ECON 710 Midterm: ECON 710 UW Madison Midterm 2013a

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31 Jan 2019

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Econometrics 710

Midterm Exam

March 12, 2013

Sample Answers

This exam concerns the model

yi=m(xi) + ei(1)

m(x) = 0+1x+2x2+ +pxp(2)

E(ziei) = 0 (3)

zi= (1; xi; :::; xp

i)0(4)

g(x) = d

dxm(x)(5)

with iid observations (yi; xi); i = 1; :::; n: The order of the polynomial pis known.

1. How should we interpret the function m(x)given the projection assumption (3)? How should we

interpret g(x)? (Brie‡y)

The model does not specify that m(x)is the conditional mean. Rather, equation (3) speci…es that

it is a projection model. Thus m(x)is the best linear predictor of yigiven linear functions of zi:

Equivalently, it is the best predictor in the class of pth order polynomials in xi:It is also the best

mean-square approximation to the conditional mean, in the class of pth order polynomials in xi:The

function g(x)is the derivative of the best linear predictor, and equals

g(x) = 1+ 22x+ 33x2+ +ppxp1

=h(x)0

where = (0; :::; p)0and h(x) = (0;1;2x; 3x2; :::; pxp1)0:

2. Describe an estimator ^g(x)of g(x):

Since g(x) = h(x)0is linear in ; the plug-in approach suggests replacing with the e¢cient estimator

for : Under the projection assumption (3) OLS is the asymptotically e¢cient estimator. It equals

= (Z0Z)1(Z0Y)where Yand Zare the stacked observations on yiand zi:Then the estimator for

g(x)is

^g(x) = h(x)0^



1+ 2^

2x+ 3^

3x2+ +p^

pxp1

3. Find the asymptotic distribution of pn(^g(x)g(x)) as n! 1:

Under the projection assumption (3) plus regularity conditions, we know that as n! 1;pn^

!d

N(0; V)where Vb=Q1Q1with Q=E(ziz0

i)and  = Eziz0

ie2

i. Then as n! 1

pn(^g(x)g(x)) = pnh(x)0^

h(x)0

=h(x)0pn^



!dh(x)0N(0; V) = N(0; h(x)0Vh(x)):

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Document Summary

This exam concerns the model yi = m(xi) + ei m(x) = (cid:12)0 + (cid:12)1x + (cid:12)2x2 + (cid:1)(cid:1)(cid:1) + (cid:12)pxp. The model does not specify that m(x) is the conditional mean. Rather, equation (3) speci(cid:133)es that it is a projection model. Thus m(x) is the best linear predictor of yi given linear functions of zi: = h(x)0(cid:12) where (cid:12) = ((cid:12)0; :::; (cid:12)p)0 and h(x) = (0; 1; 2x; 3x2; :::; pxp(cid:0)1)0: describe an estimator ^g(x) of g(x): Since g(x) = h(x)0(cid:12) is linear in (cid:12); the plug-in approach suggests replacing (cid:12) with the e cient estimator for (cid:12): under the projection assumption (3) ols is the asymptotically e cient estimator. ^(cid:12) = (z 0z)(cid:0)1(z 0y ) where y and z are the stacked observations on yi and zi: then the estimator for g(x) is. = ^(cid:12)1 + 2^(cid:12)2x + 3^(cid:12)3x2 + (cid:1)(cid:1)(cid:1) + p^(cid:12)pxp(cid:0)1: find the asymptotic distribution of pn (^g(x) (cid:0) g(x)) as n !

