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Chapter 4: Problem 114

Let \(X\) have the uniform distribution on \([0,1]\). Find the pdf of \(Y=-\ln (X)\).

Short Answer

Expert verified
The pdf of \(Y\) is \(f_Y(y) = e^{-y}\) for \(y \geq 0\).

Step by step solution

01

Identify the original probability density function (pdf)

Given that the random variable \(X\) is uniformly distributed over the interval \([0, 1]\), we have the pdf of \(X\) as \( f_X(x) = 1 \) for \( 0 \leq x \leq 1 \) and \( f_X(x) = 0 \) otherwise.
02

Define the transformation

We are given \(Y = -\ln(X)\). This transformation is a one-to-one function from \( (0, 1) \) to \( (0, \infty) \), since the natural logarithm and the negation are both monotonically increasing in the specified ranges.
03

Find the inverse transformation

To find the inverse, start with \(y = -\ln(x)\). Solving for \(x\), we get \( x = e^{-y} \). This step will help us later in obtaining the form of \( f_Y(y) \).
04

Compute the derivative for the change of variables

Calculate the derivative of the inverse transformation \( x = e^{-y} \). The derivative is \( \frac{dx}{dy} = -e^{-y} \).
05

Apply the change of variables formula

The pdf of \(Y\), \(f_Y(y)\), is given by \( f_Y(y) = f_X(x) \cdot \left| \frac{dx}{dy} \right| \) where \( x = e^{-y} \). Thus,\[ f_Y(y) = 1 \times e^{-y} = e^{-y}\]for \( y \geq 0 \) because that is the range of \( Y \).
06

Define the support of the new pdf

Since \( y = -\ln(x) \) implies \( y \geq 0 \) when \( x \) is in \( (0, 1) \), the pdf \( f_Y(y) = e^{-y} \) is valid for \( y \geq 0 \).
07

Verification

Verify that \( f_Y(y) = e^{-y} \) integrates to 1 over its support. Calculate \( \int_{0}^{\infty} e^{-y} \, dy = 1 \), confirming it is a valid pdf.

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Key Concepts

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

Uniform Distribution
The uniform distribution is one of the simplest and most intuitive types of probability distributions in statistics. When we say a random variable is uniformly distributed, it means that each outcome in a given range is equally likely. In our exercise, the variable
  • \( X \) is uniformly distributed over the interval \([0,1]\).
  • This implies every number between 0 and 1 is equally probable.
This interval not only defines the range of our random variable, but it also helps establish the probability density function (pdf). The pdf for a uniformly distributed variable over \([0,1]\) is constant, specifically, \(f_X(x) = 1\) for \(0 \leq x \leq 1\), meaning that each point in this range has the same likelihood.This simplicity is what makes the uniform distribution a fundamental concept in probability theory.
Random Variables
A random variable is a fascinating concept in probability and statistics. It is essentially a ___variable that can take different values due to some random phenomena___.In the scenario of this exercise:
  • \(X\) is our initial random variable, and it is defined as uniformly distributed over \([0,1]\).
  • \(Y\) is another random variable derived as a function of \(X\), specifically by using a logarithmic transformation.
The transformation of \(X\) to \(Y\) means that for each outcome of \(X\), we compute a corresponding value of \(Y\) by the rule \(Y = -\ln(X)\).Effectively, random variables allow us to create new variables from existing ones, based on a formula or function that describes the transformation.
Transformation of Variables
Transforming a random variable introduces us to new probability distributions and insights.In mathematical terms, a transformation redefines what values a new random variable (in our case, \(Y\)) can take.
  • For the given exercise, the transformation used is the natural logarithm, \(Y = -\ln(X)\).
  • This transformation is monotonic, preserving the order of values, and it is invertible, meaning we can find the original value if we know \(Y\).
A crucial part of this process is finding the inverse transformation.
  • Here, we start with \( y = -\ln(x) \) and solve for \(x\), getting \( x = e^{-y} \).
  • This allows us to continue to the calculation of the pdf of \(Y\) using the change of variables technique.
Through these steps, we change the landscape of possibilities for our random variable \(Y\), leading us to explore new properties and applications.
Exponential Distribution
The exponential distribution is a widely used probability distribution in statistics, especially in processes that involve time until an event occurs, like waiting times.In our exercise, through a transformation,
  • the random variable \(Y = -\ln(X)\) results in the exponential distribution.
  • The derived pdf for \(Y\), \(f_Y(y) = e^{-y}\) for \(y \geq 0\), represents this.
The exponential distribution is important because:- It is used to model time between independent events that happen at a constant average rate.- It is characterized by its constant hazard rate and is memoryless.When we apply the transformation on \(X\), we show that \(Y\) adheres to an exponential distribution 鈥搕his signifies a strong relationship between transformations and different types of distributions in statistical models.

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Most popular questions from this chapter

Let \(X\) have the geometric distribution with pmf \(p_{X}(x)=(1-p)^{2} p, x=0,1,2, \ldots\). Find the pmf of \(Y=X+1\). The resulting distribution is also referred to as geometric (see Example 3.10).

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within \(2 \mathrm{~min}\) after the hour. Let \(X=\) the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of \(X\) is $$ f(x)=\left\\{\begin{array}{cc} k x^{2} & 0 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Find the value of \(k\). [Hint: Total area under the graph of \(f(x)\) is 1.] b. What is the probability that the lecture ends within \(1 \mathrm{~min}\) of the end of the hour? c. What is the probability that the lecture continues beyond the hour for between 60 and \(90 \mathrm{~s}\) ? d. What is the probability that the lecture continues for at least \(90 \mathrm{~s}\) beyond the end of the hour?

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly ten defectives in the batch?

A theoretical justification based on a material failure mechanism underlies the assumption that ductile strength \(X\) of a material has a lognormal distribution. Suppose the parameters are \(\mu=5\) and \(\sigma=.1\). a. Compute \(E(X)\) and \(V(X)\). b. Compute \(P(X>125)\). c. Compute \(P(110 \leq X \leq 125)\). d. What is the value of median ductile strength? e. If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength of at least 125 ? f. If the smallest \(5 \%\) of strength values were unacceptable, what would the minimum acceptable strength be?

The grade point averages (GPA's) for graduating seniors at a college are distributed as a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} k\left[1-(x-3)^{2}\right] & 2 \leq x \leq 4 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Find the value of \(k\). c. Find the probability that a GPA exceeds \(3 .\) d. Find the probability that a GPA is within \(.25\) of 3 . e. Find the probability that a GPA differs from 3 by more than \(5 .\)
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