Problem 168 - Applied Statistics and Probability for Engineers

Question

An article in Financial Markets Institutions and Instruments ["Pricing Reinsurance Contracts on FDIC Losses" (2008, Vol. 17(3)\(]\) modeled average annual losses (in billions of dollars) of the Federal Deposit Insurance Corporation (FDIC) with a Weibull distribution with parameters \(\delta=1.9317\) and \(\beta=0.8472\). Determine the following: (a) Probability of a loss greater than \(\$ 2\) billion (b) Probability of a loss between \(\$ 2\) and \(\$ 4\) billion (c) Value exceeded with probability 0.05 (d) Mean and standard deviation of loss

Step-by-Step Solution

Answer

Use the Weibull CDF for probabilities. The values needed depend on given parameters.

Step 1: Understanding the Weibull Distribution

The Weibull distribution is a continuous probability distribution, commonly used in reliability analysis and risk assessment. It is described by two parameters: the shape parameter \(\beta\) and the scale parameter \(\delta\). For a Weibull-distributed random variable \(X\), the probability density function (pdf) is given by:\[ f(x; \beta, \delta) = \begin{cases} \frac{\beta}{\delta} \left(\frac{x}{\delta}\right)^{\beta-1} e^{-(x/\delta)^\beta} & x \geq 0 \ 0 & x < 0 \end{cases} \]

Step 2: Calculating Probability for Part (a)

To find the probability of a loss greater than \(2\) billion dollars, use the complementary cumulative distribution function (CCDF):\[ P(X > 2) = 1 - F(2; \beta, \delta) \]where \(F(x; \beta, \delta)\) is the cumulative distribution function (CDF) for the Weibull distribution:\[ F(x; \beta, \delta) = 1 - e^{-(x/\delta)^\beta} \]Substitute \(x = 2\), \(\beta = 0.8472\), and \(\delta = 1.9317\):\[ F(2; 0.8472, 1.9317) = 1 - e^{-(2/1.9317)^{0.8472}} \]Calculate \(P(X > 2)\).

Step 3: Calculating Probability for Part (b)

The probability that the loss is between \(2\) and \(4\) billion dollars is given by:\[ P(2 < X < 4) = F(4; \beta, \delta) - F(2; \beta, \delta) \]Substitute values:\[ F(4; 0.8472, 1.9317) = 1 - e^{-(4/1.9317)^{0.8472}} \]Calculate \(F(2; 0.8472, 1.9317)\) from Step 2, then find:\[ P(2 < X < 4) = (1 - e^{-(4/1.9317)^{0.8472}}) - (1 - e^{-(2/1.9317)^{0.8472}}) \]

Step 4: Finding Value Exceeded with Probability 0.05 (Part c)

The value \(x\) such that \(P(X > x) = 0.05\) is found using the CDF:\[ P(X > x) = 1 - F(x; \beta, \delta) = 0.05 \]Rearrange to find \( x \):\[ F(x; \beta, \delta) = 0.95 \]Set \(1 - e^{-(x/\delta)^\beta} = 0.95\), then solve:\[ e^{-(x/\delta)^\beta} = 0.05 \]\[ -(x/\delta)^\beta = \ln(0.05) \]\[ (x/\delta)^\beta = -\ln(0.05) \]Substitute \(\beta\) and \(\delta\), then solve for \(x\).

Step 5: Calculating Mean and Standard Deviation for Part (d)

For a Weibull distribution, the mean \(\mu\) and variance \(\sigma^2\) are calculated as follows:Mean: \( \mu = \delta \Gamma\left(1 + \frac{1}{\beta}\right) \)Variance: \( \sigma^2 = \delta^2 \left[ \Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2 \right] \)Calculate the mean and variance using the beta function values and then find the standard deviation \(\sigma = \sqrt{\sigma^2}\).

Key Concepts

Probability Calculation

When working with probability calculations in the Weibull distribution, we often deal with finding probabilities for specific intervals of loss or other real-world variables. Here's an approachable way to tackle these calculations using the Weibull distribution.

The first step in this process is to use the cumulative distribution function (CDF), denoted as \( F(x) \). For the Weibull distribution, it is given by:

• \( F(x; \beta, \delta) = 1 - e^{-(x/\delta)^\beta} \)

This function provides us with the probability that a random variable \( X \) is less than or equal to a certain value.

To find the probability of a loss greater than a certain value, such as \(2 billion, we use the complementary cumulative distribution function (CCDF):

• \( P(X > x) = 1 - F(x; \beta, \delta) \)

This calculates the probability that \( X \) exceeds a specified threshold. When dealing with different ranges, say between \)2 and $4 billion, you calculate the probability for each value separately and subtract one from the other.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is an integral tool in understanding Weibull distribution behavior. It gives the cumulative probability that a random variable \( X \) will take a value less than or equal to \( x \). This is particularly useful in risk assessment and reliability analysis, where you need to understand the likelihood of events occurring within certain bounds.

For the Weibull distribution, the formula is constructed as:

• \( F(x; \beta, \delta) = 1 - e^{-(x/\delta)^\beta} \)

The parameters \( \beta \) (shape) and \( \delta \) (scale) directly influence the distribution's behavior. By manipulating these parameters, you can see how probabilities of certain events might increase or decrease.

If you need a specific percentage chance, like calculating a threshold exceeded only 5% of the time, use the inverse CDF. Here, you set \( F(x; \beta, \delta) \) to your desired probability (e.g., 0.95 for a 5% exceedance) and solve for \( x \). This gives you a clear value or limit that corresponds with your reliability requirements.

Reliability Analysis

Reliability analysis is a critical application of the Weibull distribution, especially in predicting the performance and durability of components or systems. By utilizing the shape and scale parameters of the distribution, we can predict the lifespan or failure rates of various items.

The key to Weibull reliability analysis is understanding how the cumulative distribution function \( F(x; \beta, \delta) \) describes failure behavior. In contexts such as engineering or finance, a lower probability of failure within a certain time or event range can mean higher reliability and vice-versa.

When calculating reliability, the complementary CDF is particularly useful as it provides the probability of survival (or non-failure) beyond a specified point:

• \( R(x) = P(X > x) = 1 - F(x; \beta, \delta) \)

In practical applications, choosing appropriate \( \beta \) and \( \delta \) values to model past data can help forecast future reliability scenarios, assisting in decision-making processes like maintenance scheduling or financial risk management.

Statistical Mean and Variance

Understanding the statistical mean and variance of a Weibull distribution helps in quantifying the "average" expected performance and the spread or variability of outcomes. Such analyses are immensely beneficial in contexts like finance and risk management, where making informed predictions is crucial.

The mean (or expected value) of a Weibull distribution is given by:

• \( \mu = \delta \Gamma\left(1 + \frac{1}{\beta}\right) \)

Here, \( \Gamma \) represents the Gamma function, which extends the factorial function to non-integer values. This component addresses how the average loss or time to event is modeled based on \( \beta \) and \( \delta \).

Similarly, variance measures the distribution's variability:

• \( \sigma^2 = \delta^2 \left[ \Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2 \right] \)

The square root of variance gives the standard deviation \( \sigma \), denoting how much the actual outcomes typically differ from the mean. Insights from these calculations are invaluable for developing strategies to manage uncertainty and risk effectively.