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Chapter 4: Problem 14

Starting from a uniform random variable \(U \sim\) Uniform \((0,1)\), it is possible to construct many random variables through transformations. (a) Show that \(-\log U \sim \exp (1)\). (b) Show that \(-\sum_{i=1}^{n} \log U_{i} \sim \operatorname{Gamma}(n, 1)\), where \(U_{1}, \ldots, U_{n}\) are iid as \(U(0,1)\). (c) Let \(X \sim \operatorname{Exp}(a, b)\). Write \(X\) as a function of \(U\). (d) Let \(X \sim \operatorname{Gamma}(n, \beta), n\) an integer. Write \(X\) as a function of \(U_{1}, \ldots, U_{n}\), iid as \(U(0,1)\).

Short Answer

Expert verified
(a) \(-\log U \sim \exp(1)\). (b) \(-\sum_{i=1}^{n} \log U_i \sim \text{Gamma}(n,1)\). (c) \(X = a - b\log U\). (d) \(X = -\beta \sum_{i=1}^{n} \log U_i\).

Step by step solution

01

Exponential Transformation

Given that \( U \sim \text{Uniform}(0,1) \), the probability density function (PDF) of \( U \) is constant: \( f_U(u) = 1 \) for \( 0 < u < 1 \). To show \( Y = - \log U \sim \exp(1) \), we find the CDF of \( Y \). The transformation gives \( P(Y \leq y) = P(- \log U \leq y) = P(U \geq e^{-y}) = 1 - e^{-y} \). The PDF is the derivative: \( f_Y(y) = e^{-y} \) for \( y > 0 \), which is the PDF of an \( \exp(1) \) distribution.
02

Gamma Transformation Sum

To show \( -\sum_{i=1}^{n} \log U_i \sim \text{Gamma}(n, 1) \), we note that each \( -\log U_i \sim \exp(1) \) (from Step 1). Since \( \text{Gamma}(n, \beta) \) is the distribution of a sum of \( n \) independent \( \text{Exp}(\beta) \) random variables, and \( \sum_{i=1}^{n} -\log U_i \) is a sum of \( n \) independent \( \exp(1) \) random variables, by definition, this sum is \( \text{Gamma}(n, 1) \).
03

Writing Exponential Random Variable as Function of U

For \( X \sim \text{Exp}(a, b) \), the PDF is \( f_X(x) = \frac{1}{b}e^{-\frac{x-a}{b}} \). The standard transformation is \( X = a - b\log U \) where \( U \sim \text{Uniform}(0,1) \). This transformation matches the \( \exp(1) \) PDF through variable scaling and shifting.
04

Writing Gamma Random Variable as Function of Us

For \( X \sim \text{Gamma}(n, \beta) \), since \( M = -\sum_{i=1}^{n} \log U_i \sim \text{Gamma}(n, 1) \), and we need \( X \sim \text{Gamma}(n, \beta) \) which scales the sum of exponential random variables by \( \beta \). Therefore, \( X = \beta \left(-\sum_{i=1}^{n} \log U_i\right) = -\beta \sum_{i=1}^{n} \log U_i \) with \( U_i \sim \text{Uniform}(0,1) \).

