/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 If \(f\) is the density function... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 30

If \(f\) is the density function of a normal random variable with mean \(\mu\) and variance \(\sigma^{2}\), show that the tilted density \(f_{t}\) is the density of a normal random variable with mean \(\mu+\sigma^{2} t\) and variance \(\sigma^{2}\).

Short Answer

Expert verified
The tilted density function \(f_t(x)\) is given by: \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2 - 2(\mu+\sigma^2 t)x + (\mu+\sigma^2 t)^2 - \sigma^4 t^2}{2\sigma^2}\right)\] After comparing this expression to a standard normal probability density function, we can conclude that the tilted density function of a normal random variable is also a normal random variable with mean \(\mu+\sigma^2 t\) and variance \(\sigma^2\).

Step by step solution

01

Define the normal density function f

The probability density function of a normal random variable with mean \(\mu\) and variance \(\sigma^2\) is given by: \[f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\]
02

Define the tilted density function f_t

Now we find the tilted density \(f_t(x)\): \[f_t(x) = \frac{1}{Z} f(x) \exp(tx)\] Substitute the expression for \(f(x)\): \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2} + tx\right)\]
03

Simplify the exponent in the function f_t

To simplify the exponent in \(f_t(x)\), we need to group the terms involving \(x\): \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2 - 2\mu x + \mu^2}{2\sigma^2} + tx\right)\] Now, we can write this as: \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2 - 2(\mu+\sigma^2 t)x + (\mu+\sigma^2 t)^2 - \sigma^4 t^2}{2\sigma^2}\right)\]
04

Compare f_t to a standard normal probability density function

We can now compare the expression for \(f_t(x)\) to the standard normal probability density function: \[g(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x - (\mu + \sigma^2 t))^2}{2\sigma^2}\right)\] We can see that the exponent in \(f_t(x)\) and \(g(x)\) is the same, except for the term \(-\sigma^4 t^2\) that doesn't involve \(x\). This extraneous constant term will be absorbed into the normalization constant \(Z\). Now that we have simplified the expression for \(f_t(x)\), we can conclude that it indeed has a mean \(\mu+\sigma^2 t\) and variance \(\sigma^2\), showing that the tilted density function of a normal random variable is also a normal random variable with the given mean and variance.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X_{1,+\cdots}, X_{n}\) be independent random variables with \(E\left[X_{i}\right]=\theta\), \(\operatorname{Var}\left(X_{i}\right)=\sigma_{i}^{2}, i=1, \ldots, n\), and consider estimates of \(\theta\) of the form \(\sum_{i=1}^{n} \lambda_{i} X_{i}\) where \(\sum_{i=1}^{n} \lambda_{i}=1 .\) Show that \(\operatorname{Var}\left(\sum_{i=1}^{n} \lambda_{i} X_{i}\right)\) is minimized when \(\lambda_{i}=\left(1 / \sigma_{i}^{2}\right) /\left(\sum_{j=1}^{n} 1 / \sigma_{j}^{2}\right), i=1, \ldots, n\) Possible Hint: If you cannot ?o this for general \(n\), try it first when \(n=2\) The following two problems are concerned with the estimation of \(\int_{0}^{1} g(x) d x=E[g(U)]\) where \(U\) is uniform \((0,1) .\)

Let \(X_{1}, \ldots, X_{k}\) be independent with $$ P\left[X_{i}=j \mid=\frac{1}{n}, \quad j=1, \ldots, n, i=1, \ldots, k\right. $$ If \(D\) is the number of distinct values among \(X_{1}, \ldots, X_{k}\) show that $$ \begin{aligned} E[D] &=n\left[1-\left(\frac{n-1}{n}\right)^{k}\right] \\ &=k-\frac{k^{2}}{2 n} \quad \text { when } \frac{k^{2}}{n} \text { is small } \end{aligned} $$

Consider the technique of simulating a gamma \((n, \lambda)\) random variable by using the rejection method with \(g\) being an exponential density with rate \(\lambda / n .\) (a) Show that the average number of iterations of the algorithm needed to generate a gamma is \(n^{\prime \prime} e^{1-n} /(n-1) !\). (b) Use Stirling's approximation to show that for large \(n\) the answer to (a) is approximately equal to \(e[(n-1) /(2 \pi)]^{1 / 2}\). (c) Show that the procedure is equivalent to the following: Step 1: Generate \(Y_{1}\) and \(Y_{2}\), independent exponentials with rate 1 . Step 2: If \(Y_{1}<(n-1)\left[Y_{2}-\log \left(Y_{2}\right)-1\right]\), return to step 1 . Step 3: Set \(X=n Y_{2} / \lambda\). (d) Explain how to obtain an independent exponential along with a gamma from the preceding algorithm.

Let \(R\) denote a region in the two-dimensional plane. Show that for a two- dimensional Poisson process, given that there are \(n\) points located in \(R\), the points are independently and uniformly distributed in \(R\) -that is, their density is \(f(x, y)=c,(x, y) \in R\) where \(c\) is the inverse of the area of \(R\).

Stratified Sampling: Let \(U_{1}, \ldots, U_{n}\) be independent random numbers and set \(\bar{O}_{t}=\left(U_{1}+i-1\right) / n, t=1, \ldots, n\). Hence, \(O_{i}, i \geq 1\), is uniform on \(((i-1) / n, i / n) . \sum_{i-1}^{n} g\left(\bar{U}_{i}\right) / n\) is called the stratified sampling estimator of \(\int_{0}^{1} g(x) d x .\) (a) Show that \(E\left[\sum_{i=1}^{n} g\left(\bar{O}_{i}\right) / n\right]=\int_{0}^{1} g(x) d x\) (b) Show that \(\operatorname{Var}\left[\sum_{i=1}^{n} g\left(O_{i}\right) / n\right] \leq \operatorname{Var}\left[\sum_{i=1}^{n} g\left(U_{i}\right) / n\right]\) Hint: Let \(U\) be uniform \((0,1)\) and define \(N\) by \(N=i\) if \((i-1) / n
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.