Christopher Dougherty EC220 - Introduction to econometrics (chapter 10) Slideshow: introduction to maximum likelihood estimation Original citation: Dougherty,

Christopher Dougherty

EC220 - Introduction to econometrics (chapter 10)Slideshow: introduction to maximum likelihood estimation

Original citation:

Dougherty, C. (2012) EC220 - Introduction to econometrics (chapter 10). [Teaching Resource]

© 2012 The Author

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Available in LSE Learning Resources Online: May 2012

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1

INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION

This sequence introduces the principle of maximum likelihood estimation and illustrates it with some simple examples.

L

p

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

2


Suppose that you have a normally-distributed random variable X with unknown population mean and standard deviation , and that you have a sample of two observations, 4 and 6. For the time being, we will assume that is equal to 1.

L

p

0.0

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0.3

0.4

0 1 2 3 4 5 6 7 8

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

3


Suppose initially you consider the hypothesis = 3.5. Under this hypothesis the probability density at 4 would be 0.3521 and that at 6 would be 0.0175.

L

p

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

p(4) p(6)

3.5 0.3521 0.0175

0.3521

0.0175

4


The joint probability density, shown in the bottom chart, is the product of these, 0.0062.

p(4) p(6) L

3.5 0.3521 0.0175 0.0062

L

p

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

0.3521

0.0175

5


Next consider the hypothesis = 4.0. Under this hypothesis the probability densities associated with the two observations are 0.3989 and 0.0540, and the joint probability density is 0.0215.

p(4) p(6) L

3.5 0.3521 0.0175 0.0062

4.0 0.3989 0.0540 0.0215

L

p

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

0.3989

0.0540

6


Under the hypothesis = 4.5, the probability densities are 0.3521 and 0.1295, and the joint probability density is 0.0456.

p(4) p(6) L

3.5 0.3521 0.0175 0.0062

4.0 0.3989 0.0540 0.0215

4.5 0.3521 0.1295 0.0456L

p

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

0.3521

0.1295

7


Under the hypothesis = 5.0, the probability densities are both 0.2420 and the joint probability density is 0.0585.

p(4) p(6) L

3.5 0.3521 0.0175 0.0062

4.0 0.3989 0.0540 0.0215

4.5 0.3521 0.1295 0.0456

5.0 0.2420 0.2420 0.0585L

p

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

0.24200.2420

8


Under the hypothesis = 5.5, the probability densities are 0.1295 and 0.3521 and the joint probability density is 0.0456.

p(4) p(6) L

3.5 0.3521 0.0175 0.0062

4.0 0.3989 0.0540 0.0215

4.5 0.3521 0.1295 0.0456

5.0 0.2420 0.2420 0.0585

5.5 0.1295 0.3521 0.0456

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

L

p

0.3521

0.1295

9


The complete joint density function for all values of has now been plotted in the lower diagram. We see that it peaks at = 5.

p(4) p(6) L

3.5 0.3521 0.0175 0.0062

4.0 0.3989 0.0540 0.0215

4.5 0.3521 0.1295 0.0456

5.0 0.2420 0.2420 0.0585

5.5 0.1295 0.3521 0.0456

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

p

L

0.1295

0.3521

10


Now we will look at the mathematics of the example. If X is normally distributed with mean and standard deviation , its density function is as shown.

2

21

21

)(

X

eXf

11


For the time being, we are assuming is equal to 1, so the density function simplifies to the second expression.

2

21

21

)(

X

eXf

2

21

21

)(

X

eXf

12


Hence we obtain the probability densities for the observations where X = 4 and X = 6.

2

421

21

)4(

ef

26

21

21

)6(

ef

2

21

21

)(

X

eXf

2

21

21

)(

X

eXf

13


The joint probability density for the two observations in the sample is just the product of their individual densities.

