The monthly profit of Company I can be modeled by a continuous random variable with density function [math]f[/math]. Company II has a monthly profit that is twice that of Company I. Let [math]g[/math] be the density function for the distribution of the monthly profit of Company II.

Determine [math]g(y)[/math] where it is not zero.

• [math]\frac{1}{2}f(\frac{y}{2})[/math]
• [math]f(\frac{y}{2})[/math]
• [math]2f(\frac{y}{2})[/math]
• [math]2f(y)[/math]
• [math]2f(2y)[/math]

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

Step 1: Define Variables and Relationships

Let [math]X[/math] be the monthly profit of Company I, with probability density function (PDF) [math]f(x)[/math] and cumulative distribution function (CDF) [math]F(x)[/math]. Let [math]Y[/math] be the monthly profit of Company II, with PDF [math]g(y)[/math] and CDF [math]G(y)[/math]. We are given that the monthly profit of Company II is twice that of Company I. Therefore, we can express [math]Y[/math] in terms of [math]X[/math] as:

[[math]]Y = 2X[[/math]]

Step 2: Relate the Cumulative Distribution Functions (CDFs)

To find the PDF [math]g(y)[/math], we first need to determine the CDF [math]G(y)[/math]. The CDF [math]G(y)[/math] is defined as the probability that [math]Y[/math] is less than or equal to [math]y[/math]. Using the relationship [math]Y = 2X[/math], we can write:

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

Substitute [math]Y = 2X[/math] into the expression:

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

Divide both sides of the inequality by 2:

[[math]]G(y) = P\left[X \leq \frac{y}{2}\right][[/math]]

By definition, [math]P\left[X \leq \frac{y}{2}\right][/math] is the CDF of [math]X[/math] evaluated at [math]\frac{y}{2}[/math]:

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

Step 3: Determine the Probability Density Function (PDF)

The PDF [math]g(y)[/math] is the derivative of the CDF [math]G(y)[/math] with respect to [math]y[/math]. Differentiate [math]G(y) = F\left(\frac{y}{2}\right)[/math] with respect to [math]y[/math]:

[[math]]g(y) = G'(y) = \frac{d}{dy} F\left(\frac{y}{2}\right)[[/math]]

Applying the chain rule, where the outer function is [math]F[/math] and the inner function is [math]\frac{y}{2}[/math]:

[[math]]g(y) = F'\left(\frac{y}{2}\right) \cdot \frac{d}{dy}\left(\frac{y}{2}\right)[[/math]]

We know that [math]F'(x) = f(x)[/math] and [math]\frac{d}{dy}\left(\frac{y}{2}\right) = \frac{1}{2}[/math]. Substitute these back into the equation:

[[math]]g(y) = f\left(\frac{y}{2}\right) \cdot \frac{1}{2}[[/math]]

[[math]]g(y) = \frac{1}{2}f\left(\frac{y}{2}\right)[[/math]]

This is the density function for Company II's monthly profit.

Step 4: Final Result

The density function [math]g(y)[/math] for the monthly profit of Company II is [math]\frac{1}{2}f\left(\frac{y}{2}\right)[/math].

Key Insights

• Transformation of Random Variables: When a random variable [math]Y[/math] is a function of another random variable [math]X[/math] (e.g., [math]Y = aX + b[/math]), its PDF can be derived by first finding the CDF of [math]Y[/math] in terms of the CDF of [math]X[/math], and then differentiating the result.
• CDF to PDF Relationship: The PDF of a continuous random variable is the derivative of its CDF. Conversely, the CDF is the integral of the PDF.
• Chain Rule in Differentiation: The chain rule is crucial when differentiating a composite function like [math]F(y/2)[/math] to find the PDF [math]g(y)[/math]. The derivative of [math]F(h(y))[/math] is [math]F'(h(y)) \cdot h'(y)[/math].
• Scaling Effect: If [math]Y = aX[/math], then for continuous random variables, the PDF [math]g(y) = \frac{1}{|a|} f\left(\frac{y}{a}\right)[/math]. In this problem, [math]a=2[/math], so we get [math]g(y) = \frac{1}{2} f\left(\frac{y}{2}\right)[/math]. This demonstrates how scaling the variable affects the form of its density function.

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Solution: A

Let X and Y be the monthly profits of Company I and Company II, respectively. We are given that the pdf of X is f . Let us also take g to be the pdf of Y and take F and G to be the distribution functions corresponding to f and g . Then

[[math]]G(y) = \operatorname{P}[Y ≤ y] = \operatorname{P}[2X ≤ y] = \operatorname{P}[X ≤ y/2] = F(y/2)[[/math]]

and

[[math]]g(y) = G^{'}(y) = d/dy F(y/2) =1/2 F^{′}(y/2) =1/2 f(y/2) .[[/math]]

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