Question

Suppose y ~ Binomial (n, ?), with n known, and the prior for ? is ?(?) = Beta (?, ?). Here Bin stands for the Binomial distribution and P(y|?) = (n over y) ?^y (1-?)^{n-y}, y ? {0, 1, ..., n}, ?(?) = 1/B(?, ?) ?^{?-1}(1-?)^{?-1}, ? ? (0, 1). (1) What is the posterior density of ?? ?(?|y) = Beta(? - y, ? + n - y) ?(?|y) = Beta(? - y, ? + n + y) ?(?|y) = Beta(? + y, ? + n - y) ?(?|y) = Beta(? + y, ? + n + y) (2) What is the marginal likelihood of y? m(y) = (n over y) B(?-y, ?+n-y)/B(?, ?) m(y) = (n+y over y) B(?-y, ?+n-y)/B(?, ?) m(y) = (n over y) B(?+y, ?+n-y)/B(?, ?) m(y) = (n+y over y) B(?+y, ?+n-y)/B(?, ?) (3) Suppose the above model is labelled M0. There is a possible alternative model M1 which is y ~ Bin (n, 1/2). Find the Bayes Factor B01. B01 = 0.5^{(n-y)}/B(?-y, ?+n-y) B01 = 0.5^{(n-y)}B(?, ?)/B(?+y, ?+n-y) B01 = 0.5^n/B(?-y, ?+n-y) B01 = 0.5^n B(?, ?)/B(?+y, ?+n-y) Use the same data set from Homework 2 "HW2_data.csv", but only consider the y values. It is assumed (yi) i=1 to n iid ~ N(?, ?^2), with ? and ? unknown. Let ? = 1/?^2 and take the prior for (?, ?) as ?(?|?) = N(0, (0.01?)^{-1}) and ?(?) = Ga(0.01, 0.01). (4) What's the posterior distribution of ? given ?? ?_n(?|?) = N(1.58, 1/15.01?) ?_n(?|?) = N(1.58, 1/30.01?) ?_n(?|?) = N(1.67, 1/15.01?) ?_n(?|?) = N(1.67, 1/30.01?) (5) What's the posterior distribution of ?? ?_n(?) = Ga(15.01, 102.99) ?_n(?) = Ga(15.01, 12.43) ?_n(?) = Ga(30.01, 102.99) ?_n(?) = Ga(30.01, 12.43)

          Suppose y ~ Binomial (n, ?), with n known, and the prior for ? is ?(?) = Beta (?, ?). Here Bin stands for the Binomial distribution and P(y|?) = (n over y) ?^y (1-?)^{n-y}, y ? {0, 1, ..., n}, ?(?) = 1/B(?, ?) ?^{?-1}(1-?)^{?-1}, ? ? (0, 1). (1) What is the posterior density of ?? ?(?|y) = Beta(? - y, ? + n - y) ?(?|y) = Beta(? - y, ? + n + y) ?(?|y) = Beta(? + y, ? + n - y) ?(?|y) = Beta(? + y, ? + n + y) (2) What is the marginal likelihood of y? m(y) = (n over y) B(?-y, ?+n-y)/B(?, ?) m(y) = (n+y over y) B(?-y, ?+n-y)/B(?, ?) m(y) = (n over y) B(?+y, ?+n-y)/B(?, ?) m(y) = (n+y over y) B(?+y, ?+n-y)/B(?, ?) (3) Suppose the above model is labelled M0. There is a possible alternative model M1 which is y ~ Bin (n, 1/2). Find the Bayes Factor B01. B01 = 0.5^{(n-y)}/B(?-y, ?+n-y) B01 = 0.5^{(n-y)}B(?, ?)/B(?+y, ?+n-y) B01 = 0.5^n/B(?-y, ?+n-y) B01 = 0.5^n B(?, ?)/B(?+y, ?+n-y) Use the same data set from Homework 2 "HW2_data.csv", but only consider the y values. It is assumed (yi) i=1 to n iid ~ N(?, ?^2), with ? and ? unknown. Let ? = 1/?^2 and take the prior for (?, ?) as ?(?|?) = N(0, (0.01?)^{-1}) and ?(?) = Ga(0.01, 0.01). (4) What's the posterior distribution of ? given ?? ?_n(?|?) = N(1.58, 1/15.01?) ?_n(?|?) = N(1.58, 1/30.01?) ?_n(?|?) = N(1.67, 1/15.01?) ?_n(?|?) = N(1.67, 1/30.01?) (5) What's the posterior distribution of ?? ?_n(?) = Ga(15.01, 102.99) ?_n(?) = Ga(15.01, 12.43) ?_n(?) = Ga(30.01, 102.99) ?_n(?) = Ga(30.01, 12.43)

Suppose y Binomial (n, ?), with n known, and the prior for ? is ?(?) = Beta (?, ?). Here Bin stands for the Binomial distribution and P(y|?) = (n over y) ?^y (1-?)^n-y, y ? 0, 1, ..., n, ?(?) = 1/B(?, ?) ?^?-1(1-?)^?-1, ? ? (0, 1). (1) What is the posterior density of ?? ?(?|y) = Beta(? - y, ? + n - y) ?(?|y) = Beta(? - y, ? + n + y) ?(?|y) = Beta(? + y, ? + n - y) ?(?|y) = Beta(? + y, ? + n + y) (2) What is the marginal likelihood of y? m(y) = (n over y) B(?-y, ?+n-y)/B(?, ?) m(y) = (n+y over y) B(?-y, ?+n-y)/B(?, ?) m(y) = (n over y) B(?+y, ?+n-y)/B(?, ?) m(y) = (n+y over y) B(?+y, ?+n-y)/B(?, ?) (3) Suppose the above model is labelled M0. There is a possible alternative model M1 which is y Bin (n, 1/2). Find the Bayes Factor B01. B01 = 0.5^(n-y)/B(?-y, ?+n-y) B01 = 0.5^(n-y)B(?, ?)/B(?+y, ?+n-y) B01 = 0.5^n/B(?-y, ?+n-y) B01 = 0.5^n B(?, ?)/B(?+y, ?+n-y) Use the same data set from Homework 2 "HW2data.csv", but only consider the y values. It is assumed (yi) i=1 to n iid N(?, ?^2), with ? and ? unknown. Let ? = 1/?^2 and take the prior for (?, ?) as ?(?|?) = N(0, (0.01?)^-1) and ?(?) = Ga(0.01, 0.01). (4) What's the posterior distribution of ? given ?? ?n(?|?) = N(1.58, 1/15.01?) ?n(?|?) = N(1.58, 1/30.01?) ?n(?|?) = N(1.67, 1/15.01?) ?n(?|?) = N(1.67, 1/30.01?) (5) What's the posterior distribution of ?? ?n(?) = Ga(15.01, 102.99) ?n(?) = Ga(15.01, 12.43) ?n(?) = Ga(30.01, 102.99) ?n(?) = Ga(30.01, 12.43)

Added by Johnathan B.

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