Problem 174 - Applied Statistics and Probability for Engineers

Question

Suppose that \(X\) has a lognormal distribution and that the mean and variance of \(X\) are 100 and \(85,000,\) respectively. Determine the parameters \(\theta\) and \(\omega^{2}\) of the lognormal distribution. [Hint: define \(x=\exp (\theta)\) and \(y=\exp \left(\omega^{2}\right)\) and write two equations in terms of \(x\) and \(y .]\)

Step-by-Step Solution

Answer

\(\theta \approx \ln(73.45)\), \(\omega^2 = \ln(1.85)\).

Step 1: Understanding the Lognormal Distribution

A lognormal distribution arises when a random variable's logarithm is normally distributed. This means that if \(Y \sim \mathcal{N}(\mu, \sigma^2)\), then \(X = e^Y\) is lognormally distributed, with parameters \(\theta = \mu\) and \(\omega^2 = \sigma^2\).

Step 2: Express Mean in Terms of Parameters

The mean of a lognormally distributed variable \(X\) is given by the formula: \(\mathbb{E}[X] = e^{\theta + \frac{\omega^2}{2}}\). Given that the mean is 100, our equation becomes \(e^{\theta + \frac{\omega^2}{2}} = 100\).

Step 3: Express Variance in Terms of Parameters

The variance of a lognormal distribution is given by \(\text{Var}(X) = (e^{\omega^2} - 1) e^{2\theta + \omega^2}\). Given that the variance is 85,000, our second equation is \((e^{\omega^2} - 1) e^{2\theta + \omega^2} = 85,000\).

Step 4: Substitute for Mean to Simplify

Solving the mean equation \(e^{\theta + \frac{\omega^2}{2}} = 100\), we express \(e^{\theta}\) and \(e^{\omega^2}\) in terms of known quantities: if \(x = e^{\theta}\) and \(y = e^{\omega^2}\), then \(x y^{1/2} = 100\). Thus, \(x = \frac{100}{y^{1/2}}\).

Step 5: Substitute and Simplify the Variance Equation

Substitute \(x = \frac{100}{y^{1/2}}\) from Step 4 into the variance equation, which expressed in terms of \(x\) and \(y\) is \((y - 1)x^2 y = 85,000\). This gives us \((y - 1) \cdot \left(\frac{100}{y^{1/2}}\right)^2 \cdot y = 85,000\).

Step 6: Solve for \(y\)

Simplifying further, we have \((y - 1) \cdot \frac{10000}{y} = 85,000\), which leads to the equation: \(10000(y - 1) = 85,000 y\). Solve this to find \(y = 1.85\).

Step 7: Solve for \(x\)

With \(y = e^{\omega^2} = 1.85\), substitute back into the expression for \(x\): \(x = \frac{100}{y^{1/2}} = \frac{100}{\sqrt{1.85}}\). Calculate to find \(x \approx 73.45\).

Step 8: Determine \(\theta\) and \(\omega^2\) from \(x\) and \(y\)

Since \(x = e^{\theta}\) and \(y = e^{\omega^2}\), \(\theta = \ln(x) \approx \ln(73.45)\) and \(\omega^2 = \ln(y) = \ln(1.85)\).

Key Concepts

Mean of Lognormal Distribution

The mean of a lognormal distribution is an essential concept because it gives a measure of the central tendency of the variable. For a lognormally distributed variable, say \(X\), the mean is calculated using this formula:

• \(\mathbb{E}[X] = e^{\theta + \frac{\omega^2}{2}}\)

Here, \(\theta\) is the mean of the associated normal distribution of the variable's logarithm, and \(\omega^2\) is the variance of the normal distribution. It's crucial to understand that this formula forms the link between the parameters of the normal distribution and the observed mean of the lognormal distribution.
To solve a problem in practice, substitute known values such as the mean of 100. This turns the formula into a specific equation to solve, like \(e^{\theta + \frac{\omega^2}{2}} = 100\). This step helps in progressing towards finding the unknown parameters of the distribution.

Variance of Lognormal Distribution

The variance of a lognormal distribution provides insight into how spread out the values of the variable are. It is computed using the following formula:

• \(\text{Var}(X) = (e^{\omega^2} - 1) e^{2\theta + \omega^2}\)

This formula showcases the relationship between the normal distribution parameters \(\theta\) and \(\omega^2\), and the variance of the lognormal distribution. For a given variance, like 85,000 in our problem, this formula transforms into a specific equation: \((e^{\omega^2} - 1) e^{2\theta + \omega^2} = 85,000\). Understanding and using this equation is vital for deducing the lognormal parameters. The variance provides a quantifiable measure of variability, making it an essential aspect of descriptive statistics.

Parameters of Lognormal Distribution

The parameters \(\theta\) and \(\omega^2\) are foundational for characterizing a lognormal distribution. These are derived by taking the natural logarithm to connect with the normal distribution from which a lognormal distribution is formed.

• \(\theta\) (or the location parameter) represents the mean of the normal distribution of the logarithm of the variable.
• \(\omega^2\) or \(\sigma^2\) (known as the scale parameter) signifies the variance of that normal distribution.

To find these parameters, define transformations like \(x = e^{\theta}\) and \(y = e^{\omega^2}\). From the equations derived from mean and variance, solving for these transformations gives the values of \(x\) and \(y\) as intermediary steps.
Subsequently, using inverse logarithms allows determination of \(\theta\) and \(\omega^2\): \(\theta = \ln(x)\) and \(\omega^2 = \ln(y)\). The understanding of these parameters enables one to fully describe the variability and distribution shape, preparing you to handle real-world data exhibiting skewness akin to lognormal behavior.

Mathematical Problem Solving

Mathematical problem solving is a skill that involves breaking complex problems into manageable steps. When addressing exercises like finding parameters of a lognormal distribution, it's crucial to:

• Comprehend the problem and identify the known (mean and variance) and unknown elements (\(\theta\) and \(\omega^2\)).
• Transform abstract mathematical expressions into applicable equations based on real values, simplifying the problem.
• Use algebraic principles to manipulate equations, leading to the substitution method, where new variables (here, \(x\) and \(y\)) help untangle intricate relationships.
• Solve for intermediary variables and backtrack to original unknowns using calculated values.

Practicing such problem-solving strategies not only refines your existing skills but also enhances analytical thinking and confidence when dealing with complex theoretical or applied statistics problems. By honing these approaches, one becomes adept at navigating the bridge from theoretical distributions to practical data interpretation.