Eric Slud
Fall 2007
Solutions to Stat 710 Problem Set 2
#19.3. Z(t) is a standard Brownian motion, which implies that for 0 t 1,
Z(t) tZ(1) Y (t) is a process Gaussian nite dimensional distributions with
mean-0 and covariances for 0 s t 1 given by Cov(Y (
Stat 710 HW2 Solutions
# 1 in-class. We are considering iid N (, 2 ) r.v.s Xi , and =
(, ), with generalized method of moments estimators dened in terms of
e() = E (I[X1 1] , I[X1 1] ) = (1 + )/), (1 )/). These
moment estimators (, ) are uniquely determin
Eric Slud
Fall 2007
Solutions to Stat 710 Problem Set 3
#6.1. Here we make explicit use of the likelihood ratio exp(Wn ) (dQn /dPn )
= exp(2 /2 + X n ), where X N (0, 1) under Pn . We know tightness of
n
eWn under Pn which implies that every innite sequen
Stat 710
11/13/02
Solutions to Problem Set 3
Ch. 5 #15. (Xi , Yi ) is iid with joint distribution given by Yi = f0 (Xi )+ei .
Least squares says argmin n1 n (Yi f (Xi )2 . What we are
i=1
minimizing is asymptotically E(Y1 f (X1 )2 , so if this min is uniq
Stat 710
12/19/02
Solutions to Selected Problems, HW 5
Ch. 11, #4. The key point in this problem is to use only the projection
denition, with suitable limiting operations, to obtain the usual denition.
It is a backward approach:
(i) Since E(X M | Y ) is a
Eric Slud
Fall 2007
Solutions to Stat 710 Problem Set 4
#7.1. Here the dominating measure on N = cfw_0, 1, 2, . . . is counting measure.
For each x 0, the function
s(x, ) p(x, ) = exp(x log )/2)/ x!
is obviously continuously dierentiable on (0, ). Then p
1.
A2 ~ N ( 30, 9 )
A1 ~ N ( 0, 12 )
9
12
B ~ N ( 70, 20 )
20
T = Time (mins)
elapsed from 1:00
1:00
2:00
3:00
(a) Determine the distribution of the variable A = Amount of time for Al to reach the connecting
gate (in total minutes elapsed from 1:00). That
Solutions for HW1, Stat 710, F07
#5.12. The condition is: there exists a unique pth quantile xp , i.e. a number
such that P (X1 xp ) = p and for all > 0,
P (X1 xp ) < p,
P (X1 xp + ) < 1 p
The Lemma needed to prove consistency under this condition is Lemm
1. (a) If = .05, then /2 = .025, so that z.025 = 1.96. If 1 = .999, then = .001, so that z.001 = 3.09.
2
| 3000 2500 |
1.96 + 3.09
Moreover,
= = 0.625 , so that n =
= 65.29, so take n 66.
800
0.625
(b) 95% margin of error = (z.025)(800 / 64 ) = (1.
Stat 710 HW1 Solutions
(2). The density of the tn distributed random variable Xn = Zn / Vn
has density of the form Cn (1 + x2 /n)(n+1)/2 , where Zn N (0, 1) and
Vn Gamma(n/2, n/2) are independent. We know that the distributional
convergence of Xn to N (0,