Springer Texts in Business and Economics

The Binomial Distribution

a. Pr[X = 5 or 6] = Pr[X = 5] + Pr[X = 6]

= b(n = 20,X = 5, 0 = 0.1) C b(n = 20,X = 6, 0 = 0.1)

j (0.1)5(0.9)15 C (0.1)6(0.9)14

= 0.0319 C 0.0089 = 0.0408.

This can be easily done with a calculator, on the computer or using the Binomial tables, see Freund (1992).

Hence,

n (1 - 0)n-X (1 - 1 C 0)n-n+X n - X

= n(1 - 0)n-X0X = b(n, X, 0).

X

c. Using the MGF for the Binomial distribution given in problem 2.14a, we get MX(t) = [(1 - 0) C 0e‘]n.

Differentiating with respect to t yields MX(t) = n[(1 - 0) C 0e‘]n-10e‘. Therefore, MX (0) = n0 = E(X).

Differentiating MX(t) again with respect to t yields

MX(t) = n(n - 1)[(1 - 0) C 0e‘]n-2(0e*)2 C n[(1 - 0) C 0e‘]n-10e‘.

Therefore MX(0) = n(n - 1)02 C n0 = E(X2).

Hence var(X) = E(X2) — (E(X))2 = n0 + n202 — n02 — n202 = n0(1 — 0).

n

An alternative proof for E(X) = Xb(n, X, 0). This entails factorial

X=0

moments and the reader is referred to Freund (1992).

d. The likelihood function is given by

X Xi n-x Xi

L(0) = f(Xi,..,Xn; 0) = 0i=i (1 — 0) i=i

P X^ log0 C (n — P X^ log(1 — 0)

n

X

Э log L(0) i=1 1

90 0

Solving for 0 one gets

n n n

J^Xi — 0J2Xi — 0n C 0J2Xi = 0

i= 1 i=1 i=1

n

so that 0mle = Xi/n = X.

i=1

n

e. E(X) = J2 E(Xi)/n = n0/n = 0.

i=1

Hence, X is unbiased for 9. Also, var(X) = var(X) / n = 0(1 — 0)/ n which goes to zero as n! 1. Hence, the sufficient condition for X to be consistent for 0 is satisfied.

f. The joint probability function in part d can be written as

f(Xb..,Xn; 0) = 0nX(1 — 0)n-nX = h(X, 0)g(X1,... ,Xn)

where h(X, 0) = 0n2X(1 — 0)n-nX and g(Xi,.., Xn) = 1 for all X/s. The latter function is independent of 0 in form and domain. Hence, by the factorization theorem, X is sufficient for 0. g. Xi was shown to be MVU for 0 for the Bernoulli case in Example 2 in the text.

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

,,, P Xi n - P Xi

1 1=1 2 i=1 < k inside C. Taking logarithms of both sides

- X xi(log 3) + n - X xi log2 < logk

i=1

solving - ^XXy log 6 < K or > K

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

nn

X Xi ~ b(n, 0) and under Ho ; 0 = 0.2. Therefore, X Xi ~ b(n = 20, 0 =

i=1 i=1

0.2). Hence, a = Pr[b(n = 20,0 = 0.2) > K] < 0.05.

From the Binomial tables for n = 20 and 0 = 0.2, K = 7 gives Pr[b(n = 20,

n

0 = 0.2) > 7] = 0.0322. Hence, X Xi > 7 is our required critical region.

i=1

i. The likelihood ratio test is

L(0.2)

L(0mle)

so that LR = —2 logL(0.2) + 2 logL(0mle)

i=1

X (1 — X )/n

and the LM statistic is given by LM =

(0.2)(0.8)/n

Although, the three statistics, LR, LM and W look different, they are all based on |X — 21 > k and for a finite n, the same exact critical value could be obtained from the binomial distribution.