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Chapter 4: Problem 206

Suppose that \(X\) has a lognormal distribution with parameters \(\theta=0\) and \(\omega^{2}=4 .\) Determine the following: (a) \(P(10

Short Answer

Expert verified
(a) Use standard normal table: P(1.151 < Z < 1.956). (b) Find x using x=e^{2(-1.645)}. (c) Mean: e^2, Variance: (e^4-1)e^4.

Step by step solution

01

Understanding the Lognormal Distribution

A lognormal distribution indicates that if a random variable is log-normally distributed, its natural logarithm is normally distributed. Here, if \(X\) is log-normally distributed with parameters \(\theta = 0\) and \(\omega^2 = 4\), then \(\ln X\) is normally distributed with mean \(\mu = 0\) and variance \(\sigma^2 = 4\).
02

Step (a): Calculate the Probabilities

To find \(P(10<X<50)\), convert \(X\) to the normal distribution by using \(Z = \frac{\ln(X) - \mu}{\sigma}\). For \(X = 10\), \(\ln(10)\) is approximately 2.302. The standard normal variable is \(Z = \frac{2.302 - 0}{2} = 1.151\). Similarly, for \(X = 50\), \(\ln(50)\) is approximately 3.912. The standard normal variable is \(Z = \frac{3.912 - 0}{2} = 1.956\). Use the standard normal distribution table to find \(P(1.151 < Z < 1.956)\).
03

Step (b): Find the Quantile for 5%

To find the \(x\) such that \(P(X<x) = 0.05\), find the 5th percentile of the standard normal distribution, which is approximately \(-1.645\). Use the relationship \(x = e^{\sigma Z + \mu}\) to find the value, \(x = e^{2(-1.645) + 0}\). Simplify this to get \(x\).
04

Step (c): Calculate Mean and Variance

The mean and variance of a lognormal distribution are given by \(E[X] = e^{\mu + \frac{\sigma^2}{2}}\) and \(\text{Var}(X) = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2}\). Compute these using \(\mu = 0\) and \(\sigma^2 = 4\): \(E[X] = e^{\frac{4}{2}} = e^2\), and \(\text{Var}(X) = (e^4 - 1)e^4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculations
Understanding probability calculations in the context of a lognormal distribution involves converting the lognormal variable into a standard normal variable. This is because while direct calculation with a lognormal distribution can be complex, the transformation to a normal distribution simplifies things. Here鈥檚 how we solve this:
  • Given that you need to find the probability of a range, such as \(P(10 < X < 50)\), you begin by using the transformation \(Z = \frac{\ln(X) - \mu}{\sigma}\).
  • For \(X = 10\), calculate \(\ln(10)\), which results in approximately 2.302, and then standardize it: \(Z = \frac{2.302 - 0}{2} = 1.151\).
  • Similarly, for \(X = 50\), \(\ln(50)\) is about 3.912, giving \(Z = \frac{3.912 - 0}{2} = 1.956\).
  • Finally, use a standard normal distribution table or calculator to determine the probability \(P(1.151 < Z < 1.956)\).
By transforming the variable and referring to the probability table, you can effectively determine the likelihood of the random variable falling within a specified range.
Quantile Calculation
Finding a quantile in a lognormal distribution is essentially identifying a value \(x\) such that a certain percentage of the data lies below this value. In this example, we seek the 5th percentile, meaning we are looking for a value where 5% of the probability distribution lies below it. Here's the procedure:
  • Locate the equivalent quantile in the standard normal distribution for which \(P(Z < z) = 0.05\). From standard normal tables, this \(z\)-value is approximately \(-1.645\).
  • Use the conversion formula for a lognormal distribution: \(x = e^{\sigma z + \mu}\).
  • Substitute in the values for \(\sigma = 2\) and \(\mu = 0\), giving: \(x = e^{2(-1.645) + 0}\).
  • Calculate \(x\) to find the value where 5% of the distribution lies below it.
This step effectively links the standard normal distribution to the lognormal, allowing you to find precise probability percentiles needed for decision-making or analysis.
Mean and Variance of Lognormal Distribution
Understanding the mean and variance of a lognormal distribution involves using specific formulas derived from its properties. Lognormal distributions can be tricky because they are inherently skewed, but deterministic calculations help quantify their central tendency and spread:
  • The mean of a lognormal distribution \(E[X]\) is calculated using: \(E[X] = e^{\mu + \frac{\sigma^2}{2}}\). Plug in \(\mu = 0\) and \(\sigma^2 = 4\): \(E[X] = e^{0 + 2} = e^2\).
  • Variance \(\text{Var}(X)\) is determined by: \(\text{Var}(X) = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2}\). With the given parameters, solve: \(\text{Var}(X) = (e^4 - 1)e^4\).
By using these equations, you can precisely calculate the mean, which gives a measure of expected value, and the variance, which reflects the data's spread around the mean. Such calculations are crucial in assessing the behavior of any lognormally distributed variable.

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