Strategic Practice and Homework 10 - Projects at Harvard

Stat 110 Penultimate Homework, Fall 2011Prof. Joe Blitzstein (Department of Statistics, Harvard University)1. Judit plays in a total of N ⇠ Geom(s) chesstournamentsinhercareer.Supposethat in each tournament she has probability p of winning the tournament, independently.Let T be the number of tournaments she wins in her career.(a) Find the mean and variance of T .(b) Find the MGF of T . What is the name of this distribution (with its parameters)?2. Let X 1 ,X 2 be i.i.d., and let ¯X = 1 2 (X 1 + X 2 ). In many statistics problems, it isuseful or important to obtain a conditional expectation given ¯X. As an example ofthis, find E(w 1 X 1 + w 2 X 2 | ¯X), where w 1 ,w 2 are constants with w 1 + w 2 =1.3. A certain stock has low volatility on some days and high volatility on other days.Suppose that the probability of a low volatility day is p and of a high volatility dayis q =1 p, andthatonlowvolatilitydaysthepercentchangeinthestockpriceis2 2N (0, 1 ), while on high volatility days the percent change is N (0, 2 ), with 1 < 2 .Let X be the percent change of the stock on a certain day. The distribution issaid to be a mixture of two Normal distributions, and a convenient way to representX is as X = I 1 X 1 + I 2 X 2 where I 1 is the indicator r.v. of having a low volatility day,2I 2 =1 I 1 , X j ⇠N(0, j ), and I 1 ,X 1 ,X 2 are independent.(a) Find the variance of X in two ways:Cov(I 1 X 1 + I 2 X 2 ,I 1 X 1 + I 2 X 2 )directly.using Eve’s Law, and by calculating(b) The kurtosis of a r.v. Y with mean µ and standard deviationis defined byKurt(Y )=E(Y µ)443.This is a measure of how heavy-tailed the distribution of Y . Find the kurtosis of X(in terms of p, q,21 ,22 ,fullysimplified). Theresultwillshowthateventhoughthekurtosis of any Normal distribution is 0, the kurtosis of X is positive and in fact canbe very large depending on the parameter values.4. We wish to estimate an unknown parameter ✓, basedonar.v.X we will getto observe. As in the Bayesian perspective, assume that X and ✓ have a jointdistribution. Let ˆ✓ be the estimator (which is a function of X). Then ˆ✓ is said to beunbiased if E(ˆ✓|✓) =✓, andˆ✓ is said to be the Bayes procedure if E(✓|X) =ˆ✓.1