Problem Set #3: Estimation and finite sample inference 1 Averages Economics 835: Econometrics

Transcription

Problem Set #3: Estimation and finite sample inference 1 Averages Economics 835: Econometrics
Problem Set #3: Estimation and finite sample inference
Economics 835: Econometrics
Fall 2012
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Averages
Let x1 , x2 , . . . , xn be a random sample of size n on the random variable x. Let µ = E(x) and σ 2 = var(x).
Let a1 , a2 , . . . , an be a sequence of n numbers.
a) Let:
Wn = a1 x1 + a2 x2 + · · · + an xn
Find a condition on the a’s that imply that Wn is an unbiased estimator of µ.
b) Find var(Wn ) as a function of the a’s and of σ 2 .
c) For any set of numbers a1 , a2 , . . . , an it is the case that:
(a1 + a2 + · · · + an )2
≤ a21 + a22 + · · · + a2n
n
¯ n ), where X
¯ n is the usual
Use this to show that if Wn is an unbiased estimator of µ, then var(Wn ) ≥ var(X
sample average.
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Inference on the median
One advantage of the median over the mean is that yields exact small-sample inference even without assuming
x is normally distributed.
Let x be a random variable with an unknown continuous probability distribution, and suppose that we have
a random sample of size n on that random variable. We would like to construct a 95% confidence interval
for the median M
a) We start by constructing a hypothesis test. Suppose we are testing the null that the median of x is M
against the alternative that it is not.
H0 :
Pr(x ≤ M ) = 0.5
H1 :
Pr(x ≤ M ) 6= 0.5
Let the test statistic be:
tn =
n
X
I(xi ≤ M )
i=1
This test statistic has a simple interpretation: it is the number of observations in the sample less than our
candidate median. What is the exact distribution of tn under the null?
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ECON 835, Fall 2012
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b) Suppose n = 100. Calculate the critical values at which you would reject the null at 5% significance.
You can use just about any statistical package to calculate percentiles of the binomial distribution, or you
could use the Excel function CRITBINOM.
c) Use the previous results to construct a formula for the exact 95% confidence interval for the median when
n = 100.
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The joint normal distribution
We say that the k-vector x has a joint normal distribution with mean µ and covariance matrix Σ (or
x ∼ N (µ, Σ)) if its PDF takes the form:
1
1
0 −1
(x
−
µ)
Σ
(x
−
µ)
exp
f (x) =
−
2
(2π)k/2 |Σ|1/2
The joint normal distribution has several features that make it useful:
1. If x ∼ N (µ, Σ) then Ax + b ∼ N (Aµ + b, AΣA0 ) for any j-vector b and j-by-k matrix A.
2. Let x1 , x2 , . . . xk be a sequence of k independent N (0, 1) random variables, and let x = [x1 x2 · · · xk ].
Then x ∼ N (0, I) where I is the k-by-k identity matrix.
3. Suppose x can be partitioned into two submatrices x1 and x2 , where x1 is length j and x2 is length
k, and that:
x1
µ1
Σ11 Σ12
x=
∼N
,
x2
µ2
Σ21 Σ22
Then:
−1
x1 |x2 ∼ N µ1 + Σ12 Σ−1
22 (x2 − µ2 ), Σ11 − Σ12 Σ22 Σ21
Suppose that Dn = x1 , x2 , . . . xn is a random sample of size n independent N (µ, σ 2 ) random variables. We
said in class that under the null hypothesis H0 : µ = µ0 , the test statistic:
x
¯−µ0
√
√ x
¯ − µ0
σ/ n
t= n
=
s
s/σ
had the Student’s t distribution with n − 1 degrees of freedom. Now “the Student’s t distribution with d
degrees of freedom” is defined as the distribution of the ratio a/b where a and b are independent random
variables, a ∼ N (0, 1) and b is the square root of a χ2 (d) random variable. We won’t go through the full
proof that our test statistic has the tn−1 distribution ,but we will prove a few of the pieces.
Pn
x
¯−µ
√ ∼ N (0, 1).
a) Let x
¯ = n1 i=1 xi . Prove1 that x
¯ ∼ N (µ, σ 2 /n) and that σ/
n
Pn
1
b) Prove that xi − x
¯ is normal and independent of x
¯ for all i. Does this imply that s2 = 1−n
¯)2
i=1 (xi − x
√
2
and s/σ = s /σ are independent of x
¯?
1 The hard/important part here is not showing that E(¯
x) = µ or that var(¯
x) = σ 2 /n, it’s showing that x
¯ is normally
distributed.
ECON 835, Fall 2012
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More on the joint normal distribution
a) Suppose that
y
x
∼N
µy
µx
2
σy
,
σxy
σxy
σx2
where y and x are scalar random variables. Find the conditional distribution of y given x. You can express
this distribution by a PDF or simply in the same sort of way I’m describing their joint distribution.
b) We will use a simple counterexample to prove that two normal random variables are not necessarily
jointly normal. Let x ∼ N (0, 1) and let
x if |x| ≤ 1
y=
−x if |x| > 1
Prove that y ∼ N (0, 1) but that x and y are not jointly normal.