2

421

21

)4(

ef

26

21

21

)6(

ef

2

21

21

)(

X

eXf

2

21

21

)(

X

eXf

2

62

124

2

1

21

21

eejoint density

14


In maximum likelihood estimation we choose as our estimate of the value that gives us the greatest joint density for the observations in our sample. This value is associated with the greatest probability, or maximum likelihood, of obtaining the observations in the sample.

2

21

21

)(

X

eXf

2

21

21

)(

X

eXf

2

421

21

)4(

ef

26

21

21

)6(

ef

2

62

124

2

1

21

21

eejoint density

15


In the graphical treatment we saw that this occurs when is equal to 5. We will prove this must be the case mathematically.

p(4) p(6) L

3.5 0.3521 0.0175 0.0062

4.0 0.3989 0.0540 0.0215

4.5 0.3521 0.1295 0.0456

5.0 0.2420 0.2420 0.0585

5.5 0.1295 0.3521 0.0456

0.00

0.02

0.04

0.06

0 1 2 3 4 5 6 7 8

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

p

L

0.1295

0.3521

16


To do this, we treat the sample values X = 4 and X = 6 as given and we use the calculus to determine the value of that maximizes the expression.

2

6212

421

21

21

)6,4|(

eeL

17


When it is regarded in this way, the expression is called the likelihood function for , given the sample observations 4 and 6. This is the meaning of L(| 4,6).

2

6212

421

21

21

)6,4|(

eeL

18


To maximize the expression, we could differentiate with respect to and set the result equal to 0. This would be a little laborious. Fortunately, we can simplify the problem with a trick.

2

6212

421

21

21

)6,4|(

eeL

19


log L is a monotonically increasing function of L (meaning that log L increases if L increases and decreases if L decreases).

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

20


It follows that the value of which maximizes log L is the same as the one that maximizes L. As it so happens, it is easier to maximize log L with respect to than it is to maximize L.

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

21


The logarithm of the product of the density functions can be decomposed as the sum of their logarithms.

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

22


Using the product rule a second time, we can decompose each term as shown.

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

23


Now one of the basic rules for manipulating logarithms allows us to rewrite the second term as shown.

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

abab loglog

2

2)4(21

)4(21

log)4(21

log2

X

eXeX

24


log e is equal to 1, another basic logarithm result. (Remember, as always, we are using natural logarithms, that is, logarithms to base e.)

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

abab loglog

2

2)4(21

)4(21

log)4(21

log2

X

eXeX

25


Hence the second term reduces to a simple quadratic in X. And so does the fourth.

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

abab loglog

2

2)4(21

)4(21

log)4(21

log2

X

eXeX

26


We will now choose so as to maximize this expression.

2

6212

421

21

21

)6,4|(

eeL

22

26

212

421

26

212

421

26

212

421

621

421

21

log2

log21

loglog21

log

21

log21

log

21

21

loglog

ee

ee

eeL

27


Quadratic terms of the type in the expression can be expanded as shown.

22 621

421

21

log2log

L

22222

21

21

221

21 aaaaa

28


Thus we obtain the differential of the quadratic term.

22 621

421

21

log2log

L

22222

21

21

221

21 aaaaa

aa

dd 2

21

29


Applying this result, we obtain the differential of log L with respect to . (The first term in the expression for log L disappears completely since it is not a function of .)

22 621

421

21

log2log

L

22222

21

21

221

21 aaaaa

aa

dd 2

21

)6()4(log

dLd

30


Thus from the first order condition we confirm that 5 is the value of that maximizes the log-likelihood function, and hence the likelihood function.

22 621

421

21

log2log

L

22222

21

21

221

21 aaaaa

aa

dd 2

21

)6()4(log

dLd

5ˆ0log

dLd

31


Note that a caret mark has been placed over , because we are now talking about an estimate of , not its true value.

22 621

421

21

log2log

L

22222

21

21

221

21 aaaaa

aa

dd 2

21

)6()4(log

dLd

5ˆ0log

dLd

32


Note also that the second differential of log L with respect to is -2. Since this is negative, we have found a maximum, not a minimum.

