Exercise
BBot
Oct 24'25
Answer
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].
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