The lifetime of a light bulb has density function, [math]f[/math], where [math]f(x)[/math] is proportional to

[[math]] \frac{x^2}{1+x^3}, \, 0 \lt x \lt5, \, \textrm{and}, \, 0 \, \textrm{otherwise}. [[/math]]

Calculate the mode of this distribution.

• 0.00
• 0.79
• 1.26
• 4.42
• 5.00

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

Step 1: Understand the PDF and Goal

The probability density function (PDF) of the light bulb's lifetime is given as proportional to [math]\frac{x^2}{1+x^3}[/math] for [math]0 \lt x \lt 5[/math], and [math]0[/math] otherwise. We need to find the mode of this distribution. The mode is the value of [math]x[/math] where the PDF reaches its maximum. To find this, we will calculate the first derivative of the PDF, set it to zero, and solve for [math]x[/math]. Let the constant of proportionality be [math]c[/math]. The function can be written as:

[[math]]f(x) = c \frac{x^2}{1+x^3} = c x^2 (1+x^3)^{-1}[[/math]]

The constant [math]c[/math] does not affect the location of the mode, so it is not necessary to determine its value and it can be treated as a non-zero constant during differentiation and solving.

Step 2: Calculate the First Derivative of the PDF

We differentiate [math]f(x) = c x^2 (1+x^3)^{-1}[/math] with respect to [math]x[/math] using the product rule, which states that if [math]f(x) = u(x)v(x)[/math], then [math]f'(x) = u'(x)v(x) + u(x)v'(x)[/math]. Let [math]u(x) = cx^2[/math] and [math]v(x) = (1+x^3)^{-1}[/math]. First, find the derivatives of [math]u(x)[/math] and [math]v(x)[/math]:

• [math]u'(x) = \frac{d}{dx}(cx^2) = 2cx[/math]
• [math]v'(x) = \frac{d}{dx}((1+x^3)^{-1}) = -1(1+x^3)^{-2} \cdot (3x^2) = -3x^2(1+x^3)^{-2}[/math]

Now, apply the product rule to find [math]f'(x)[/math]:

[[math]] f'(x) = (2cx)(1+x^3)^{-1} + (cx^2)(-3x^2(1+x^3)^{-2}) [[/math]]

This simplifies to:

[[math]] f'(x) = 2cx(1+x^3)^{-1} - 3cx^4(1+x^3)^{-2} [[/math]]

Step 3: Set the Derivative to Zero and Solve for x

To find the mode, we set the first derivative [math]f'(x)[/math] equal to zero:

[[math]] 0 = 2cx(1+x^3)^{-1} - 3cx^4(1+x^3)^{-2} [[/math]]

To solve for [math]x[/math], we can factor out the common term [math]cx(1+x^3)^{-2}[/math]:

[[math]] 0 = cx(1+x^3)^{-2} \left[ 2(1+x^3) - 3x^3 \right] [[/math]]

Since [math]c \ne 0[/math], [math]x \ne 0[/math] (as [math]x[/math] is in the domain [math]0 \lt x \lt 5[/math]), and [math](1+x^3)^{-2}[/math] is also non-zero for [math]x \in (0,5)[/math], we can divide both sides by [math]cx(1+x^3)^{-2}[/math]. This means the expression inside the square brackets must be zero:

[[math]] 2(1+x^3) - 3x^3 = 0 [[/math]]

Expand and simplify the equation:

[[math]] 2 + 2x^3 - 3x^3 = 0 [[/math]]

[[math]] 2 - x^3 = 0 [[/math]]

Solve for [math]x^3[/math]:

[[math]] x^3 = 2 [[/math]]

Finally, solve for [math]x[/math]:

[[math]] x = 2^{1/3} [[/math]]

Calculating the numerical value:

[[math]] x \approx 1.259921... [[/math]]

Rounding to two decimal places, the mode is [math]x \approx 1.26[/math]. This value is within the specified domain [math]0 \lt x \lt 5[/math].

Key Insights

• The mode of a continuous probability distribution is found by identifying the [math]x[/math]-value where the probability density function (PDF) reaches its maximum, which involves setting the first derivative of the PDF to zero.
• A constant of proportionality in a PDF (or any function being maximized/minimized) does not influence the location of the extremum and can be disregarded when finding the mode.
• Accurate application of differentiation rules, such as the product rule and chain rule, is fundamental for correctly calculating the derivative of complex functions.
• Careful algebraic manipulation, including factoring and simplification, is essential to solve for the variable after setting the derivative to zero.
• Always verify that the calculated mode falls within the defined domain of the probability distribution.

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

Solution: C

It is not necessary to determine the constant of proportionality. Let it be c. To determine the mode, set the derivative of the density function equal to zero and solve.

[[math]] \begin{align*} 0 &= f^{'}(x) = \frac{d}{dx} cx^2(1+x^3)^{-1} = 2cx(1+x^3)^{-1} + cx^2[-(1+x^3)^{-2}]3x^2 \\ &= 2cx(1+x^3)-3cx^4 \\ &= 2cx + 2cx^4 -3cx^4 = 2cx - cx^4 \\ &= 2- x^3 \Rightarrow x = 2^{1/3} = 1.26. \end{align*} [[/math]]

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