Method of Moments

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Introduction to the Science of Statistics The Method of Moments To investigate the method of moments on simulated data using R, we consider 1000 repetitions of 100 independent observations of a (0.23, 5.35) random variable. > xbar x2bar for (i in 1:1000){x mean(betahat) [1] 6.315644 > sd(betahat) [1] 2.203887 To obtain a sense of the distribution of the estimators ˆ↵ and ˆ, we give histograms. > hist(alphahat,probability=TRUE) > hist(betahat,probability=TRUE) Histogram of alphahat Histogram of betahat Density 0 1 2 3 4 5 6 Density 0.00 0.05 0.10 0.15 0.1 0.2 0.3 0.4 0.5 alphahat 0 5 10 15 betahat As we see, the variance in the estimate of is quite large. We will revisit this example using maximum likelihood estimation in the hopes of reducing this variance. The use of the delta method is more difficult in this case because it must take into account the correlation between ¯X and X 2 for independent gamma random variables. Indeed, from the simulation, we have an estimate. > cor(xbar,x2bar) [1] 0.8120864 Moreover, the two estimators ˆ↵ and ˆ are fairly strongly positively correlated. Again, we can estimate this from the simulation. > cor(alphahat,betahat) [1] 0.7606326 In particular, an estimate of ˆ↵ and ˆ are likely to be overestimates or underestimates in tandem. 202
Introduction to the Science of Statistics 13.4 Answers to Selected Exercises 13.2. Let T be the random variable that is the angle between the positron trajectory and the µ + -spin µ 2 = E ↵ T 2 = 1 Z ⇡ t 2 (1 + ↵ cos t)dt = ⇡2 2⇡ ⇡ 3 ⇡ 2 /3)/2. This leads to the method of mo- ˆ↵ = 1 ✓t 2 2 ⇡ 2 ◆ 3 Thus, ↵ =(µ 2 ments estimate 2↵ where t 2 is the sample mean of the square of the observations. y -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 The Method of Moments Figure 13.2: Densities f(t|↵) for the values of ↵ = 1 (yellow). 1/3 (red), 0 (black), 1/3 (blue), 1 (light blue). -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 13.4. Let X be the random variable for the number of tagged fish. Then, X is a hypergeometric random variable with mean µ X = kt N and variance 2 X = k t N t N k N N N 1 N = g(µ X )= kt . Thus, g 0 (µ X )= kt µ X µ 2 . X The variance of ˆN ✓ ◆ 2 Var( ˆN) kt ⇡ g 0 (µ) 2 X 2 = µ 2 k t ✓ ◆ 2 N t N k kt X N N N 1 = t kt/µ X t kt/µ X k µ 2 k X kt/µ X kt/µ X kt/µ X 1 ✓ ◆ 2 kt = k µ ✓ ◆ 2 Xt kt µ X t kt kµ X kt = k µ X k µ X k(t µ X ) kt kt kt µ X k k kt µ X µ 2 X = k2 t 2 µ 3 X (k µ X )(t µ X ) kt µ X Now if we replace µ X by its estimate r we obtain 2ˆN ⇡ k2 t 2 µ 2 X (k r)(t r) r 3 kt r For t = 200,k = 400 and r = 40, we have the estimate ˆN = 268.4. This compares to the estimate of 276.6 from simulation. For t = 1709,k = 6375 and r = 138, we have the estimate ˆN = 6373.4. . x 203
Page 1 and 2: Topic 13 Method of Moments 13.1 Int
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