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

Short Answer

Expert verified
The t-distribution with one degree of freedom and the Cauchy distribution are the same as their respective probability density functions are identical when compared.

Step by step solution

01

Define the distributions

The probability density function (PDF) of the t-distribution with \(r\) degrees of freedom is given by: \[f(t) = \dfrac{\Gamma(\frac{r+1}{2})}{\sqrt{r\pi}\Gamma(\frac{r}{2})}\left(1+\frac{t^2}{r}\right)^{-\frac{r+1}{2}}\]where \(\Gamma\) is the Gamma function. The PDF of the Cauchy distribution is given by: \[g(x) = \dfrac{1}{\pi(1 + x^2)}\].
02

Set r=1 for t-distribution

Now we need to plug \(r=1\) into the formula of the t-distribution and simplify it. Thus, it becomes: \[f(t) = \frac{\Gamma(1)}{\sqrt{\pi}\Gamma(0.5)}\left(1+t^2)^{-1}\]. We know that \(\Gamma(1) = 1\) as per the definition of the Gamma function. Also, \(\Gamma(0.5) = \sqrt{\pi}\). Substituting these values in, we have: \[f(t) = \frac{1}{\pi(1 + t^2)}\].
03

Compare the distributions

Comparing the equation derived in Step 2 and the original equation for the Cauchy distribution, we notice that they are the same. As the equations are identical, the t-distribution with one degree of freedom is equivalent to the Cauchy distribution.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, X_{3}\) be iid random variables each having a standard normal distribution. Let the random variables \(Y_{1}, Y_{2}, Y_{3}\) be defined by $$ X_{1}=Y_{1} \cos Y_{2} \sin Y_{3}, \quad X_{2}=Y_{1} \sin Y_{2} \sin Y_{3}, \quad X_{3}=Y_{1} \cos Y_{3} $$ where \(0 \leq Y_{1}<\infty, 0 \leq Y_{2}<2 \pi, 0 \leq Y_{3} \leq \pi .\) Show that \(Y_{1}, Y_{2}, Y_{3}\) are mutually independent.

A certain job is completed in three steps in series. The means and standard deviations for the steps are (in minutes): $$ \begin{array}{ccc} \hline \text { Step } & \text { Mean } & \text { Standard Deviation } \\ \hline 1 & 17 & 2 \\ 2 & 13 & 1 \\ 3 & 13 & 2 \\ \hline \end{array} $$Assuming independent steps and normal distributions, compute the probability that the job will take less than 40 minutes to complete.

. Let \(X\) and \(Y\) have the joint \(\operatorname{pmf} p(x, y)=e^{-2} /[x !(y-x) !], y=0,1,2, \ldots ;\) \(x=0,1, \ldots, y\), zero elsewhere. (a) Find the mgf \(M\left(t_{1}, t_{2}\right)\) of this joint distribution. (b) Compute the means, the variances, and the correlation coefficient of \(X\) and \(Y\). (c) Determine the conditional mean \(E(X \mid y)\). Hint: Note that $$ \sum_{x=0}^{y}\left[\exp \left(t_{1} x\right)\right] y ! /[x !(y-x) !]=\left[1+\exp \left(t_{1}\right)\right]^{y} $$ Why?

Let \(X\) and \(Y\) have a bivariate normal distribution with parameters \(\mu_{1}=\) 20, \(\mu_{2}=40, \sigma_{1}^{2}=9, \sigma_{2}^{2}=4\), and \(\rho=0.6\). Find the shortest interval for which \(0.90\) is the conditional probability that \(Y\) is in the interval, given that \(X=22\).

Let \(F\) have an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2} .\) Prove that \(1 / F\) has an \(F\) -distribution with parameters \(r_{2}\) and \(r_{1}\).

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