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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 probability distributions. It is often used as a starting point for generating other random variables through transformations. In a uniform distribution over the interval (0, 1), each number between 0 and 1 is equally probable.The probability density function (PDF) for a uniform distribution is:
  • Uniform (0, 1):\[ f_U(u) = 1, \quad \text{for } 0 < u < 1\]
This distribution is critical in probability because it provides a base for building more complex distributions, such as the exponential or gamma distributions, through transformations. The inherent symmetry and equal probability make the uniform distribution a foundational concept in statistics and probability theory.
Exponential Distribution
The exponential distribution is a continuous probability distribution that describes the time between events in a process where events occur continuously and independently at a constant average rate. A key feature of this distribution is the memoryless property, meaning that the future probability does not depend on any past outcomes.When starting with a uniform random variable, a classic transformation to achieve an exponential distribution is:
  • \(-\log U\), where \(U\sim \text{Uniform}(0,1)\)
This transformation results in a new random variable that is distributed exponentially with rate 1, denoted as \(\exp(1)\). The PDF of an exponential distribution with rate \(\lambda = 1\) is:
  • \[f_Y(y) = e^{-y}, \quad \text{for } y > 0\]
This transformation is utilized in many practical applications, such as modeling lifetimes of objects and times until a given event.
Gamma Distribution
The gamma distribution is a two-parameter family of continuous probability distributions. It's an extension of the exponential distribution, useful for modeling the sum of multiple independent exponential random variables.To derive a gamma distribution from a uniform distribution, consider the following transformation:
  • \(-\sum_{i=1}^{n} \log U_i\), where each \(U_i\sim \text{Uniform}(0,1)\) and are independent
This sum follows a gamma distribution with shape parameter \(n\) and rate 1, denoted as \(\operatorname{Gamma}(n, 1)\). The gamma distribution's shape and scale parameters give it the flexibility to model a wide array of situations, particularly in queuing models and insurance risk theory.The PDF of a gamma distribution is:
  • \[f_X(x) = \frac{x^{n-1}e^{-x}}{(n-1)!}, \quad \text{for } x > 0\]
The transformation highlights the power and versatility of probability transformations in expanding upon the behavior modeled by an exponential distribution.
Random Variable Transformations
Random variable transformations are essential techniques in probability and statistics, enabling the modification of random variables to derive new distributions. This allows the modeling of complex real-world situations starting from simple random variables.For example, by applying transformations to a uniform random variable, we can produce various other distributions. Let's consider a few transformations:
  • **Exponential Transformation**: \( X = a - b\log U \) for \( X \sim \text{Exp}(a, b) \) and \( U \sim \text{Uniform}(0,1)\).
  • **Gamma Transformation**: \[ X = -\beta \sum_{i=1}^{n} \log U_i \] for \( X \sim \text{Gamma}(n, \beta) \) with \( U_i \sim \text{Uniform}(0,1)\).
These transformations help convert uniformly distributed random variables into exponentially and gamma-distributed ones, respectively. Understanding these transformations is crucial because they provide insight into generating different types of distributions from uniformicity, allowing statisticians and practitioners to model the stochastic behavior seen in various domains.

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Most popular questions from this chapter

If \(\Lambda\) is a left-invariant measure over \(G\), show that \(\Lambda^{*}\) defined by \(\Lambda^{*}(B)=\Lambda\left(B^{-1}\right)\) is right invariant, where \(B^{-1}=\left\\{g^{-1}: g \in B\right\\}\). [Hint: Express \(\Lambda^{*}(B g)\) and \(\Lambda^{*}(B)\) in terms of \(\Lambda\).]

The Taylor series approximation to the estimator ( \(5.5 .8\) ) is carried out in a number of steps. Show that: (a) Using a first-order Taylor expansion around the point \(\bar{x}\), we have $$ \begin{aligned} \frac{1}{\left(1+\theta^{2} / v\right)^{(v+1) / 2}}=& \frac{1}{\left(1+\bar{x}^{2} / \nu\right)^{(v+1) / 2}} \\ &-\frac{v+1}{v} \frac{\bar{x}}{\left(1+\bar{x}^{2} / v\right)^{(v+3) / 2}}(\theta-\bar{x})+R(\theta-\bar{x}) \end{aligned} $$ where the remainder, \(R(\theta-\bar{x})\), satisfies \(R(\theta-\bar{x}) /(\theta-\bar{x})^{2} \rightarrow 0\) as \(\theta \rightarrow \bar{x}\). (b) The remainder in part (a) also satisfies $$ \int_{-\infty}^{\infty} R(\theta-\bar{x}) e^{-\frac{p}{2 \sigma^{2}}(\theta-\bar{x})^{2}} d \theta=O\left(1 / p^{3 / 2}\right) $$ (c) The numerator and denominator of \((5.5 .8)\) can be written $$ \int_{-\infty}^{\infty} \frac{1}{\left(1+\theta^{2} / \nu\right)^{(\nu+1) / 2}} e^{-\frac{p}{2 \sigma^{2}}(\theta-\bar{x})^{2}} d \theta=\frac{\sqrt{2 \pi \sigma^{2} / p}}{\left(1+\bar{x}^{2} / \nu\right)^{(\nu+1) / 2}}+O\left(\frac{1}{p^{3 / 2}}\right) $$ and $$ \begin{aligned} \int_{-\infty}^{\infty} & \frac{\theta}{\left(1+\theta^{2} / \nu\right)^{(v+1) / 2}} e^{-\frac{p}{2 \sigma^{2}}(\theta-\bar{x})^{2}} d \theta \\ &=\frac{\sqrt{2 \pi \sigma^{2} / p}}{\left(1+\bar{x}^{2} / \nu\right)^{(v+1) / 2}}\left[1-\frac{(v+1) / v}{\left(1+\bar{x}^{2} / v\right)}\right] \bar{x}+O\left(\frac{1}{p^{3 / 2}}\right) \end{aligned} $$ which yields \((5.6 .32)\).

