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Statistical Method In Economics

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PROBLEMS

Problem 1 Let X1 , ..., X n be a random sample (i.i.d.) of size n from a population with distribution ƒ (x) with mean µ and variance σ2 (both finite). Consider a statistic formed by taking a linear combination of the X i s, c1 X1 +... +c n X n , where ci ≥ 0. For example, the sample mean, X , is the linear combination with ci = 1 for all i. n

a. of µ?

b.

Under what condition(s) is c1 X1 +... +c n X n an unbiased estimator

What is the variance of c1 X 1 + ... + cn X n ?

c. Find the linear combination (derive the c's) that minimizes the variance without violating the condition(s) you derived in part a. (i.e. find the most efficient linear unbiased estimator of µ).

Problem 2 Suppose that you have a sample of size n, X1 , ..., X n , from a population with mean µ and variance σ2. You know that your draws from the popula- tion, the X i , are identically distributed, however, they are not independent 2

and have Cov (X , X i ) =( gσ > 0 , for all i = (. / a.

What is E X ? Is the sample mean an unbiased estimator of µ? b.

What is the MSE of X ? You can use the fact that V ar Σ Σ Σ V ar (X i ) + Cov (X i , X ( ). i

c.

0?

X

=

Σ i

i

i (/=i

Is the sample mean a consistent estimator of µ? consistent if g =

Would it be

1

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Problem 3 2

Suppose that Z1, Z2, ..., Zn is a random sample from an exponential dis-

tribution with parameter Z, so ƒ (x) = a. b. c.

Ze — λx Find the Method of Moments estimator for Z. 0

x≥0 elserhere

Find the Maximum Likelihood estimator for Z. Find the Maximum Likelihood estimator for

√

Z.

d.

Is the4 estimator in part b unbiased? Is it consistent?

Problem Assume a sample of continuous random variables: X1 , X2 , ..., X n , where E [X ]i = µ, V ar [X ]i = σ2 > 0. Consider the following estimators: µ 1,n Σ 1 n X n, µ^2,n= X i. n+1

a. b.

^

=

i=1

Are µ^1,n and µ^2,n unbiased?

Are µ^ and ^ consistent?

1,n µ 2,n

c.What do you conclude about the relation between unbiased and consistent estimators? d.

For what values of n does µ^2,n have a lower mean square error than µ 1,n ? ^ Problem 5 You have a sample of size n, X 1 , ..., X n from a U [8l, 8u] distribution. a. Assume that it is known that 8l = 0.

Find the MM and MLE estimators of 8u. b.

8u . c.

Now assume that both 8l and 8u are unknown. estimators of 8l and

Find the MM

Determine whether the estimators in parts a and b are unbiased and consistent.

Problem 6 Let X1 , ..., X n be a random sample of size n from a U [0, 8] population.

Consider the following estimators of 8: ^ 81 = k 1 X 2 ^ 82 = k 2 X ^ 83 =

k3 X (n)

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where the ki are constants and

3

X (n) = max (X 1 , ..., X n ) X (n ) is known as the nth order statistic. a.

Calculate Pr X distribution. b.

(n)

≤ x , given our knowledge of the population

Calculate the pdf of X (n ) .

h i c.

Choose k1 ,k^21, ^and k3 such^that all three estimators are^ unbiased; 2 3 1 1 call these values k ,k , and k (In other words, solve E k X = 8, and so forth). d.

Calculate the following variances and compare them:

V ar ^k1 X 2 V ar ^k2 X V ar ^k3 X (n)

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ANSWERS

Problem 1 a.

For the statistic to be unbiased we must have ! Σn µ = E ci Xi i=1

=

E (X ) i

=

µ

Σn c

Σn c i

i=1

i=1

Given that all of the X i are independent, the variance of the sta! n Σ Σn V ar ci Xi = c2i V ar (X i ) i=1

i=1

= c.

Σn

ci = 1.

So the statistic will be unbiased if and only if b. tistic is

i

i=1

We want to minimize σ2

subject to the constraint

Σn i=1

Σn

2Σ

n

c

i=1

2

i

c2 i

i=1

ci = 1

The Lagrangian of this optimization problem is Σn 2 G = σ2 c — i Z i=1

σ

1

Σn i=1

ci — 1

!

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2

And we have the first-order conditions 6G = 6ci

σ 22c — i Z= 0

Z 2σ2 ci = This implies that all of the ci are the same, that is c1 = c2 = ... = cn = c. We can then solve for this common Σn value c by substituting back into the 1 constraint

i=1

c = 1, which implies that c =

n.

