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Lecture 2

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MGSC 372 Lecture 2: Fall 2016 Semester Notes 2
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McGill University

Management Science

MGSC 372

Brian Smith

Fall

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Solution
The MLE of π is:
230 + 275 + 317 + 285 + 303 1410
= = 282
5 5
A discrete example MLE for binomial distribution
π π₯ πβπ₯
Recall that π π = π₯π= ( π₯π π(1 β π) π
Likelihood function for three values of 1:{π₯2,π₯3,π₯ }
π π π
πΏ π = ( )π π₯11 β π )πβπ₯1Γ( )ππ₯2(1 β π )πβπ₯2Γ( )π π₯3(1 β π )πβπ₯3)
π₯ 1 π₯2 π₯3
π π π
πΏ π = ( )( )( )π (π₯1+π₯2+π₯3Γ 1 β π )(πβπ₯1+ πβπ₯ 2 πβπ₯ 3)
π₯1 π₯2 π₯3
π π π
πΏ π = ( )( )( )π π₯1+π₯2+π₯3Γ 1 β π )(3πβ π₯1+π₯2+π₯3])
π₯1 π₯2 π₯3
Taking Logarithms
π π π
lnπΏ π = ln( )( )( ) + lnππ₯1+π₯2+π₯ 3+ ln 1 β π)(3πβ π₯1+π₯2+π₯3])
π₯1 π₯2 π₯3
π π π
lnπΏ π = ln( )( )( )+ π₯ +1π₯ + 2 3)lnπ + 3π β π₯ + 1 + π₯2 3])ln 1 β π)
π₯1 π₯2 π₯3
Differentiating,
π π₯ + π₯ + π₯ 3π β [π₯ + π₯ + π₯ ]
lnπΏ π = 0 + 1 2 3 + 1 2 3
ππ π 1 β π
π π₯ 1 π₯ 2 π₯ 3 3π β [π₯1+ π₯ 2 π₯ ]3
β΄ ππ lnπΏ π = 0 β π = 1 β π
Solving for the MLE of π
So that:
1 β π π₯ +1π₯ + 2 3) = π 3π β π₯ +1π₯ + 2 3])
(π₯ + π₯ + π₯ ) β π π₯ + π₯ + π₯ )= 3ππ β π π₯ + π₯ + π₯ )
1 2 3 1 2 3 1 2 3
π₯ 1 π₯ +2π₯ = 3ππ π₯1+ π₯2+ π₯ 3 1 π₯1+ π₯2+ π₯ 3 1
β΄ π = = Γ = Γπ₯Μ
3π π 3 π
The General Result
Generalizing to the case of π observations we get:
π₯ + π₯ + π₯ + β―+ π₯ β π π₯ 1
π = 1 2 3 π = π=1 π= Γπ₯Μ
ππ ππ π
Binomial Example #1
A coin is tossed 10 times and the number of heads is recorded. This process is repeated 8 times. The
number of heads obtained in each of the 8 repetitions of the experiment is as follows:
π₯1= 3,π₯ 2 5,π₯ 3 8,π₯ =46,π₯ = 5,π₯ = 76π₯ = 57π₯ = 4 8
Does this appear to be a fair coin?
Solution
βπ₯ 3 + 5 + 8 + 6 + 4 + 7 + 5 + 4 42
ππΏπΈ = π= = = 0.525
ππ 10 β 8 80
The probability of heads is close to 50% so the coin appears to be fair.
Binomial Example #2
An assembly line produces a fixed, but unknown, percentage of defective items. Five random samples of
100 items are selected and the number (%) of defective items items in each sample is recorded as
follows:
Sample # 1 2 3 4 5
% Defectives 4 7 5 3 4
Solution
The MLE for the percentage of defective items produced by the assembly line is given by:
βπ₯ π 4 + 7 + 5 + 3 + 4 23
ππΏπΈ = ππ = 100 β 5 = 500 = 0.046
We conclude that πΜ = 0.046 so approximately 4.6% of items produced.
Some comments on the method of maximum likelihood estimation Note 1: For estimating a population mean for normal distributions, the least squares estimate (LSE) and
the maximum likelihood estimate (MLE) are identical. Both methods result in the sample mean as the
estimator. In general, if the error terms in a model are normally and independently distributed with
constant variance, then the MLE and LSE are identical.
Note 2: Least squares estimation is purely a descriptive procedure β it is independent of the probability
distribution of the variable. Maximum likelihood estimation explicitly requires knowledge of the
probability distribution.
Note 3: MLE is a preferred method of parameter estimation in statistics and is an indispensable tool for
many statistical modeling techniques, in particular in non-linear modeling with non-normal data. An
example is logistic regression.
Note 4: MLE has many desirable properties in estimation: sufficiency (complete information about the
parameter of interest contained in its MLE estimator); consistency (estimate improves with increasing
sample size); efficiency (lowest possible variance of parameter estimates). In contrast, LSE is not a
sufficient estimator since it ignores the probability distribution. Note, however, that LSE estimators are
unbiased whereas MLE estimators are biased.
Note 5: In nonlinear, heteroskedastic (unequal variance) models, such as ARCH and GARCH models for
volatility, LSE is an inappropriate method of estimation and MLE is required.
The Lognormal Distribution and Application
The Lognormal Distribution
A continuous random variable π₯ follows a lognormal distribution if its natural logarithm, lnβ‘(π₯), follows a
normal distribution.
In other words, if lnβ‘(π₯) is normal, then π₯ is lognormal.
Properties of the lognormal distribution:
β’ Skewed to the right
β’ Strictly positive (i.e. bounded below by 0)
The lognormal distribution may be used to model data on asset prices (note that prices are bound below
by 0).
Histograms of lognormal variable π and normaldistribution π₯π§ π ( ) The Lognormal Distribution
The lognormal distribution is described by two parameters, its mean and variance, as in the case of a
normal distribution.
The mean of a lognormal distribution is given by:
1
πΈ π₯ = π (π+ 2 )
The variance of a lognormal distribution is given by:
2 2
πππ π₯ = π (2π+π )(ππ β 1)
2
Where π and π are the mean and variance of the normal distribution of the lnβ‘(π₯) variable and π β
2.718 is the natural base for logarithms.
π
The median of a lognormal distribution π₯ is given by π
Recall that the exponential and logarithmic functions are inverse functions so that
ln π₯ = π¦ β π₯ = π π¦
This relationship is used to switch between the lognormal variable π₯ and the normal variable π¦ = lnβ‘(π₯).
Lognormal Distribution Example
The material failure mechanism for a specific metal alloy has determined that the tensile strength π₯ of
2
the material, measured as the force per unit of cross section area π/π , has a lognormal distribution
with parameter

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