Problem 88
Question
Derive the probability density function for a lognormal random variable \(Y\) from the relationship that \(Y=\exp (W)\) for a normal random variable \(W\) with mean \(\theta\) and variance $\omega^{2}.
Step-by-Step Solution
Verified Answer
The pdf of a lognormal random variable is \( f_Y(y) = \frac{1}{y\sqrt{2\pi}\omega} e^{-\frac{(\ln(y)-\theta)^2}{2\omega^2}}, \) for \( y > 0.\)
1Step 1: Understand the Transformations
Recall that a lognormal random variable, \( Y \), is defined as \( Y = \exp(W) \), where \( W \) is a normal random variable with mean \( \theta \) and variance \( \omega^2 \). First, we understand that \( W \sim N(\theta, \omega^2) \).
2Step 2: Determine the PDF of the Normal Variable
The probability density function (pdf) of the normal variable \( W \) is given by:\[ f_W(w) = \frac{1}{\sqrt{2\pi}\omega} e^{-\frac{(w-\theta)^2}{2\omega^2}}.\]
3Step 3: Use Change of Variables to Find PDF of Y
Since \( Y = e^W \), the relationship for the probability densities is given by the change of variables technique. Let \( W = \ln(Y) \) so that:\[ f_Y(y) = f_W(\ln(y)) \left| \frac{d}{dy} (\ln(y)) \right|.\]The derivative \( \frac{d}{dy}(\ln(y)) \) is \( \frac{1}{y} \).
4Step 4: Plug in the Values to the Density Function
Substitute \( \ln(y) \) into the pdf of \( W \):\[ f_Y(y) = \frac{1}{\sqrt{2\pi}\omega} e^{-\frac{(\ln(y)-\theta)^2}{2\omega^2}} \cdot \frac{1}{y}.\]
5Step 5: Express the Final PDF of the Lognormal Distribution
Thus, the probability density function for the lognormal random variable \( Y \) is:\[ f_Y(y) = \frac{1}{y\sqrt{2\pi}\omega} e^{-\frac{(\ln(y)-\theta)^2}{2\omega^2}}, \quad y > 0.\] This final formula represents the lognormal distribution for the variable \( Y \).
Key Concepts
Probability Density FunctionNormal DistributionChange of Variables Method
Probability Density Function
In the realm of probability and statistics, the probability density function (pdf) serves a pivotal role in determining the likelihood of a continuous random variable assuming a particular value. Unlike discrete distributions, where probabilities of specific values can be directly identified, continuous distributions use the pdf to obtain probabilities over intervals. The pdf, therefore, acts as a function that describes how dense the probability is at different points in a continuous sample space. This density function provides us the structure to compute probabilities, as the probability of a variable falling within a certain range involves integrating the pdf over that interval.
- The area under the pdf curve across an interval equates to the probability of the variable falling within that range.
- The total area under the pdf over its entire domain must be exactly 1, ensuring that the probability sums up to a complete certainty.
- A specific value for a continuous random variable, theoretically, has a zero probability since the area at a single point is zero; hence, we focus on ranges.
Normal Distribution
The normal distribution is a cornerstone in the field of statistics and probability, recognized by its bell-shaped and symmetric curve, often referred to as the Gaussian distribution. It is entirely defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). The pdf of a normal distribution is pivotal in many statistical methods and theories because it naturally occurs in many real-life situations, such as heights, test scores, and measurement errors.
- The mean (\( \mu \)) indicates the central location of the distribution, showing where the peak is located.
- The standard deviation (\( \sigma \)) measures the spread or dispersion of the distribution, indicating how "wide" the bell curve is.
- Approximately 68% of data within a normal distribution lies within one standard deviation from the mean, and about 95% within two standard deviations.
Change of Variables Method
The change of variables method is a powerful technique in calculus and statistics that allows us to find the distribution of a transformed random variable. By changing the variable, we express one random variable in terms of another, thus enabling us to determine the new distribution.
- Start by identifying the relationship between the old and new variables. For instance, if we have \( Y = e^W \), where \( W \sim N(\theta, \omega^2) \), the function of interest transforms the variable \( W \) into \( Y \).
- Calculate the derivative of this relationship to help adjust the density function. This involves differentiating the new variable with respect to the old one, such as \( \frac{d}{dy}(\ln(y)) = \frac{1}{y} \).
- Use these derivatives to adjust the probability density function by scaling with the absolute value of the derivative, ensuring that the function remains correctly normalized.
Other exercises in this chapter
Problem 86
The random variable \(X\) has the probability distribution $$ f_{X}(x)=\frac{x}{8}, \quad 0 \leq x \leq 4 $$ Determine the probability distribution of \(Y=(X-2)
View solution Problem 87
An aircraft is flying at a constant altitude with velocity magnitude \(r_{1}\) (relative to the air) and angle \(\theta_{1}\) (in a twodimensional coordinate sy
View solution Problem 89
The computational time of a statistical analysis applied to a data set can sometimes increase with the square of \(N,\) the number of rows of data. Suppose that
View solution Problem 90
Power meters enable cyclists to obtain power measurements nearly continuously. The meters also calculate the average power generated over a time interval. Profe
View solution