Question

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

Short answer

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

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: \[g(x) = \dfrac{1}{\pi(1 + x^2)}\].

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: \[f(t) = \frac{1}{\pi(1 + t^2)}\].

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.