Problem 2 a.

E X

!

1Σ Xi n i=1 n

= E

1 Σ E (X ) i n i=1 n 1Σ µ = n i=1 = µ n

=

So X is an unbiased estimator of µ. b.

M SE X

=

2

E X —µ = 0 + V ar 1 = = = =

n2

1 ,@

+ V arn X !

1Σ n i=1 X i ! Σn

V ar Σ

n2

i=1 Xi

Σ Σ V ar (X )i +

i , Σ Σ Σ 1 @ σ2+

n2 σ2

i

i

(n — 1) n n +

gσ2

i

gσ

Cov (X 1 , iX )(

(=/ i

,

1

2,

(/=i

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c. To determine whether the sample mean is a consistent estimator, we need to see whether the MSE goes to zero as n approaches infinity. σ2 n→∞

lim M SE X

=

n= →∞

limn

n→∞

+

(n — 1) gσ n

2 lim σ+ n→ ∞ n

lim = 0 + gσ2 = gσ2

2

(n n — 1) gσ

2

Thus (assuming a positive variance) the sample mean is not a consistent estimator of the population mean unless g = 0.

Problem 3 a.

We know that the first moment of Z ∫is ∞ E (Z) = zƒ (z) dz ∫ O∞ λx = zZe —dz O 1 = Z Then, we calculate our method of moments estimator by setting E (Z) = Z and solving for Z. 1 = 1 Σn Z i Z n i=1 1 Σ nn = = Z^MM z i i=1 Z b.

The exponential pdf is ƒ (z) = Ze—λx . Thus, the likelihood function for Z is Y n G(Z; z) = i=1 Ze— λxi .

Then, the log likelihood function is Y n ln G(Z; z) = ln i=1 Ze—λxi Σ n = (—Zz )i + n ln Z i=1 Differentiating the function with respect to Z and set it to 0, Σ n 6 ln G(Z; z) n =— i=1 (z i) + Z= 0 6Z =⇒ Σ n n i=1 zi = Z

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So we find that the MLE estimator is the same as the MM estimator: n 1 ^ Z M LE = Σn i=1 z i = Z √

Using the invariance property, the MLE for Z is simply √ q 1 r n ^ n ^Z = √ = Z ML E = M LE i=1 Z Σ

c.

i

.

Z ^ d.

We know t hat Z will be consistent because all MLE estimators M LE ^M LE ) = are consistent.

But is it unbiased? A direct calculation of E(Z 1 1 1 E is untractable (note that E 7 / E 7 , in general). But we can 7 ( ) use Jensens's inequality to help us out, which tells us = that for any random variable X , if g (x) is a convex function, then E (g (X )) ≥ g (E (X )) with a strict inequality if g (x) is strictly convex. Thus, using Z as our 1 random variable and the strictly convex function g Z = , we have 7 1 1 1 E =Z >E Z = Z E (Z) and we know that our estimator is biased.

Note that if we had instead used the version of the exponetial pdf that replaces Z with 1 , and β estimated the parameter β by either method, we ^ which is both unbiased (since E (Z) = = β) and would have found β = Z, consistent (by the law of large numbers).

Problem 4 a.

Bias µ ^ 1,n = E µ^ unbiased for every n.

Bias µ^2,n = E µ2,n ^ n n+1

µ—µ=

—µ = n+1

0/so µ

1,n

—µ=E

1 λ

— µ = E [X ] n— µ = µ — µ = 0 so µ "

1 n+1

Σn i=1

#i X

^ 2,n is biased for for every n.

—µ=

Σn

1 n+1 i=1

^ 1,n is

E [X i ] — µ =

b. lim n → ∞ P .µ^ — µ <. o = lim P (|Xn — µ| < o) = P (µ — o < X n→∞ 1,n P (X n ≤ µ + o)—P (X n ≤ µ — o) (because the R.V. are continuous)= F (µ + o)— F (µ — o) /= 0 for some o > 0. Thus µ consistent. " ^ 1,n is not # n Σ 2 Σn 1 1 As for µ ^ 2,n : V ar µ^ 2,n = V ar Xi = V ar [X ]i = n+1 n+1 1 n+1

i=1

2

n

< µ + o) =

i=1

nσ2 so:

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= lim V ar µ^ +lim Bias µ^ 2,n n→ ∞ limn → ∞ M S E µ ^ 2,n 0, proving that µ ^ 2,n is2,n consistent.n → ∞ c. An unbiased estimator is not necessarily consistent; a consistent estimator is not necessarily unbiased.

