Question

n expression (3.4.13), the normal location model was presented. Often real data, though, have more outliers than the normal distribution allows. Based on Exercise 3.6.5, outliers are more probable for \(t\) -distributions with small degrees of freedom. Consider a location model of the form $$ X=\mu+e $$ where \(e\) has a \(t\) -distribution with 3 degrees of freedom. Determine the standard deviation \(\sigma\) of \(X\) and then find \(P(|X-\mu| \geq \sigma)\).

Short answer

The standard deviation \(\sigma\) of \(X\) is \(\sqrt{3}\) and \(P(|X-\mu| \geq \sigma)\) is approximately 0.646.

Step-by-step solution

Compute Standard Deviation

The standard deviation of a t-distribution with 3 degrees of freedom (df > 2), can be computed as \(\sigma = \sqrt{\frac{df}{df-2}}\). Therefore, plugging in the value of df, \(\sigma\) can be computed as \(\sigma = \sqrt{\frac{3}{3-2}} = \sqrt{3}\).

Compute Cumulative Probability

The cumulative distribution function for t-distribution will be used to compute the probability \(P(|X-\mu| \geq \sigma)\). By symmetry of the t-distribution, we can reflect the part of the distribution curve on the right side of our standard deviation to the left. So, \(P(|X-\mu| \geq \sigma)=2*P(X-\mu \geq \sigma)\). Using the t-distribution table with degrees of freedom 3, the probability \(P(X-\mu \geq \sigma)\) corresponding to \(\sigma = \sqrt{3}\) is found to be approximately 0.323. Therefore, \(P(|X-\mu| \geq \sigma)\) = 2 * 0.323 = 0.646.