22 621

421

21

log2log

L

22222

21

21

221

21 aaaaa

aa

dd 2

21

)6()4(log

dLd

5ˆ0log

dLd

33


We will generalize this result to a sample of n observations X1,...,Xn. The probability density for Xi is given by the first line.

2

21

21

)(

iX

i eXf

34


The joint density function for a sample of n observations is the product of their individual densities.

2

212

21

21

...21 1

nXX

ee

2

21

21

)(

iX

i eXf

35


Now treating the sample values as fixed, we can re-interpret the joint density function as the likelihood function for , given this sample. We will find the value of that maximizes it.

2

21

21

)(

iX

i eXf

2

212

21

1 21

...21

),...,|(1

nXX

n eeXXL

36


We will do this indirectly, as before, by maximizing log L with respect to . The logarithm decomposes as shown.

2

212

21

1 21

...21

),...,|(1

nXX

n eeXXL

221

2

212

21

2

212

21

21

...21

21

log

21

log...21

log

21

...21

loglog

1

1

n

XX

XX

XXn

ee

eeL

n

n

2

21

21

)(

iX

i eXf

37


We differentiate log L with respect to .

)(...)(log

1

nXXdLd

221 2

1...

21

21

loglog

nXXnL

38


The first order condition for a minimum is that the differential be equal to zero.

0ˆ0log

nX

dLd

i

)(...)(log

1

nXXdLd

221 2

1...

21

21

loglog

nXXnL

39


Thus we have demonstrated that the maximum likelihood estimator of is the sample mean. The second differential, -n, is negative, confirming that we have maximized log L.

221 2

1...

21

21

loglog

nXXnL

)(...)(log

1

nXXdLd

0ˆ0log

nX

dLd

i

XXn i

1̂

40


So far we have assumed that , the standard deviation of the distribution of X, is equal to 1. We will now relax this assumption and find the maximum likelihood estimator of it.

2

21

21

)(

iX

i eXf

41


We will illustrate the process graphically with the two-observation example, keeping fixed at 5. We will start with equal to 2.

0.0

0.2

0.4

0.6

0.8

0 1 2 3 4 5 6 7 8 9

L

p

0

0.02

0.04

0.06

0 1 2 3 4

42


With equal to 2, the probability density is 0.1760 for both X = 4 and X = 6, and the joint density is 0.0310.

0.0

0.2

0.4

0.6

0.8

0 1 2 3 4 5 6 7 8 9

L

p

p(4) p(6) L

2.0 0.1760 0.1760 0.0310

0

0.02

0.04

0.06

0 1 2 3 4

43


Now try equal to 1. The individual densities are 0.2420 and so the joint density, 0.0586, has increased.

0.0

0.2

0.4

0.6

0.8

0 1 2 3 4 5 6 7 8 9

L

p

p(4) p(6) L

2.0 0.1760 0.1760 0.0310

1.0 0.2420 0.2420 0.0586

0

0.02

0.04

0.06

0 1 2 3 4

44


Now try putting equal to 0.5. The individual densities have fallen and the joint density is only 0.0117.

0.0

0.2

0.4

0.6

0.8

0 1 2 3 4 5 6 7 8 9

L

p

p(4) p(6) L

2.0 0.1760 0.1760 0.0310

1.0 0.2420 0.2420 0.0586

0.5 0.1080 0.1080 0.0117

0

0.02

0.04

0.06

0 1 2 3 4

45


The joint density has now been plotted as a function of in the lower diagram. You can see that in this example it is greatest for equal to 1.

0

0.02

0.04

0.06

0 1 2 3 4

0.0

0.2

0.4

0.6

0.8

0 1 2 3 4 5 6 7 8 9

p(4) p(6) L

2.0 0.1760 0.1760 0.0310

1.0 0.2420 0.2420 0.0586

0.5 0.1080 0.1080 0.0117L

p

46


We will now look at this mathematically, starting with the probability density function for X given and .

2

21

21

)(

iX

i eXf

47


The joint density function for the sample of n observations is given by the second line.

