/*! 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 43 The density function of a chi-sq... [FREE SOLUTION] | 91Ó°ÊÓ

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The density function of a chi-squared random variable having \(n\) degrees of freedom can be shown to be $$ f(x)=\frac{\frac{1}{2} e^{-x / 2}(x / 2)^{\frac{\pi}{2}-1}}{\Gamma(n / 2)}, \quad x>0 $$ where \(\Gamma(t)\) is the gamma function defined by $$ \Gamma(t)=\int_{0}^{\infty} e^{-x} x^{t-1} d x, \quad t>0 $$ Integration by parts can be employed to show that \(\Gamma(t)=(t-1) \Gamma(t-1)\), when \(t>1\). If \(Z\) and \(\chi_{n}^{2}\) are independent random variables with \(Z\) having a standard normal distribution and \(\chi_{n}^{2}\) having a chi-square distribution with \(n\) degrees of freedom, then the random variable \(T\) defined by $$ T=\frac{Z}{\sqrt{\chi_{n}^{2} / n}} $$ is said to have a \(t\) -distribution with \(n\) degrees of freedom. Compute its mean and variance when \(n>2\).

Short Answer

Expert verified
The mean of the t-distribution is 0. The variance of the t-distribution with n degrees of freedom (n > 2) can be computed as \(Var[T] = \frac{n}{n-2}\).

Step by step solution

01

T density function

\(g(t) \, dt = f(z) \, dz \times h(\chi_{n}^{2}) \, d\chi_{n}^{2}\) where f(z) is the probability density function of Z, and h(χ²) is the probability density function of \(\chi_{n}^{2}\). #Step 2: Calculate the density function of T# To calculate g(t), we need to express f(z) and h(χ²) in terms of t and use the Jacobian method to find the transformation.
02

Express f(z) and h(χ²) in terms of t

From the definition of T, we get \(Z = t\sqrt{\frac{\chi_{n}^{2}}{n}}\) and thus \(z = t\sqrt{\frac{x}{n}} \Rightarrow dz = \frac{\sqrt{x}}{2\sqrt{n}} dt\). Also, \(\chi_{n}^{2} = x\). Now, we can express f(z) in terms of t using Z distribution: \(f(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}}\) And, h(x) using the chi-squared distribution: \(h(x) = \frac{\frac{1}{2} e^{-x/2}(x/2)^{\frac{n}{2}-1}}{\Gamma(n/2)}\) #Step 3: Calculate g(t) using the Jacobian method# Now, substitute f(z) and h(x) in the expression for g(t) and use the Jacobian method to find the transformation.
03

Calculate g(t) with the Jacobian method

\(g(t) \, dt = f(z) \, dz \times h(x) \, dx = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \frac{\sqrt{x}}{2\sqrt{n}} dt \times \frac{\frac{1}{2} e^{-x/2}(x/2)^{\frac{n}{2}-1}}{\Gamma(n/2)}\) Substitute z and x into the above expression: \(g(t) \, dt = \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2x}{2n}} \frac{\sqrt{x}}{2\sqrt{n}} dt \times \frac{\frac{1}{2} e^{-x/2}(x/2)^{\frac{n}{2}-1}}{\Gamma(n/2)}\) Simplify the expression: \(g(t) = \frac{e^{-\frac{t^2x}{2n}}(x)^{\frac{n}{2}-1} e^{-x/2}}{2\sqrt{2\pi n}\Gamma(n/2)}\) #Step 4: Compute the mean and variance of T# Now that we have the density function of T, we can compute its mean and variance. The mean of a t-distribution is 0.
04

Compute the variance of T

The variance of T can be computed using its n degrees of freedom, as follows: \(Var[T] = \frac{n}{n-2}\), for n > 2. Hence, the mean of T is 0 and the variance is \(\frac{n}{n-2}\) when n > 2.

