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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
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 \(X_{1}, X_{2}\) be two independent random variables having gamma distributions with parameters \(\alpha_{1}=3, \beta_{1}=3\) and \(\alpha_{2}=5, \beta_{2}=1\), respectively. (a) Find the mgf of \(Y=2 X_{1}+6 X_{2}\) (b) What is the distribution of \(Y ?\)

Show that the graph of a pdf \(N\left(\mu, \sigma^{2}\right)\) has points of inflection at \(x=\mu-\sigma\) and \(x=\mu+\sigma\).

. Using the computer, obtain plots of the pdfs of chi-squared distributions with degrees of freedom \(r=1,2,5,10,20\). Comment on the plots.

. 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 \(X\) be \(N\left(\mu, \sigma^{2}\right)\) so that \(P(X<89)=0.90\) and \(P(X<94)=0.95\). Find \(\mu\) and \(\sigma^{2}\).
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