [math]f_Y(y)[/math]:

[[math]]f_Y(y) = 0.125(0.1y)^{0.25} \exp[-(0.1y)^{1.25}][[/math]]

This matches option E.

"Key Insights"

• Method of Transformations (CDF Method): To find the PDF of a transformed random variable [math]Y = g(X)[/math], first find its CDF [math]F_Y(y) = P(Y \leq y)[/math] by expressing [math]X[/math] in terms of [math]Y[/math] (i.e., [math]X = g^{-1}(Y)[/math]) and using the known CDF of [math]X[/math]. Then, differentiate [math]F_Y(y)[/math] with respect to [math]y[/math] to obtain [math]f_Y(y)[/math].
• Chain Rule Application: Differentiating [math]F_Y(y) = F_X(g^{-1}(y))[/math] requires careful application of the chain rule. If [math]u = g^{-1}(y)[/math], then [math]\frac{d}{dy}F_X(u) = f_X(u) \cdot \frac{du}{dy}[/math].
• Exponential Distribution Properties: For an exponential random variable [math]X[/math] with mean [math]\lambda[/math], its CDF is [math]F_X(x) = 1 - e^{-x/\lambda}[/math]. In this problem, the mean is 1, so [math]\lambda = 1[/math], simplifying the CDF to [math]F_X(x) = 1 - e^{-x}[/math].
• Algebraic Manipulation: Proficiency in manipulating fractional exponents and simplifying expressions is crucial to arrive at the final form of the PDF, especially when matching it with multiple-choice options.

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

Solution: E

[[math]] F ( y ) = \operatorname{P}[Y ≤ y ] = \operatorname{P}[10 X 0.8 ≤ y] = \operatorname{P}[ X \leq (Y/10)^{10/8}] = 1-\exp(-(Y/10)^{10/8}). [[/math]]

Therefore,

[[math]] f(y) = F^{'}(y) = \frac{1}{8} \left ( \frac{Y}{10}\right )^{1/4} \exp(-(Y/10)^{5/4}). [[/math]]

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