Springer Texts in Business and Economics
The Exponential Distribution
a. Using the MGF for the exponential distribution derived in problem 2.14e, we get
1
(1 — 0t).
Differentiating with respect to t yields
MX(t) =
Therefore
MX0(O) = O = E(X).
Differentiating MX(t) with respect to t yields 2O2 (1 - Ot) 2O2
(1 - Ot)4 (1 - Ot)3'
Therefore
MXX (0) = 2O2 = E(X2).
Hence
var(X) = E(X2) - (E(X))2 = 2™2 - ™2 = ™2.
b. The likelihood function is given by
n
_ / 1 Y - P Xi/O
L (O)= 0e 'd1
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so that
n
V X'
э logL (O) - n = 0
@O O C O2
solving for O one gets
n
Xi - nO D 0
i=1
so that
Omle = X.
c. The method of moments equates E(X) = X. In this case, E(X) = O, hence O = X is the same as MLE.
n
d. E(X) = E(Xi)/n = n0/n = 0. Hence, X is unbiased for 0. Also,
i=1
var(X) = var(X)/n = 02/n which goes to zero as n! 1. Hence, the sufficient condition for Xto be consistent for 0 is satisfied.
e.
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The joint p. d.f. is given by
where h(X; 0) = e-nX/e (1)n and g(X1,... ,Xn) = 1 independent of 0 in form and domain. Hence, by the factorization theorem, X is a sufficient statistic for 0.
f. logf(X; 0) = - log 0 - X and
9 logf(X; 0) -1 i X X - 0
90 = T + 02 = 02
92 logf(X; 0) 1 2X0 0 - 2X
90 = 02 - “0^ = 03
The Cramer-Rao lower bound for any unbiased estimator 0 of 0 is given by
-1 -03 _ 02
nE ^ 92 logf(X;0)^ nE (0 - 2X) _ n.
But var(X) = 02/n, see part (d). Hence, X attains the Cramer-Rao lower bound.
g.
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The likelihood ratio is given by
- £ X/2
2n e ‘=1 < k inside C.
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Taking logarithms of both sides and rearranging terms, we get
or
n
EXi > K
i=1
where K is determined by making the size of C = a < 0.05. In this case,
n
^2 Xi is distributed as a Gamma p. d.f. with " = 0 and a = n. Under
i=i
Ho; 0 = 1. Therefore,
n
Xt ~ Gamma(a = n, = 1/.
i=i
Hence, Pr[Gamma(a = n, " = 1/ > K] < 0.05
K should be determined from the integral Л щ/xn-1e-x dx = 0.05 for n = 20.
h. The likelihood ratio test is
L(1/ L(X/ '
so that n
2J2 Xi - 2nlogX — 2n.
i=1
In this case,
C (0/
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and
The Wald statistic is based upon
'2 _n_
X2
using the fact that 0mle = Xi. The LM statistic is based upon LM = S2(1/I-1(D = Xi — nj І = n (X — 1)2 .
All three test statistics are based upon |X — 11 > k and, for finite n, the same
n
exact critical value could be obtained using the fact that Xi is Gamma
i=1
(a = n, and " = 1) under Ho, see part (g).