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Key Concepts

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

Chi-Square Distribution
The Chi-Square Distribution is an important concept in statistics, especially when dealing with hypothesis testing and confidence intervals for variance. This distribution arises when you sum the squares of independent standard normal random variables. It is defined by a parameter called degrees of freedom, denoted by \(n\). The density function for a chi-square distribution is given by:\[f(x) = \frac{\frac{1}{2} e^{-x/2}(x/2)^{\frac{n}{2}-1}}{\Gamma(n/2)}, \quad x > 0\]This formula includes the Gamma Function, which plays a pivotal role in computing the values precisely. The shape of the chi-square distribution changes with different degrees of freedom:- With low \(n\), it is skewed to the right.- As \(n\) increases, it becomes more similar to a normal distribution.Understanding this distribution helps in analyzing variance and also in its applications like the chi-square test.
T-Distribution
The T-Distribution is another essential statistical distribution used mainly when making estimates from small sample sizes. It is similar to the standard normal distribution but has heavier tails. This means it can account for variability more efficiently in smaller samples.A notable property of the t-distribution is that it's defined by degrees of freedom \(n\). The probability density function of a t-distribution is not as straightforward as the chi-square but is derived from it. A t-distribution can be visualized when a standard normal random variable Z and an independent chi-square random variable \(\chi_{n}^{2}\) are considered. The formula is:\[T = \frac{Z}{\sqrt{\chi_{n}^{2}/n}}\]This formula shows that the t-distribution approaches the normal distribution as the degrees of freedom increase. It's crucial for conducting t-tests and calculating confidence intervals around a sample mean when the population standard deviation is unknown.
Gamma Function
The Gamma Function, denoted as \(\Gamma(t)\), is a continuous extension to the factorial function, used extensively in various statistical distributions like the chi-square and t-distributions. It's defined for \(t > 0\) by the integral:\[\Gamma(t) = \int_{0}^{\infty} e^{-x} x^{t-1} \, dx\]A key property of the Gamma Function is its recursive nature:\[\Gamma(t) = (t-1)\Gamma(t-1), \quad t>1\]This makes it incredibly useful in deriving the distribution components and calculations. For example, in a chi-square distribution with degrees of freedom \(n\), the value \(\Gamma(n/2)\) is crucial in its density function. Recognizing the role of the gamma function aids in simplifying complex statistical calculations and understanding probability density functions of various distributions.
Mean and Variance Calculation
Calculating the mean and variance of a statistical distribution helps in understanding the central tendency and the spread of data points.For the t-distribution with \(n\) degrees of freedom:- The mean is always 0, which implies symmetry around the central point.- The variance is calculated by the formula:\[Var[T] = \frac{n}{n-2}, \quad \text{for } n > 2\]This shows that as the degrees of freedom increase, the variance approaches 1, aligning closely with the standard normal distribution. These calculations are key for performing statistical analyses such as hypothesis testing, where you want to measure the reliability and variability of estimates drawn from sample data.

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Most popular questions from this chapter

\(A, B\), and \(C\) are evenly matched tennis players. Initially \(A\) and \(B\) play a set, and the winner then plays \(C\). This continues, with the winner always playing the waiting player, until one of the players has won two sets in a row. That player is then declared the overall winner. Find the probability that \(A\) is the overall winner.

Polya's urn model supposes that an urn initially contains \(r\) red and \(b\) blue balls. At each stage a ball is randomly selected from the urn and is then returned along with \(m\) other balls of the same color. Let \(X_{k}\) be the number of red balls drawn in the first \(k\) selections. (a) Find \(E\left[X_{1}\right]\) (b) Find \(E\left[X_{2}\right]\). (c) Find \(E\left[X_{3}\right]\). (d) Conjecture the value of \(E\left[X_{k}\right]\), and then verify your conjecture by a conditioning argument. (e) Give an intuitive proof for your conjecture. Hint: Number the initial \(r\) red and \(b\) blue balls, so the urn contains one type \(i\) red ball, for each \(i=1, \ldots, r ;\) as well as one type \(j\) blue ball, for each \(j=1, \ldots, b\). Now suppose that whenever a red ball is chosen it is returned along with \(m\) others of the same type, and similarly whenever a blue ball is chosen it is returned along with \(m\) others of the same type. Now, use a symmetry argument to determine the probability that any given selection is red.

A and B play a series of games with A winning each game with probability \(p\). The overall winner is the first player to have won two more games than the other. (a) Find the probability that \(\mathrm{A}\) is the overall winner. (b) Find the expected number of games played.

Two players take turns shooting at a target, with each shot by player \(i\) hitting the target with probability \(p_{i}, i=1,2\). Shooting ends when two consecutive shots hit the target. Let \(\mu_{i}\) denote the mean number of shots taken when player \(i\) shoots first, \(i=1,2\) (a) Find \(\mu_{1}\) and \(\mu_{2}\). (b) Let \(h_{i}\) denote the mean number of times that the target is hit when player \(i\) shoots first, \(i=1,2\). Find \(h_{1}\) and \(h_{2}\).

Suppose there are \(n\) types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of the \(n\) types. Suppose one continues collecting until a complete set of at least one of each type is obtained. (a) Find the probability that there is exactly one type \(i\) coupon in the final collection. Hint: Condition on \(T\), the number of types that are collected before the first type \(i\) appears. (b) Find the expected number of types that appear exactly once in the final collection.

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