Problem 5 a.

MM: moment equation:

2

=

Since we know that 8l = 0, we only need to use the first

0 + 8h E(X) = , 2 Then, the MM estimator is obtained by solving ^ Σ1 n 8h = X i= X 2 n Σ i=1 ^ 8h = n X i = 2X 2 i=1 n MLE: The uniform pdf is 1 1 ƒ(x) = = 8h — 8l 8h and the likelihood function and log likelihood functions are given (respec- tively) by G(0, 8h ; x1 , x2 , ..., xn ) = ln G(0, 8h ; x) =

n 1 8h —n ln 8h

In order to maximize the log likelihood function above, we need to minimize 8h subject to the constraint xi ≤ 8h 6 xi .

Then, it must be that b.

^8h are The first two moments = max(xi )

8l + 8h 2 , 2 8 + 8h2 + 8l 8h . l E(X 2 ) = 3 Then, the MM estimators are obtained by solving n ^ 8l + ^8 h 1 Σ = i=1 X i = X, 2 n 2 2 ^ n 8l + 8^ h+ 8 ^ 8l^h = 1 Σ i=1 X 2i = X 2 .

3 n E(X) =

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They are given by ^l 8 ^h c.

8

=

X —

=

X +

q q

2

3(X 2 — X ),

2

3(X 2 — X ).

We begin with the MM estimator for part a:

^ 8 ^h =8h 2 So this estimator is unbiased.

Then we consider the variance. Σ 4 n 4 8 21+ 8 2— 2 28 81 2 V ar 8^ h = V ar 2X = V ar (X i ) = n 12 i=1 Because the bias is zero and the variance n2 approaches zero as n gets large, the MSE also approaches zero, and the estimator is consistent.

E

8^ h = E 2X = 2

For the MLE estimator, we use the fact (shown in problem 6) that, for n—1 this estimator, ƒ (x) = nx . Then we can see that the MLE estimator is 8n biased: ∫ h 8 nxn n E 8^h = dx = 8 O hn 8 n+1 h n However, as n gets large, —→ 1, so t he bias approaches zero, and we n+1 know that in general, MLE estimators are consistent.

For the MM estimators in part b, finding the expected value is not par- ticularly tractable nor particularly interesting, so I will retract this portion of the question.

Problem 6 Note that for a U [0, 8] distribution, ƒ (x) = 1 , F (x) = x , E (X) = 8 , 8

8

and V ar (X ) = 8 .12 a.

It will be important that the sample draws are independent (and identical). Then, 2

Pr X (n) ≤ x

=

2

Pr (max(X i ) ≤ x) = П Pr (X ≤i x) i=1 = F (x)n x 8 =

n

n

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b.

In part a, we found the cdf of X (n ) , so the pdf is just the derivaive of this: d

F(n) (x) n dx d x dx 8

ƒ(n) (x) = =

nxn— 1

=

8n

c. We want to choose constants k1, k2, and k3 such that our estimators are unbiased. For the first estimator, k^1 E E (Xk2^)1X2 8 k^1 2 k^1

Then, for the second estimator k^2E X E k^28 X k^ 2

2 k^2

And for the third estimator ∫

k^3 O

8

^ n k3 n 8

(n)

xnxn—1

8nn+1

8

=

8

=

8

=

2

= = =

8 8 8

=

2

= 8

E ^ k^3X(n)

k 3E X

=

= 8

dx = 8

8 n+1 k^3

n+1 = n = 8

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d. Now we calculate the variances of our estimators, using the con- stants that we found in part c.

82 V ar (2X 2 ) 8 = 4V ar (X 2 ) = 3 V ar 2X = 4V ar X = 3n 2 n+1 n+1 V ar = V ar2 X (n) n X (n) n To find V ar X (n) , we will need E (n) ∫ X 2 : 8 nxn+1 n X 2 = dx = 8 2 E (n) O n 8 n+2 so 2 n n n n82 2 V ar X = 82 = n+2 8 — (n) +1 (n + 2) (n + 1)2 and 2 n+1 n+1 n82 V ar = 82 X (n) n n 2 = n (n + 2) (n + 2) (n + 1) Thus, whenever n > 1, V ar k^3 X (n)

^

< V ar k 2 X

< V ar k^1 X 2

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