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Chapter 3: Problem 11

Show that the \(t\) -distribution with \(r=1\) degree of freedom and the Cauchy distribution are the same.

Short Answer

Expert verified
By substituting r=1 into the formula for t-distribution and comparing it with the formula for Cauchy distribution, we can prove that the two distributions are the same when r=1.

Step by step solution

01

- Define t-distribution

The t-distribution with \(r\) degree of freedom is defined by the probability density function:\[f(x) = \frac{1}{\sqrt{r} B(\frac{1}{2}, \frac{r}{2})} (1 + \frac{x^2}{r})^{-\frac{r+1}{2}}\]where \(B(\frac{1}{2}, \frac{r}{2})\) is the beta function.
02

- Define Cauchy distribution

The Cauchy distribution is defined by the probability density function:\[f(x) = \frac{1}{\pi (1+x^2)}\]
03

- Compare the distributions

We can see that if we put \(r = 1\) in the t-distribution function, we would get:\[f(x) = \frac{1}{\sqrt{1} B(\frac{1}{2}, \frac{1}{2})} (1 + x^2)^{-\frac{1+1}{2}} = \frac{1}{B(0.5, 0.5)(1 + x^2)}\]The beta function, \(B(0.5, 0.5)\), is equal to \(\pi\). So, replacing this, we would get \[f(x) = \frac{1}{\pi (1+x^2)}\]The distribution exactly matches the Cauchy Distribution.

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Most popular questions from this chapter

Readers may have encountered the multiple regression model in a previous course in statistics. We can briefly write it as follows. Suppose we have a vector of \(n\) observations \(\mathbf{Y}\) which has the distribution \(N_{n}\left(\mathbf{X} \boldsymbol{\beta}, \sigma^{2} \mathbf{I}\right)\), where \(\mathbf{X}\) is an \(n \times p\) matrix of known values, which has full column rank \(p\), and \(\beta\) is a \(p \times 1\) vector of unknown parameters. The least squares estimator of \(\boldsymbol{\beta}\) is $$ \widehat{\boldsymbol{\beta}}=\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime} \mathbf{Y} $$ (a) Determine the distribution of \(\widehat{\boldsymbol{\beta}}\). (b) Let \(\hat{\mathbf{Y}}=\mathbf{X} \hat{\boldsymbol{\beta}}\). Determine the distribution of \(\widehat{\mathbf{Y}}\) (c) Let \(\widehat{\mathbf{e}}=\mathbf{Y}-\hat{\mathbf{Y}}\). Determine the distribution of \(\widehat{\mathbf{e}}\). (d) By writing the random vector \(\left(\widehat{\mathbf{Y}}^{\prime}, \widehat{\mathbf{e}}^{\prime}\right)^{\prime}\) as a linear function of \(\mathbf{Y}\), show that the random vectors \(\hat{\mathbf{Y}}\) and \(\widehat{\mathbf{e}}\) are independent. (e) Show that \(\widehat{\beta}\) solves the least squares problem; that is, $$\|\mathbf{Y}-\mathbf{X} \widehat{\boldsymbol{\beta}}\|^{2}=\min _{\mathbf{b} \in R^{p}}\|\mathbf{Y}-\mathbf{X} \mathbf{b}\|^{2}$$

Assuming a computer is available, investigate the probabilities of an "outlier" for a \(t\) -random variable and a normal random variable. Specifically, determine the probability of observing the event \(\\{|X| \geq 2\\}\) for the following random variables: (a) \(X\) has a standard normal distribution. (b) \(X\) has a \(t\) -distribution with 1 degree of freedom. (c) \(X\) has a \(t\) -distribution with 3 degrees of freedom. (d) \(X\) has a \(t\) -distribution with 10 degrees of freedom. (e) \(X\) has a \(t\) -distribution with 30 degrees of freedom.

Let \(X\) be a random variable such that \(E\left(X^{2 m}\right)=(2 m) ! /\left(2^{m} m !\right), m=\) \(1,2,3, \ldots\) and \(E\left(X^{2 m-1}\right)=0, m=1,2,3, \ldots\) Find the mgf and the pdf of \(X\).

Determine the constant \(c\) in each of the following so that each \(f(x)\) is a \(\beta\) pdf: (a) \(f(x)=c x(1-x)^{3}, 0

Let \(X\) have a Poisson distribution with parameter \(m .\) If \(m\) is an experimental value of a random variable having a gamma distribution with \(\alpha=2\) and \(\beta=1\), compute \(P(X=0,1,2)\). Hint: Find an expression that represents the joint distribution of \(X\) and \(m\). Then integrate out \(m\) to find the marginal distribution of \(X\).
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