1. Estimation Maximum Likelihood Estimates Industrial Engineering

2. Discrete Case p x ( ) L q = × p x ( )
Suppose we have hypothesized a discrete distribution from which our data which has some unknown parameter Let denote the probability mass function for this distribution. The likelihood function is q p x q ( ) p x n ( ) L q = × 1 2

3. Discrete Case p x ( ) L q = × L ( ) q p x ( )
Suppose we have hypothesized a discrete distribution from which our data which has some unknown parameter Let denote the probability mass function for this distribution. The likelihood function is q p x q ( ) p x n ( ) L q = × 1 2 L ( ) q
is just the joint probability mass function

4. Discrete Case L for all possible ( $ ) q ³ L ( ) q $ q
Since is just the joint probability, we want to choose some which maximizes this joint probability mass function. $ q L
for
all
possible ( $ ) q

5. Continuous Case
Suppose we have a set of nine observations x1, x2, X9 which have underlying distribution exponential (in this case scale parameter l = 2.0).
0.053
0.458
0.112
0.602
0.178
0.805
0.255
1.151
0.347

6. Continuous Case L f x ( ) l = ×
Suppose we have a set of nine observations x1, x2, X9 which have underlying distribution exponential (in this case scale parameter l = 2.0). Our object is to estimate the true but unknown parameter l. L f x n ( ) l = × 1 2

7. MLE (Exponential) L f x ( ) l = × = × l e = å l e - x - x n n 1 2 n 1

8. MLE (Exponential) = l e L f x ( ) l = × = × l e = å l e - 9 3 961 . -2 = × - l e x n 1 2 = å - l e n x i = - l 9 3
961 e .

9. MLE (Exponential)

10. MLE (Exponential) ¶ l L ( ) =
We can use the plot to graphically solve for the best estimate of l. Alternatively, we can find the maximum analytically by using calculus. Specifically, l L ( ) =

11. Log Likelihood Ln LN L ( ) q =
The natural log is a monotonically increasing function. Consequently, maximizing the log of the likelihood function is the same as maximizing the likelihood function itself. Ln LN L ( ) q =

12. MLE (Exponential) L f x n ( ) l = × 1 2 = å - l e n x i

13. å MLE (Exponential) Ln nln x ( ) l = - L f x ( ) l = × = å l e i - x n1 2 = å - l e n x i Ln
nln x i ( ) l = - å

14. å å å MLE (Exponential) Ln nln x ( ) l = - ¶ l ¶l Ln n x ( ) ln( = - =
= 0 = - å n x i l

15. MLE (Exponential) n x i l - = å n x i l = å

16. MLE (Exponential) n x i l - = å n x i l = å x l = 1

17. MLE (Exponential) n x i l - = å n x i l = å x l = 1 x l = 1 $

18. MLE (Exponential) x l = 1 $ i X 1 0.053 2 0.112 3 0.178 4 0.255 5
0.347 6
0.458 7
0.602 8
0.805 9
1.151
Sum =
3.961
X-bar =
0.440 x l = 1 $

19. MLE (Exponential) $ . l = 1 44 2 27 x l = 1 $ i X 1 0.053 2 0.112 3
0.178 4
0.255 5
0.347 6
0.458 7
0.602 8
0.805 9
1.151
Sum =
3.961
X-bar =
0.440 x l = 1 $ $ . l = 1 44 2 27

20. Experimental Data
Suppose we wish to make some estimates on time to fail for a new power supply. 40 units are randomly selected and tested to failure. Failure times are recorded follow: X 1 . 19 =

21. Failure Data