/*! 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 17 Find the uniform distribution of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 17

Find the uniform distribution of the continuous type on the interval \((b, c)\) that has the same mean and the same variance as those of a chi-square distribution with 8 degrees of freedom. That is, find \(b\) and \(c\).

Short Answer

Expert verified
The uniform distribution that has the same mean and the same variance as those of a chi-square distribution with 8 degrees of freedom is the interval \( (4, 12) \).

Step by step solution

01

Set up the equation for the mean

For a uniform distribution, the mean is \( \frac{b+c}{2} \). Since the mean of a chi-square distribution with 8 degrees of freedom is 8, set these equal to each other: \( \frac{b+c}{2} = 8 \). Solve this equation for one of the variables to obtain \(c = 16 - b\).
02

Set up the equation for the variance

For a uniform distribution, the variance is \( \frac{(c-b)^2}{12} \). Known that the variance of a chi-square distribution with 8 degrees of freedom is 16, we set \( \frac{(c-b)^2}{12} = 16\). Substitute \(c = 16 - b\) into the equation to obtain \( \frac{(16-2b)^2}{12} = 16 \).
03

Solve the equation for \(b\)

Solve the equation \( \frac{(16-2b)^2}{12} = 16 \) for \(b\). This will yield two possible values for \(b\), namely 4 and 12.
04

Calculate \(c\)

Substitute these two possible values of \(b\) into the equation for \(c\) to obtain the corresponding values of \(c\). When \(b = 4, c = 12\), and when \(b = 12, c = 4\).
05

Determine the valid solution

For the interval \(b, c\), \(b < c\). Therefore, only the solution \(b = 4, c = 12\) is valid.

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

Compute the measures of skewness and kurtosis of a gamma distribution which has parameters \(\alpha\) and \(\beta\).

Show that the moment generating function of the negative binomial distribution is \(M(t)=p^{r}\left[1-(1-p) e^{t}\right]^{-r}\). Find the mean and the variance of this distribution. Hint: In the summation representing \(M(t)\), make use of the MacLaurin's series for \((1-w)^{-r}\)

Let \(X\) have an exponential distribution. (a) Show that $$ P(X>x+y \mid X>x)=P(X>y) $$ Hence, the exponential distribution has the memoryless property. Recall from (3.1.6) that the discrete geometric distribution had a similar property. (b) Let \(F(x)\) be the cdf of a continuous random variable \(Y\). Assume that \(F(0)=0\) and \(00\). Suppose property (3.3.7) holds for \(Y\). Show that \(F_{Y}(y)=1-e^{-\lambda y}\) for \(y>0\) Hint: Show that \(g(y)=1-F_{Y}(y)\) satisfies the equation $$ g(y+z)=g(y) g(z) $$

Let \(X\) and \(Y\) be independent random variables, each with a distribution that is \(N(0,1) .\) Let \(Z=X+Y .\) Find the integral that represents the cdf \(G(z)=\) \(P(X+Y \leq z)\) of \(Z .\) Determine the pdf of \(Z\). Hint: We have that \(G(z)=\int_{-\infty}^{\infty} H(x, z) d x\), where $$ H(x, z)=\int_{-\infty}^{z-x} \frac{1}{2 \pi} \exp \left[-\left(x^{2}+y^{2}\right) / 2\right] d y $$ Find \(G^{\prime}(z)\) by evaluating \(\int_{-\infty}^{\infty}[\partial H(x, z) / \partial z] d x\).

. Let \(X_{1}, X_{2}\), and \(X_{3}\) be three independent chi-square variables with \(r_{1}, r_{2}\), and \(r_{3}\) degrees of freedom, respectively. (a) Show that \(Y_{1}=X_{1} / X_{2}\) and \(Y_{2}=X_{1}+X_{2}\) are independent and that \(Y_{2}\) is \(\chi^{2}\left(r_{1}+r_{2}\right)\) (b) Deduce that $$ \frac{X_{1} / r_{1}}{X_{2} / r_{2}} \quad \text { and } \quad \frac{X_{3} / r_{3}}{\left(X_{1}+X_{2}\right) /\left(r_{1}+r_{2}\right)} $$ are independent \(F\) -variables.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

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.