Chapter 4: Problem 109
The weekly amount of downtime \(Y\) (in hours) for an industrial machine has approximately a gamma distribution with \(\alpha=3\) and \(\beta=2\). The loss \(L\) (in dollars) to the industrial operation as a result of this downtime is given by \(L=30 Y+2 Y^{2}\). Find the expected value and variance of \(L\).
Short Answer
Step by step solution
Understand the Gamma Distribution
Calculate Expected Value and Variance of Y
Define the Loss Function L
Find Expected Value of L
Calculate Variance of L
Conclude the Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Expected Value
- For \(\alpha=3\) and \(\beta=2\), the expected value \(E(Y)\) becomes \(\frac{3}{2} = 1.5\) hours.
- \(E(L) = 30E(Y) + 2E(Y^2)\)
- With \(E(Y^2) = \text{Var}(Y) + (E(Y))^2 = 0.75 + (1.5)^2 = 3\), hence \(E(L) = 30 \, \times \, 1.5 + 2 \, \times \, 3 = 51\) dollars.
Variance
- The variance of the downtime \(Y\) is \(\frac{3}{4} = 0.75\) hours squared.
Loss Function
- \(30Y\) accounts for a linear increase in cost per hour of downtime.
- \(2Y^2\) captures an additional quadratic penalty as the downtime increases.
Gamma Distribution Properties
- Shape parameter \(\alpha\) and Rate parameter \(\beta\): These define the form of the distribution and its scale. An increase in \(\alpha\) shapes the distribution towards a normal distribution, while \(\beta\) inversely scales the distribution.
- Flexibility: It can model various shapes, making it a versatile choice for different statistical applications.
- Memoryless property in special cases: A unique property, although primarily associated with the exponential distribution, a special case of Gamma with \(\alpha = 1\).