midIpracticalexamsol.pdf - ISyE 6412A Theoretical Statistics Fall 2019 Practice Midterm Exam I WARNING this practice midterm exam is intended to help

# midIpracticalexamsol.pdf - ISyE 6412A Theoretical...

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ISyE 6412A Theoretical Statistics Fall 2019 Practice Midterm Exam I WARNING: this practice midterm exam is intended to help you better understand the ma- terials we discussed in class. The actual format of the exam might be different!!! Suppose that X is a Binomial random variable with P θ ( X = k ) = ( n k ) θ k (1 - θ ) n - k for k = 0 , 1 , 2 . . . , n, where n is a known positive integer. It is known only that 0 θ 1 , and it is desired to guess the value of θ on the basis of X. If the guessed value is d and the true value is θ, the loss is ( θ - d ) 2 . (a) Specify S, Ω , D, and L (i.e., the sample space, the set of all possible distribution functions, the decision space, and the loss function). Answer: S = { x : x = 0 , 1 , . . . , n } , Ω = { Binomail( n, θ ) : 0 θ 1 } or simply Ω = { θ : 0 θ 1 } , D = { d : 0 d 1 } , L ( θ, d ) = ( θ - d ) 2 . (b) Compute the risk function of the 3 procedures δ i defined by δ 1 ( X ) = X n ; δ 2 ( X ) = X + n/ 2 n + n ; δ 3 ( X ) = X + 1 n + 2 . Answer: First we need to find out E θ ( X ) = and V ar θ ( X ) = (1 - θ ) . You can either simply state this as a well-known probability result (and we expect that you know all basic important probability distribution functions), or you can derive it on your own by using the fact that X is the sum of iid Bernoulli random variable. If you somehow forgot these probability results, you can also calculate them directly by definition as follows. E θ ( X ) = n X k =0 kP ( X = k ) = n X k =0 k n k ! θ k (1 - θ ) n - k = and E θ ( X 2 ) = n X k =0 k 2 P ( X = k ) = n X k =0 k 2 n k ! θ k (1 - θ ) n - k = + n ( n - 1) θ 2 , and thus V ar ( X ) = E θ ( X 2 ) - [ E θ ( X )] 2 = (1 - θ ) . If you need more technical details about E ( X ) and E ( X 2 ) , Consider the binomial formula ( a + b ) n = n k =0 ( n k ) a k b n - k , differentiating both sides with respect to a gives us n ( a + b ) n - 1 = n k =1 k ( n k ) a k - 1 b n - k . Continue to differentiate both sides with respect to a, and we have n ( n - 1)( a + b ) n - 2 = n k =2 k ( k - 1) ( n k ) a k - 2 b n - k . From these two new equations, show that n X k =0 k n k ! a k b n - k = na ( a + b ) n - 1 and n X k =0 k 2 n k ! a k b n - k = na ( a + b ) n - 1 + n ( n - 1) a 2 ( a + b ) n - 2 for any real numbers a and b. Then E ( X ) and E ( X 2 ) can be easily calculated from these two equations by letting a = θ and b = 1 - θ. Now let us compute the risk function of a general procedure of the form δ ( X ) = a + bX : R δ ( θ ) = E θ L ( θ, δ ( X )) = E θ ( a + bX - θ ) 2 = ( a + b E θ X - θ ) 2 + V ar ( a + bX ) = ( a + ( nb - 1) θ ) 2 + nb 2 θ (1 - θ ) , 1

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where we use the fact that E ( Y - θ ) 2 = ( E Y - θ ) 2 + V ar ( Y ) . In particular, R δ 1 ( θ ) = θ (1 - θ ) /n (as a = 0 , b = 1 /n ) , R δ 2 ( θ ) = 1 4( n + 1) 2 , R δ 3 ( θ ) = [1 + ( n - 4) θ (1 - θ )] / ( n + 2) 2 .
• Spring '14
• AlexanderShapiro
• Probability theory, θ2

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