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Lecture 2

MGSC 372 Lecture 2: Fall 2016 Semester Notes 2

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McGill University
Management Science
MGSC 372
Brian Smith

Solution The MLE of πœ‡ is: 230 + 275 + 317 + 285 + 303 1410 = = 282 5 5 A discrete example MLE for binomial distribution 𝑛 π‘₯ π‘›βˆ’π‘₯ Recall that 𝑃 𝑋 = π‘₯𝑖= ( π‘₯𝑖 𝑖(1 βˆ’ 𝑝) 𝑖 Likelihood function for three values of 1:{π‘₯2,π‘₯3,π‘₯ } 𝑛 𝑛 𝑛 𝐿 𝑝 = ( )𝑝 π‘₯11 βˆ’ 𝑝 )π‘›βˆ’π‘₯1Γ—( )𝑝π‘₯2(1 βˆ’ 𝑝 )π‘›βˆ’π‘₯2Γ—( )𝑝 π‘₯3(1 βˆ’ 𝑝 )π‘›βˆ’π‘₯3) π‘₯ 1 π‘₯2 π‘₯3 𝑛 𝑛 𝑛 𝐿 𝑝 = ( )( )( )𝑝 (π‘₯1+π‘₯2+π‘₯3Γ— 1 βˆ’ 𝑝 )(π‘›βˆ’π‘₯1+ π‘›βˆ’π‘₯ 2 π‘›βˆ’π‘₯ 3) π‘₯1 π‘₯2 π‘₯3 𝑛 𝑛 𝑛 𝐿 𝑝 = ( )( )( )𝑝 π‘₯1+π‘₯2+π‘₯3Γ— 1 βˆ’ 𝑝 )(3π‘›βˆ’ π‘₯1+π‘₯2+π‘₯3]) π‘₯1 π‘₯2 π‘₯3 Taking Logarithms 𝑛 𝑛 𝑛 ln𝐿 𝑝 = ln( )( )( ) + ln𝑝π‘₯1+π‘₯2+π‘₯ 3+ ln 1 βˆ’ 𝑝)(3π‘›βˆ’ π‘₯1+π‘₯2+π‘₯3]) π‘₯1 π‘₯2 π‘₯3 𝑛 𝑛 𝑛 ln𝐿 𝑝 = ln( )( )( )+ π‘₯ +1π‘₯ + 2 3)ln𝑝 + 3𝑛 βˆ’ π‘₯ + 1 + π‘₯2 3])ln 1 βˆ’ 𝑝) π‘₯1 π‘₯2 π‘₯3 Differentiating, 𝑑 π‘₯ + π‘₯ + π‘₯ 3𝑛 βˆ’ [π‘₯ + π‘₯ + π‘₯ ] ln𝐿 𝑝 = 0 + 1 2 3 + 1 2 3 𝑑𝑝 𝑝 1 βˆ’ 𝑝 𝑑 π‘₯ 1 π‘₯ 2 π‘₯ 3 3𝑛 βˆ’ [π‘₯1+ π‘₯ 2 π‘₯ ]3 ∴ 𝑑𝑝 ln𝐿 𝑝 = 0 β†’ 𝑝 = 1 βˆ’ 𝑝 Solving for the MLE of 𝒑 So that: 1 βˆ’ 𝑝 π‘₯ +1π‘₯ + 2 3) = 𝑝 3𝑛 βˆ’ π‘₯ +1π‘₯ + 2 3]) (π‘₯ + π‘₯ + π‘₯ ) βˆ’ 𝑝 π‘₯ + π‘₯ + π‘₯ )= 3𝑝𝑛 βˆ’ 𝑝 π‘₯ + π‘₯ + π‘₯ ) 1 2 3 1 2 3 1 2 3 π‘₯ 1 π‘₯ +2π‘₯ = 3𝑝𝑛 π‘₯1+ π‘₯2+ π‘₯ 3 1 π‘₯1+ π‘₯2+ π‘₯ 3 1 ∴ 𝑝 = = Γ— = Γ—π‘₯Μ… 3𝑛 𝑛 3 𝑛 The General Result Generalizing to the case of π‘˜ observations we get: π‘₯ + π‘₯ + π‘₯ + β‹―+ π‘₯ βˆ‘ π‘˜ π‘₯ 1 𝑝 = 1 2 3 π‘˜ = 𝑖=1 𝑖= Γ—π‘₯Μ… π‘˜π‘› π‘˜π‘› 𝑛 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 βˆ‘π‘₯ 3 + 5 + 8 + 6 + 4 + 7 + 5 + 4 42 𝑀𝐿𝐸 = 𝑖= = = 0.525 π‘›π‘˜ 10 βˆ— 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: βˆ‘π‘₯ 𝑖 4 + 7 + 5 + 3 + 4 23 𝑀𝐿𝐸 = π‘›π‘˜ = 100 βˆ— 5 = 500 = 0.046 We conclude that 𝑝̂ = 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 – 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, ln⁑(π‘₯), follows a normal distribution. In other words, if ln⁑(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 𝒙 and normaldistribution π₯𝐧 𝒙 ( ) 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 𝐸 π‘₯ = 𝑒 (πœ‡+ 2 ) The variance of a lognormal distribution is given by: 2 2 π‘‰π‘Žπ‘Ÿ π‘₯ = 𝑒 (2πœ‡+𝜎 )(π‘’πœŽ βˆ’ 1) 2 Where πœ‡ and 𝜎 are the mean and variance of the normal distribution of the ln⁑(π‘₯) variable and 𝑒 β‰… 2.718 is the natural base for logarithms. πœ‡ The median of a lognormal distribution π‘₯ is given by 𝑒 Recall that the exponential and logarithmic functions are inverse functions so that ln π‘₯ = 𝑦 β†’ π‘₯ = 𝑒 𝑦 This relationship is used to switch between the lognormal variable π‘₯ and the normal variable 𝑦 = ln⁑(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 𝑁/π‘š , has a lognormal distribution with parameter
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