Problem 171 - Applied Statistics and Probability for Engineers

Question

Suppose that \(X\) has a lognormal distribution with parameters \(\theta=-2\) and \(\omega^{2}=9 .\) Determine the following: (a) \(P(500

Step-by-Step Solution

Answer

(a) Calculate using the normal CDF for Z-scores. (b) Use normal inverse CDF for 10th percentile. (c) Use formulas for mean and variance of lognormal distribution.

Step 1: Understanding the Lognormal Distribution

A random variable \(X\) is lognormally distributed if \(\ln(X)\) is normally distributed. The lognormal distribution has parameters \(\theta\) (the mean of \(\ln(X)\)) and \(\omega^2\) (the variance of \(\ln(X)\)). For this problem, \(\theta = -2\) and \(\omega^2 = 9\).

Step 2: Converting the Interval

To find \(P(500

Step 3: Standardizing the Variables

Find the standard normal equivalent values (\(Z\)-scores) for \(\ln(500)\) and \(\ln(1000)\): \[ Z_1 = \frac{\ln(500) - (-2)}{3}, \quad Z_2 = \frac{\ln(1000) - (-2)}{3} \]Compute the actual \(Z\)-scores.

Step 4: Calculating the Probability

Use standard normal distribution tables (or software) to find \(P(Z_1

Step 5: Finding the 10th Percentile

To find the value \(x\) such that \(P(X

Step 6: Calculating the Mean of X

The mean \(E(X)\) of a lognormal distribution is given by:\[ E(X) = e^{\theta + \frac{\omega^2}{2}}\] Substitute the values to get \(E(X)\).

Step 7: Calculating the Variance of X

The variance \(Var(X)\) of a lognormal distribution is given by:\[ Var(X) = \left(e^{\omega^2} - 1\right) e^{2\theta + \omega^2}\] Calculate \(Var(X)\) using the given \(\theta\) and \(\omega^2\).

Key Concepts

Probability Calculations

When dealing with lognormal distributions, probability calculations often revolve around understanding the transformation that links a lognormal random variable to a normal distribution. To find probabilities for specific intervals of a lognormally distributed variable like determining \(P(500 < X < 1000)\), you start by transforming these bounds using natural logarithms. This is because a variable \(X\) is lognormally distributed if \(\ln(X)\) is normally distributed. The given parameters \(\theta = -2\) and \(\omega^2 = 9\) indicate that \(\ln(X)\) follows a normal distribution with a mean of \(-2\) and a standard deviation of \(3\) (since the standard deviation is the square root of the variance).

• Calculate \(\ln(500)\) and \(\ln(1000)\).
• Find the corresponding Z-scores using the formula \(Z = \frac{\ln(x) - \theta}{\omega}\).
• Use standard normal distribution tables to compute \(P(Z_1 < Z < Z_2)\), which gives the probability for the desired interval.

This method highlights that understanding the relationship between the lognormal distribution and its normal counterpart is essential in solving such probability problems.

Mean of Lognormal Distribution

The mean of a lognormal distribution, denoted as \(E(X)\), is slightly more complex than regular distributions due to the exponential transformation involved. For a lognormal distribution with parameters \(\theta\) and \(\omega^2\), the mean is calculated using the formula: \[ E(X) = e^{\theta + \frac{\omega^2}{2}} \] This formula arises because the natural logarithm of a lognormally distributed random variable results in a normally distributed variable. Therefore, the exponential factor accounts for both the mean and half of the variance of the associated normal distribution. Inserting the values from the exercise:

• With \(\theta = -2\) and \(\omega^2 = 9\), the mean is \(e^{-2 + \frac{9}{2}} = e^{2.5} \).
• This computation reflects how the variance of \(\ln(X)\) influences the location and dispersion of \(X\).

Understanding this concept is crucial for anyone looking to apply lognormal distributions in fields like finance or real-world modeling.

Variance of Lognormal Distribution

The variance of a lognormal distribution characterizes the spread of the distribution and can be derived from its underlying normal distribution parameters. The formula for the variance \(Var(X)\) of a lognormal distribution is: \[ Var(X) = \left(e^{\omega^2} - 1\right) e^{2\theta + \omega^2} \] This equation considers the original parameters and provides an exponentially weighted spread measure. Here's how it's applied in the context of the problem:

• Insert \(\theta = -2\) and \(\omega^2 = 9\) into the formula.
• First, compute \(e^{9}\) and subtract one from it.
• Then multiply this result by \(e^{2 \cdot (-2) + 9}\).

This systematically reveals the variance, which is usually larger due to the exponential nature of the lognormal distribution. This significant variance indicates that the distribution is skewed with a long tail on the right, which is a hallmark of lognormal behavior.

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Problem 172

Other exercises in this chapter

Problem 169

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Problem 170

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Problem 172

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Problem 173

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Key Concepts

Probability Calculations

When dealing with lognormal distributions, probability calculations often revolve around understanding the transformation that links a lognormal random variable to a normal distribution. To find probabilities for specific intervals of a lognormally distributed variable like determining \(P(500 < X < 1000)\), you start by transforming these bounds using natural logarithms. This is because a variable \(X\) is lognormally distributed if \(\ln(X)\) is normally distributed. The given parameters \(\theta = -2\) and \(\omega^2 = 9\) indicate that \(\ln(X)\) follows a normal distribution with a mean of \(-2\) and a standard deviation of \(3\) (since the standard deviation is the square root of the variance).

• Calculate \(\ln(500)\) and \(\ln(1000)\).
• Find the corresponding Z-scores using the formula \(Z = \frac{\ln(x) - \theta}{\omega}\).
• Use standard normal distribution tables to compute \(P(Z_1 < Z < Z_2)\), which gives the probability for the desired interval.

This method highlights that understanding the relationship between the lognormal distribution and its normal counterpart is essential in solving such probability problems.

Mean of Lognormal Distribution

The mean of a lognormal distribution, denoted as \(E(X)\), is slightly more complex than regular distributions due to the exponential transformation involved. For a lognormal distribution with parameters \(\theta\) and \(\omega^2\), the mean is calculated using the formula: \[ E(X) = e^{\theta + \frac{\omega^2}{2}} \] This formula arises because the natural logarithm of a lognormally distributed random variable results in a normally distributed variable. Therefore, the exponential factor accounts for both the mean and half of the variance of the associated normal distribution. Inserting the values from the exercise:

• With \(\theta = -2\) and \(\omega^2 = 9\), the mean is \(e^{-2 + \frac{9}{2}} = e^{2.5} \).
• This computation reflects how the variance of \(\ln(X)\) influences the location and dispersion of \(X\).

Understanding this concept is crucial for anyone looking to apply lognormal distributions in fields like finance or real-world modeling.

Variance of Lognormal Distribution

The variance of a lognormal distribution characterizes the spread of the distribution and can be derived from its underlying normal distribution parameters. The formula for the variance \(Var(X)\) of a lognormal distribution is: \[ Var(X) = \left(e^{\omega^2} - 1\right) e^{2\theta + \omega^2} \] This equation considers the original parameters and provides an exponentially weighted spread measure. Here's how it's applied in the context of the problem:

• Insert \(\theta = -2\) and \(\omega^2 = 9\) into the formula.
• First, compute \(e^{9}\) and subtract one from it.
• Then multiply this result by \(e^{2 \cdot (-2) + 9}\).

This systematically reveals the variance, which is usually larger due to the exponential nature of the lognormal distribution. This significant variance indicates that the distribution is skewed with a long tail on the right, which is a hallmark of lognormal behavior.