Page 279 - Applied statistics and probability for engineers

Section 7-4/Methods of Point Estimation 257

Moments Let X , X , , X n1 2 … be a random sample from the probability distribution f x) where
(
(
f x) can be a discrete probability mass function or a continuous probability density
k
k
(
2
function. The kth population moment (or distribution moment) is E X ), = 1 , ,...
(
. The corresponding kth sample moment is 1/ n)∑ n i= 1 X ,k = 1 2, ….
k
,
i
To illustrate, the i rst population moment is E X( ) = μ, and the i rst sample moment is
( 1/ n)∑ n i= 1 X i = X. Thus, by equating the population and sample moments, we i nd that ˆ μ = X .
That is, the sample mean is the moment estimator of the population mean. In the general
case, the population moments will be functions of the unknown parameters of the distribution,
,
say, θ 1 , θ …, θ .
2
m
Moment Estimators
…
Let X , X , , X n1 2 be a random sample from either a probability mass function or a
probability density function with m unknown parameters θ 1 , θ 2 , …, θ . The moment
m
∧ ∧ ∧
1,
estimators Q Q2 , … , Qm are found by equating the i rst m population moments to
the i rst m sample moments and solving the resulting equations for the unknown
parameters.
Example 7-7 Exponential Distribution Moment Estimator Suppose that X , X , … , X n is a random sample
1
2
from an exponential distribution with parameter λ. Now there is only one parameter to estimate,
so we must equate E X( ) to X. For the exponential, E X ( ) = 1 / λ. Therefore, E X ( ) = X results in 1 / λ = X, so λ = 1/ X,
is the moment estimator of λ.

As an example, suppose that the time to failure of an electronic module used in an automo-
bile engine controller is tested at an elevated temperature to accelerate the failure mecha-
nism. The time to failure is exponentially distributed. Eight units are randomly selected and
tested, resulting in the following failure time (in hours): x 1 = 11 96, x 2 = 5 03, x 3 = 67 40,
.
.
.
x 4 = 16 07, x 5 = 31 50, x 6 = 7 73, x 7 = 11 10, and x 8 = 22 38. Because x = 21 65. , the moment
.
.
.
.
.
ˆ
0
/
estimate of λ is λ = 1 / x = 1 21 .65 = .0462 .
Example 7-8 Normal Distribution Moment Estimators Suppose that X , X , … , X n is a random sample
1
2
2
(
from a normal distribution with parameters μ and σ . For the normal distribution, E X) = μ and
2
E X ) = μ 2 + È . Equating E X( ) to X and E(X ) to ∑ n i X gives
2
2
2
1
(
n i
1 n
2
2
μ = X, μ + σ = ∑ X i 2
n = i 1
Solving these equations gives the moment estimators
n n ⎞ 2 n
2 ⎛ 1 2
X
∑ X i − n ⎜ ∑ X i ⎟ ⎠ ∑ (X i − )
2
ˆ μ = X, ˆ σ = = i 1 ⎝ n i =1 = = i 1
n n
Practical Conclusion: Notice that the moment estimator of σ is not an unbiased estimator.
2