2

21

21

)(

iX

i eXf

2

212

21

21

...21 1

nXX

ee

48


As before, we can re-interpret this function as the likelihood function for and , given the sample of observations.

2

21

21

)(

iX

i eXf

2

212

21

1 21

...21

),...,|,(1

nXX

n eeXXL

49


We will find the values of and that maximize this function. We will do this indirectly by maximizing log L.

2

21

21

)(

iX

i eXf

2

212

21

21

...21

loglog1

nXX

eeL

2

212

21

1 21

...21

),...,|,(1

nXX

n eeXXL

50


We can decompose the logarithm as shown. To maximize it, we will set the partial derivatives with respect to and equal to zero.

2212

22

1

2

212

21

2

212

21

21

...211

21

log1

log

21

...21

21

log

21

log...21

log

21

...21

loglog

1

1

n

n

XX

XX

XXnn

XXn

ee

eeL

n

n

51


When differentiating with respect to , the first two terms disappear. We have already seen how to differentiate the other terms.

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

nX

XX

XXL

i

n

n

2

12

2212

1

)(...)(1

21

...211log

52


Setting the first differential equal to 0, the maximum likelihood estimate of is the sample mean, as before.

XL

ˆ0

log

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

nX

XX

XXL

i

n

n

2

12

2212

1

)(...)(1

21

...211log

53


Next, we take the partial differential of the log-likelihood function with respect to .

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

54


Before doing so, it is convenient to rewrite the equation.

abab loglog

loglog)1(log1

log 1

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

55


The derivative of log with respect to is 1/. The derivative of --2 is –2--3.

23 )(log

iXnL

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

56


Setting the first derivative of log L to zero gives us a condition that must be satisfied by the maximum likelihood estimator.

0)ˆ(ˆˆ

0log 23

iXnL

23 )(log

iXnL

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

57


We have already demonstrated that the maximum likelihood estimator of is the sample mean.

0)(ˆ 22 XXn i

0)ˆ(ˆˆ

0log 23

iXnL

23 )(log

iXnL

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

58


Hence the maximum likelihood estimator of the population variance is the mean square deviation of X.

22 )(1

ˆ XXn i

0)(ˆ 22 XXn i

0)ˆ(ˆˆ

0log 23

iXnL

23 )(log

iXnL

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

59


Note that it is biased. The unbiased estimator is obtained by dividing by n – 1, not n.

23 )(log

iXnL

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

0)ˆ(ˆˆ

0log 23

iXnL

0)(ˆ 22 XXn i

22 )(1

ˆ XXn i


23 )(log

iXnL

2

2

2212

221

loglog

21

...211

21

log1

loglog

i

n

Xnn

XXnnL

0)ˆ(ˆˆ

0log 23

iXnL

0)(ˆ 22 XXn i

However it can be shown that the maximum likelihood estimator is asymptotically efficient, in the sense of having a smaller mean square error than the unbiased estimator in large samples.

60

22 )(1

ˆ XXn i

Copyright Christopher Dougherty 2011.

These slideshows may be downloaded by anyone, anywhere for personal use.

Subject to respect for copyright and, where appropriate, attribution, they may be

used as a resource for teaching an econometrics course. There is no need to

refer to the author.

The content of this slideshow comes from Section 10.6 of C. Dougherty,

Introduction to Econometrics, fourth edition 2011, Oxford University Press.

Additional (free) resources for both students and instructors may be

downloaded from the OUP Online Resource Centre

http://www.oup.com/uk/orc/bin/9780199567089/.

Individuals studying econometrics on their own and who feel that they might

benefit from participation in a formal course should consider the London School

of Economics summer school course

EC212 Introduction to Econometrics

http://www2.lse.ac.uk/study/summerSchools/summerSchool/Home.aspx

or the University of London International Programmes distance learning course

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www.londoninternational.ac.uk/lse.

11.07.25

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Christopher Dougherty EC220 - Introduction to econometrics (chapter 10) Slideshow: introduction to maximum likelihood estimation Original citation: Dougherty,

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