A measure \(\Lambda\) over a group \(G\) is said to be right invariant if it satisfies \(\Lambda(B g)=\Lambda(B)\) and left invariant if it satisfies \(\Lambda(g B)=\Lambda(B)\). Note that if \(G\) is commutative, the two definitions agree. (a) If the elements \(g \in G\) are real numbers \((-\infty

Let \(X\) and \(Y\) be independently distributed according to distributions \(P_{\xi}\) and \(Q_{\eta}\), respectively. Suppose that \(\xi\) and \(\eta\) are real-valued and independent according to some prior distributions \(\Lambda\) and \(\Lambda^{\prime} .\) If, with squared error loss, \(\delta_{\Lambda}\) is the Bayes estimator of \(\xi\) on the basis of \(X\), and \(\delta_{\Lambda^{\prime}}^{\prime}\) is that of \(\eta\) on the basis of \(Y\), (a) show that \(\delta_{\Lambda^{\prime}}^{\prime}-\delta_{\Lambda}\) is the Bayes estimator of \(\eta-\xi\) on the basis of \((X, Y)\); (b) if \(\eta>0\) and \(\delta_{\Lambda^{\prime}}^{*}\) is the Bayes estimator of \(1 / \eta\) on the basis of \(Y\), show that \(\delta_{\Lambda} \cdot \delta_{\Lambda^{\prime}}^{*}\) is the Bayes estimator of \(\xi / \eta\) on the basis of \((X, Y)\).

The empirical Bayes estimator (7.7.27) can also be derived as a hierarchical Bayes estimator. Consider the hierarchical model $$ \begin{aligned} X_{i j} \mid \xi_{i} & \sim N\left(\xi_{i}, \sigma^{2}\right), \quad j=1, \ldots, n, \quad i=1, \ldots, s \\ \xi_{i} \mid \mu & \sim N\left(\mu, \tau^{2}\right), \quad i=1, \ldots, s \\ \mu & \sim \text { Uniform }(-\infty, \infty) \end{aligned} $$ where \(\sigma^{2}\) and \(\tau^{2}\) are known. (a) Show that the Bayes estimator, with respect to squared error loss, is $$ E\left(\xi_{i} \mid \mathbf{x}\right)=\frac{\sigma^{2}}{\sigma^{2}+n \tau^{2}} E(\mu \mid \mathbf{x})+\frac{n \tau^{2}}{\sigma^{2}+n \tau^{2}} \bar{x}_{i} $$ where \(E(\mu \mid \mathbf{x})\) is the posterior mean of \(\mu\). (b) Establish that \(E(\mu \mid \mathbf{x})=\bar{x}=\Sigma x_{i j} / n s\). [This can be done by evaluating the expectation directly, or by showing that the posterior distribution of \(\xi_{i} \mid \mathbf{x}\) is $$ \xi_{i} \mid \mathbf{x} \sim N\left[\frac{\sigma^{2}}{\sigma^{2}+n \tau^{2}} \bar{x}+\frac{n \tau^{2}}{\sigma^{2}+n \tau^{2}} \bar{x}_{i}, \frac{\sigma^{2}}{\sigma^{2}+n \tau^{2}}\left(n \tau^{2}+\frac{\sigma^{2}}{s}\right)\right] $$ Note that the \(\xi_{i}\) 's are not independent a posteriori. In fact, $$ \xi \mid \mathbf{x} \sim N_{s}\left(\frac{n \tau^{2}}{\sigma^{2}+n \tau^{2}} M, \frac{n \sigma^{2} \tau^{2}}{\sigma^{2}+n \tau^{2}} M\right) $$ where \(\left.M=I+\left(\sigma^{2} / n \tau^{2}\right) J\right]\) (c) Show that the empirical Bayes estimator (7.7.32) can also be derived as a hierarchical Bayes estimator, by appending the specification \((\alpha, \beta) \sim\) Uniform \(\left(\Re^{2}\right)\) [that is, \(\pi(\alpha, \beta)=d \alpha d \beta,-\infty<\alpha, \beta<\infty]\) to the hierarchy \((7.7 .30)\).
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