Page 281 - Applied statistics and probability for engineers

Section 7-4/Methods of Point Estimation 259

Example 7-10 Bernoulli Distribution MLE Let X be a Bernoulli random variable. The probability mass function is

⎧
⎪
x
; (

f x p) = ⎨ p (1 − p) 1 − x , x = 0 ,1
⎩ ⎪ , 0 otherwise
where p is the parameter to be estimated. The likelihood function of a random sample of size n is
L p ( ) = p (1 − p) 1 − x 1 p (1 − p) 1 − x 2 … p (1 − p) 1 − x n
x 1
x n
x 2
n n
n ∑ x i n − ∑ x i
= Π p (1 − p) 1 1−x i = p = i 1 ( 1− ) p = i 1
x i

i=1
We observe that if ˆ p maximizes L p ( ), ˆ p also maximizes ln L p ( ). Therefore,
⎛ n ⎞ ⎛ n ⎞
ln L p ( ) = ∑ x i ⎟ ln p + ⎜ n − ∑ x i ⎟ ln(1 − p)
⎜
i ⎝ =1 ⎠ ⎝ i=1 ⎠
Now,
n
n ⎛ n − ∑ ⎞
(
d ln L p) = ∑ x i − ⎜ ⎝ i=1 x i⎟ ⎠
i=1
dp p 1 − p
Equating this to zero and solving for p yields ˆ p = (1 / n)∑ n x i . Therefore, the maximum likelihood estimator of p is
i=1
n
1
ˆ
P = ∑
n X i
i =1
Suppose that this estimator were applied to the following situation: n items are selected
at random from a production line, and each item is judged as either defective (in which case
n
we set x i = 1) or nondefective (in which case we set x i = 0). Then ∑ =i x is the number of
i
1
defective units in the sample, and ˆ p is the sample proportion defective. The parameter p is
the population proportion defective, and it seems intuitively quite reasonable to use ˆ p as an
estimate of p.
Although the interpretation of the likelihood function just given is coni ned to the dis-
crete random variable case, the method of maximum likelihood can easily be extended to a
continuous distribution. We now give two examples of maximum likelihood estimation for
continuous distributions.

Example 7-11 Normal Distribution MLE Let X be normally distributed with unknown μ and known variance
σ . The likelihood function of a random sample of size n, say X X 2 ,… , X n , is
2
1 ,
n 1 −( x i −μ) ( ) 1 −1 σ 2 ∑ n n ( x i −μ) 2
2
2
σ
/ 2
L μ ( ) = Π e = n/ 2 e 2 i=1
2
i=1 σ 2 π ( 2 πσ )
Now
ln ( 2 2 −1 n ( − μ)
2
L
È 2
μ) = −(n / 2 ln ) ( 2 πσ ) −( ) ∑ x i
= i 1