Problem 2* The aim of this exercise is to deduce the p.d.f of the Student's t distribution t(n) discussed in lectures. L

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Problem 2 The Aim Of This Exercise Is To Deduce The P D F Of The Student S T Distribution T N Discussed In Lectures L 1 (74.26 KiB) Viewed 43 times

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Problem 2 The Aim Of This Exercise Is To Deduce The P D F Of The Student S T Distribution T N Discussed In Lectures L 3 (75.16 KiB) Viewed 43 times

Problem 2* The aim of this exercise is to deduce the p.d.f of the Student's t distribution t(n) discussed in lectures. Let W.~X(n) be a chi-square random variable with n degrees of freedom and let ZN (0,1) be a standard normal random variable with Z and W, independent. Then we define the following random variable, 2/n U = W. ha(1, 2) = P(U. 51|2 = z), der and say that Unt(n). (a) Using the p.d.f. of W. given in lectures, find a formula for the conditional density 1,2ER. Hint: be careful with the sign of r and 2. Start by assuming <> 0 and compute hr (1,2) separately for cases z <0 and 2 > 0. Then deal with the case r < 0 similarly. (b) Show that for any a > 0 and neN, T() dz = (c) Using the law of total probability, which here implies that PU. S 2) = 1.EPU. S1 | 2)] = E(h(2, 2), 2 use parts (a) and (b) to show that the p.d.f. of U, is T(+1) fu. (x) = Van (6) (1+) 1 ER
N U. 3.2.2 Student's t - distribution Consider two independent random variables Z ~ N(0,1) and W~x?(n). We say a random variable U ~ f(n) follows Student's t-distribution with n degrees of freedom if it can be written as 21 N(0,1) (3.2.5) VW x?(n)/n The p.d.f. can be computed to be (see exercises) T(4) fun (u) UER. (3.2.6) Van l'(n/2) (1 + 2) We can see that 1 12 n 2 n W n- E(Un) = 0 for n > 1, Var(Un) for n > 2. (3.2.7) The first equality follows by symmetry of the p.d.f. while the second follows from a direct computation from 3.2.5): 22 Var(um) º E(U3) = n() % E(2)E(W1) " UmU2 2 The last equality is left as an exercise. At this point it is alluring to check what happens to the density as n grows to infinity. Recall the following limit involving gamma functions (see if you can prove it): Is + a) 5 ** (s) lim = 1.
Then we set s = n/2 and a = 1/2 and we compute 1 lim fu.(u) T(/2+1/2) 27 n * (n/2)1/21 (n/2) n *** (1 + 37 lim lim T2 > (!) 1 1 lim 2 * > (1 + 2)41 " (!) 1 V2 e "2/2 Thus the density for large n approximates the density of a standard normal distribution. This fact can also be used to show convergence in probabil- ity of the sequence of random variables {t(n)}nen to a standard normal distribution: TL > TI SO JR lim P{U. <x} = lim to $100,7 11( x,x)(u)fu.. (u) du "S170,ruu 1( (u) lim fu. (u) du = ºz(u). T2 Pushing the limit in the integral is due to the generalized dominated con- vergence theorem, since all densities involved are bounded by a multiple of the normal density for n large enough.