SOLUTION FOR HOMEWORK 3, STAT 4352 Welcome to your third ...

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Then 6. Problem 10.66. Let X1, . . .,Xn be iid according to the pdf gθ(x) = θ −1 e −(x−δ)/θ I(x > δ). LX n(δ, θ) = θ−n e − n l=1 (Xl−δ)/θ I(X(1) > δ). Recall that X(1) = min(X1, . . .,Xn) is the minimal observation [the first ordered observation]. This is the case that I wrote you about earlier: it is absolutely crucial to take into account the indicator function (the support) because here the parameter δ defines the support. By its definition, Note that ( ˆ δMLE, ˆ θMLE) := arg max ln(LXn(δ, θ)). δ∈(−∞,∞),θ∈(0,∞) L(δ, θ) := ln(LXn(δ, θ)) = −n ln(θ) − θ−1 n (Xl − δ) + ln I(X(1) ≥ δ). Now the crucial step: you should graph the loglikelihood L as a function in δ and visualize that it takes on maximum when δ = X(1). So we get ˆ δMLE = X(1). Then by taking a derivative we get that ˆ θMLE = n −1 n l=1 (Xl − X(1)). Answer: ( ˆ δMLE ˆ θMLE) = (X(1), n −1 n l=1 (Xl − X(1)). Please note that ˆ δMLE is a biased estimator; this is a rather typical outcome. 7. Problem 10.73. Consider iid uniform observations X1, . . .,Xn with the parametric pdf l=1 fθ(x) = I(θ − 1/2 < x < θ + 1/2). As soon as the parameter is in the indicator function you should be very cautious: typically a graphic will help you to find a MLE estimator, and not a differentiation. Also, it is very helpful to figure out the nature of the parameter. Here it is obviously a location parameter, and you can write X = θ + Z, Z ∼ Uniform(−1/2, 1/2). The latter helps you to guess about a correct estimator and check a suggested one and, if necessary, simplify calculations of descriptive characteristics (mean, variance, etc.) Well, now we need to write down the likelihood function (recall that this is just a joint density only considered as a function in the parameter given a vector of observations): LXn(θ) = n I(θ − 1/2 < Xl < θ + 1/2) = I(θ − 1/2 < X(1) ≤ X(n) < θ + 1/2). l=1 Note that the latter expression implies that (X(1), X(n)) is a sufficient statistic (due to the Factorization Theorem). As a result, any good estimator, and the MLE in particular, must be a function of only these two statistics. Another remark is: it is possible to show (there exists a technique how to do this which is beyond this class objectives) that this pair of 4
extreme observations is also the minimal sufficient statistic. Please look at the situation: we have 1 parameter and need 2 univariate statistics (X(1), X(n)) to have the sufficient statistics; here this is the limit of data-reduction. Nonetheless, this is a huge data-reduction whenever n is large. Just think about this: to estimate θ you do not need any observation which is between the two extreme ones! This is not a trivial assertion. Well, now let us return to the problem at hand. If you look at the graphic of the likelihood function as a function in θ, then you may conclude that it attains its maximum on all θ such that X(n) − 1/2 < θ < X(1) + 1/2. (1) As a result, we get a very curious MLE: any point within this interval can be declared as the MLE (the MLE is not unique!). Now we can consider the particular questions at hand. (a). Let ˆ Θ1 = (1/2)(X(1) + X(n)). We need to check that this estimator satisfies (1). We just plug-in this estimator in (1) and get X(n) − 1/2 < (1/2)(X(1) + X(n)) < X(1) + 1/2. The latter relation is true because it is equivalent to the following valid inequality: X(n) − X(1) < 1. (b) Let ˆ Θ2 = (1/3)(X(1) + 2X(n)) be another candidate for the MLE. Then it should satisfy (1). In particular, if this is the MLE then (1/3)(X(1) + 2X(n)) < X(1) + 1/2 should hold. The latter inequality is equivalent to X(n) − X(1) < 3/4 which obviously may not hold. The contradictory shows that this estimator, despite being a function of the sufficient statistic, is not the MLE. 8. Problem 10.74. Here we are exploring the Bayesian approach where the parameter of interest is considered as a realization of a random variable, (it can be considered as a random variable). For the problem at hand X ∼ Binom(n, θ) and θ is a realization (which we do not directly observe) of a beta RV Θ ∼ Beta(α, β). [Please note that here your knowledge of basic/classical distributions becomes absolutely crucial: you cannot solve any problem without knowing formulae for pmf/pdf; so it is time to refresh them.] In other words, here we are observing a binomial random variable whose parameter (probability of success has a beta prior. To find a Bayesian estimator, we need to find a posterior distribution of the parameter of interest and then calculate its mean. [Please note that your knowledge of means of classical distribution becomes very handy here: as soon as you realize the underlying posterior distribution, you can use a formula for calculating its mean.] 5
Page 1 and 2: SOLUTION FOR HOMEWORK 3, STAT 4352
Page 3: Remark: We need to check that n −
Page 7 and 8: which in its turn yields Using this
Page 9: 12. Problem 10.96. Let X be a grade

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