Math 362 Problem Set #9 – Solution 27 April 2012 - Faculty web ...

as the left and right endpoints of a 95% confidence interval for σ 2 . To find a confidenceinterval for σ, we take square roots.We are 95% confident that σ lies between 5.09 and 20.01.4. The probability density function for the Student t distribution with ν degrees of freedomis given by[ (Γν+1) ] ( ) −ν+12f T (t) = √ (πν Γν1 +2)t2 2.νFor a fixed ν, the quantity in brackets is a constant, and so the “core” of the distribution) −ν+1is the function t ↦→(1 + t2 2.ν(The quantity in brackets is there in order to make the PDF integrate to 1.)Show that, as ν → ∞, the core of the Student t distribution PDF approaches the coreof the standard normal PDF.What does this tell you about the Student t distribution?Γ ( )ν+12What does it tell you about lim √ (ν→∞ πν Γν2)?( ) −ν+1We need to evaluate lim 1 + t2 2, which we could write asν→∞ νlimν→∞[(1 + t2ν) ν )] −1(1 + t2 2.νUsing continuity and the fact that the limit of the product is the product of the limits,we getlimν→∞) ν+1(1 + t2 2ν==[limν→∞) ν ] −1(1 [ ( )]+ t2 2 −1lim 1 + t2 2νν→∞ ν[e t2] − 1 2· [1] − 1 2= e − t2 2 ,which is the core of the probability density function for the standard normal distribution.