Analysis IV, Assignment 1
Nicolas Resch
January 20, 2014
[1] (1) Proof. Clearly, the function as dened takes values in R (indeed, it takes values in [0, ]). First
of all, for all x, y X , d(x, y ) = |f (x) f (y )| 0. Now, using the fact that arctan is inj
(1 ) is the condence coe
the interval.
cient of
The half-interval length is the precision
of the interval.
Sample Size. To estimate by x with a
specied error e with 100(1 )% condence
we need to have
z/2
e= p .
n
Solve for n (the necessary sample size) t
2 distribution.
Gamma and
dene
() =
Z 1
0
For > 0,
x 1e xdx.
This gives (1) = 1. Use integration by parts
to conclude
() = (
1) (
1).
Therefore if is integer we have
() = (
1)!.
Take
> 0 and use the change of variable
x = y to write
Z 1
0
y 1e
y
dy =
() .
Denition: Considering all unbiased estimators for , the one with the smallest variance
is the mininum variance unbiased estimator
and is called the most e cient estimator of
(MVUE).
Example: Let X1, ., Xn be a random sample
from a distribution with mean
Denition: If X1, ., Xn is a random sample
from a distribution F which depends on an
unknown parameter , any statistic
= h(X1, ., Xn)
used to estimate is called a point estimator
of . After the sample has been selected, the
observed values x1, ., xn are u
Solution.
P (X
204) = P
= P (Z
Z
p
36(204
15
200)
!
1.6) = 0.0548.
Example. An electronic company manufactures resistors that have a mean resistance
of 100 and a standard deviation of 10 .
Find the probability that a random sample of
n = 25 resistors will
= 0.0569.
Sampling distribution and Statistical inference
Denition: If X1, ., Xn are independent and
identically distributed with common distribution F , we call (X1, ., Xn) a random sample
from the distribution F . The sample size is
n. After the data is
SOLUTIONS TO ASSIGNMENT 2 - MATH 355
Problem 1.
Recall that,
Bn = cfw_ [0, 1] : |Sn ()| > n n ,
N = cfw_ [0, 1] : lim
n
Sn ()
= 0,
n
and
m(Bn )
We want to show that m(N c ) = 0.
c
Let > 0. We can pick 4 = n with c >
n
m(Bn )
n=1
3
.
n2 4
3/2 .
n=1 n
3
3
Analysis IV, Assignment 4
Prof. John Toth
Winter 2013
Exercise 1
Let f C 0 (R) and periodic with f (x+2) = f (x). Let an =
N
1
2
f (t)eint dt
an einx . If f is continuously dierentiable at xo [, ],
and (SN f )(x) =
n=N
then f (x0 ) = lim (SN f )(x0 ).
N
Analysis IV : Assignment 3 Solutions
John Toth, Winter 2013
(l2 (Z),
Exercise 1
2)
is a complete and separable Hilbert space.
(n)
(n)
(n)
(n)
(n)
Proof Let cfw_u(n) nN be a Cauchy sequence. Say u(n) = (., u2 , u1 , u0 , u1 , u2 , .). In particular for ev
Analysis IV, Assignment 5
Nicolas Resch
March 7, 2014
First, we will prove that if (V ) = supcfw_(f ) : f < V , then
(bn an ).
( V ) =
(1)
n
Given any f
V , since f (x) 1 for x V and f (x) = 0 for x V we have
/
(f ) =
f (x)dx =
f (x)dx
f (x)dx +
Hence, (
Analysis 4, Assignment 4
Nicolas Resch
March 7, 2014
[1] (1) Proof. First of all, we claim that either there exists x0 such that |f (x0 )| = supx[a,b] |f (x)| or
there exists y0 such that limxy0 f (x) = supx[a,b] |f (x)|, where y0 is one of the (nitely ma
Analysis IV, Assignment 3
Nicolas Resch
March 7, 2014
[1] (1) Proof. Let fn = n=1 |fk | and let f = limn fn = |fn |. Then fn fn+1 for all n and
k
n=1
fn 0. By Lebesgue Monotone Convergence Theorem and the linearity of the integral, we have
N
N
|fn |d = li
Analysis IV, Assignment 2
Nicolas Resch
January 28, 2014
[1] (a) Proof. Let x cl(L), and let (xk ) be a sequence in L converging to x. Without loss of
k=1
generality, we may assume that d(xk , x) < 21 for each k (as B (x, 21 ) L = for all k N).
k
k
(k)
Th
FACULTY OF SCIENCE
FINAL EXAMINATION
MATHEMATICS MATH 355
Analysis 4
Examiner: Professor S. W. Drury
Date: Wednesday, April 18, 2007
Associate Examiner: Professor K. N. GowriSankaran
Time: 2: 00 pm. 5: 00 pm.
INSTRUCTIONS
Attempt six questions for full cr