Springer Texts in Business and Economics

Poisson Distribution

a. Using the MGF for the Poisson derived in problem 2.14c one gets Mx(t) = ex(e‘“1).

Differentiating with respect to t yields MX(t) = ex(et-1}Xe‘.

Evaluating MX(t) at t = 0, we get Mx(0) = E(X) = X.

Similarly, differentiating MX(t) once more with respect to t, we get

M'(t) = ex(e,-r) (Xef + ex(e,-r)Xe‘

evaluating it at t = 0 gives

M"(0) = X2 + X = E(X2)

so that var(X) = E(X2) - (E(X)2) = X2 + X - X2 = X.

Hence, the mean and variance of the Poisson are both equal to X.

b. The likelihood function is

n

Подпись: e“nXXiDlP Xi

Подпись: L (X) =X1K2LXJ

so that

logL(X) = - nX + (X Xi J log X - X logXi!

Подпись: i=1i=1

n

Xi

3 log l(X) a ' „

—3Г~ = - n + — = 0

Solving for X, yields Xmle = X.

c. The method of moments equates E(X) to Xi and since E(X) = X the solution is >0 = Xi, same as the ML method.

d. E(X) = P E(Xi)/n = nX/n = X. Therefore X is unbiased for X. Also,

i=1

var(X) = var(X) = - which tends to zero as n! 1. Therefore, the

y ' n n 7

sufficient condition for X to be consistent for X is satisfied.

e. The joint probability function can be written as f(Xb..,Xn;A) = e-nAAnXX7YT = h&A) g(Xi,..,Xn)

where h(X, X) = є-”^^ and g(X1,..,Xn) = X!1X!. The latter is inde­pendent of X in form and domain. Therefore, X is a sufficient statistic for X.

f. logf(X; X) =-X + XlogX - logX! and

Э logf(X; A) X

9A + A

Э2 logf(X; A) - X

3A = "a2".

image044The Cramer-Rao lower bound for any unbiased estimator "X of X is given by 1 _ A2 _ A

nE (@2 log3Af(X;A)) = nE(X) n'

But var(X) = X/n, see part (d). Hence, X attains the Cramer-Rao lower bound.

g. The likelihood ratio is given by

n

У' X'

L(2) _ e-2n2i=i i

LW = "T^

e-4n4i=i

The uniformly most powerful critical region C of size a < 0.05 is given by

n

/1V? Xi

e2 V 2 )i 1 < k inside C

image045
Подпись: log 2 < K0

Taking logarithms of both sides and rearranging terms, we get

Vi=1

or

n

HX‘ ^K

i=1

where K is determined by making the size of C = a < 0.05. In this case,

n n

E X; ~ Poisson(nX) and under Ho; X = 2. Therefore, X; ~ Poisson

i=i i=i

(X = 18). Hence a = Pr[Poisson(18) > K] < 0.05.

From the Poisson tables, for X = 18, K = 26 gives Pr[Poisson(18) >

n

26] = 0.0446. Hence, Xi > 26 is our required critical region.

i=i

h.

Подпись: L(2) L(X ) image048

The likelihood ratio test is

so that

n

LR = -2 log L(2) + 2 log L(X) = -2n(X - 2) - 2j^ X; [log 2 - log X].

i=i

Подпись: C (A) Подпись: 92 log L (A) 9A2 image051

In this case,

Подпись: I (A) = -E Подпись: @2 logL(A) 9A2 Подпись: nA A2 Подпись: n A.

and

The Wald statistic is based upon

W = (X - 2)21 (Amk) = (X - 2)2 • XX

using the fact that Xmle = X. The LM statistic is based upon

Подпись: LM = S2(2)I-1(2) =2(X - 2)2 2 n(X - 2)2

4 n 2

Note that all three test statistics are based upon |X - 2| > K, and for finite

n

n the same exact critical value could be obtained using the fact that ^ Xi

i=1

has a Poisson distribution, see part (g).

1.4 The Geometric Distribution.

Подпись: a. Using the MGF for the Geometric distribution derived in problem 2.14d, one gets 0e* Mx(Y) [1 - (1 - 0)e*]'

Differentiating it with respect to t yields

л *, ,* 0e* [1 - (1 - 0)e‘] + (1 - 0)e*0e* 0e*

MX(t) = [1 - (1 - 0)ef = [1 - (1 - 0)ef

evaluating MX(t) at t = 0, we get 01

MX0) = E(X) = 0.

Подпись: M'(t) =Similarly, differentiating MX(t) once more with respect to t, we get 0e‘ [1 - (1 - 0)et]2 + 2 [1 - (1 - 0)et] (1 - 0)e‘0e‘

[1 - (1 - 0)e*]4

image059

image028

2 Подпись: var(X) = E(X2) - (E(X))2 =0 1 1 0

02 02 02

b. The likelihood function is given by

n

£ Xi-n

L (0) = 0n(1 - 0)i=1 so that

log L(0) = nlog 0 + Xi - nj log(1 - 0)

91ogL(0) _ iDtXi n

90 = 0 (1 - 0)

solving for 0 one gets

n

n(1 - 0) - 0J2 Xi + n0 = 0

i=1

or

n

n = Xi

i=1

which yields

л n 1 ™mle = n/ ^ ] Xi = .

Подпись: i=1-X.

The method of moments estimator equates e(x) = x so that

11

- = X or 0 =

0 X

which is the same as the MLE.

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