Related Questions

1.Write a function encipher(s, n) that takes as inputs an arbitrary string s and a non-negative integer n between 0 and 25, and that returns a new string in which the letters in s have been "rotated" by n characters forward in the alphabet, wrapping around as needed. For example:

>>> encipher('hello', 1) result: 'ifmmp'

>>> encipher('hello', 2) result: 'jgnnq'

>>> encipher('hello', 4) result: 'lipps'

Upper-case letters should be "rotated" to upper-case letters, even if you need to wrap around. For example:

>>> encipher('XYZ', 3)

result: 'ABC'

Lower-case letters should be "rotated" to lower-case letters:

>>> encipher('xyz', 3)

result: 'abc'

Non-alphabetic characters should be left unchanged:

>>> encipher('#caesar!', 2)

result: '#ecguct!'

Hints/reminders:

You can use the built-in functions ord and chr convert from single-character strings to integers and back:

>>> ord('a')

result: 97

>>> chr(97)

result: 'a'

You can use the following test to determine if a character is between 'a' and 'z' in the alphabet:

if 'a' <= c <= 'z':

A similar test will work for upper-case letters.

We recommend writing a helper function rot(c, n) that rotates a single character c forward by n spots in the alphabet. We have given you a template for this helper function in ps3pr3.py that checks to ensure that c is a single-character string. We wrote rot13(c) in lecture; rot(c, n) will be very close to rot13(c)! You can test your rot(c, n) as follows:

>>> rot('a', 1)

result: 'b'

>>> rot('y', 2)

result: 'a'

>>> rot('A', 3)

result: 'D'

>>> rot('Y', 3)

result: 'B'

>>> rot('!', 4)

result: '!'

Once you have rot(c, n), you can write a recursive encipher function.

Once you think you have everything working, here are three more examples to try:

>>> encipher('xyza', 1)

result: 'yzab'

>>> encipher('Z A', 2)

result: 'B C'

>>> encipher('Caesar cipher? I prefer Caesar salad.', 25)

result: 'Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.'

2.Write a function decipher(s) that takes as input an arbitrary string s that has already been enciphered by having its characters "rotated" by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s. For example:

>>> decipher('Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.')

result: 'Caesar cipher? I prefer Caesar salad.'

Here are two more examples:

>>> decipher('Hu lkbjhapvu pz doha ylthpuz hmaly dl mvynla lclyfaopun dl ohcl slhyulk.')

result: 'An education is what remains after we forget everything we have learned.'

>>> decipher('python')

result: 'eniwdc'

------------------------------------

def rot(c, n):
""" your docstring goes here """
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

def letter_prob(c):
""" if c is the space character (' ') or an alphabetic character,
returns c's monogram probability (for English);
returns 1.0 for any other character.
adapted from:

"""
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

if c == ' ': return 0.1904
if c == 'e' or c == 'E': return 0.1017
if c == 't' or c == 'T': return 0.0737
if c == 'a' or c == 'A': return 0.0661
if c == 'o' or c == 'O': return 0.0610
if c == 'i' or c == 'I': return 0.0562
if c == 'n' or c == 'N': return 0.0557
if c == 'h' or c == 'H': return 0.0542
if c == 's' or c == 'S': return 0.0508
if c == 'r' or c == 'R': return 0.0458
if c == 'd' or c == 'D': return 0.0369
if c == 'l' or c == 'L': return 0.0325
if c == 'u' or c == 'U': return 0.0228
if c == 'm' or c == 'M': return 0.0205
if c == 'c' or c == 'C': return 0.0192
if c == 'w' or c == 'W': return 0.0190
if c == 'f' or c == 'F': return 0.0175
if c == 'y' or c == 'Y': return 0.0165
if c == 'g' or c == 'G': return 0.0161
if c == 'p' or c == 'P': return 0.0131
if c == 'b' or c == 'B': return 0.0115
if c == 'v' or c == 'V': return 0.0088
if c == 'k' or c == 'K': return 0.0066
if c == 'x' or c == 'X': return 0.0014
if c == 'j' or c == 'J': return 0.0008
if c == 'q' or c == 'Q': return 0.0008
if c == 'z' or c == 'Z': return 0.0005
return 1.0

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ECON 710 Midterm: ECON 710 UW Madison Midterm 2013a

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