HW2solcor - Chapter 3 11 The densities are p x = exp ηx...

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Unformatted text preview: Chapter 3. 11. The densities are p η ( x ) = exp( ηx- A ( η )) sin( x )1 (0 ,π ) ( x ) , where A ( η ) = log Z R exp( ηx ) sin( x )1 (0 ,π ) ( x ) dx. By integration by parts, Z π exp( ηx ) sin( x ) dx = 1 η exp( ηx ) sin( x ) π- 1 η 2 exp( ηx ) cos( x ) π- 1 η 2 Z π exp( ηx ) sin( x ) dx = 1 + exp( ηπ ) η 2- 1 η 2 Z π exp( ηx ) sin( x ) dx, so that we obtain Z π exp( ηx ) sin( x ) dx = 1 + exp( ηπ ) 1 + η 2 > for any η ∈ R . Therefore, the natural parameter space Θ is R , and the densities are p η ( x ) = 1 + η 2 1 + exp( ηπ ) exp( ηx ) sin( x )1 (0 ,π ) ( x ) . 15. From results about a radius of convergence, we have A ( η ) = ln ∞ X x =1 e- ηx x 2 < ∞ for η > . Thus, Θ = R + . For θ ∈ Θ, by Theorem 3.1, ∞ X x =1 e- ηx x 2 = ∞ X x =1 ∂ 2 ∂η 2 e- ηx = ∂ 2 ∂η 2 ∞ X x =1 e- ηx = ∂ 2 ∂η 2 e- η 1- e- η = ∂ 2 ∂η 2 1 e η- 1 = ∂ ∂η- e η ( e η- 1) 2 = e η ( e η + 1) ( e η- 1) 3 . 1 Therefore, we obtain A ( η ) = ln e η ( e η + 1) ( e η- 1) 3 = η + ln( e η + 1)- 3 ln( e η- 1) and p η ( x ) = ( e η- 1) 3 e η ( e η + 1) e- ηx x 2 . From the argument about the cumulant generating function, we have E [ X ] = E [- T ( X )] =- ∂A ( η ) ∂η =- 1- e η e η + 1 + 3 e η e η- 1 = e 2 η + 4 e η + 1 e 2 η- 1 Var( X ) = ∂ 2 A ( η ) ∂η 2 = ∂ ∂η 1 + 1 1 + e- η- 3 1- e- η = e- η (1 + e- η ) 2 + 3 e- η (1- e- η ) 2 = 4 e- η ( e- 2 η + e- η + 1) ( e- 2 η- 1) 2 19. a) Suppose that | h | < K for some K > 0. Also, since h is differentiable, for a > 0, there exists a δ > 0 such that | h ( x ) | x = | h ( x )- h (0) | x < | h (0) | + for any 0 < x < δ. 2 From the above inequalities, we obtain that Z ∞ | h (1 /x 2 ) | dx = Z ∞ 1 / √ δ + Z 1 / √ δ ! | h (1 /x 2 ) | dx < Z ∞ 1 / √ δ ( | h (0) | + ) 1 x 2 dx + Z 1 / √ δ Kdx = √ δ ( | h (0) | + ) + K 1 √ δ < ∞ . b) By DCT which is justified by a), lim n →∞ n E h (1 / ( n 2 Z 2 ) = lim n →∞ Z ∞-∞ n √ 2 πn 2 h (1 /x 2 ) exp- x 2 2 n 2 dx = r 2 π lim n →∞ Z ∞ h (1 /x 2 ) exp- x 2 2 n 2 dx = r 2 π Z ∞ h (1 /x 2 ) lim n →∞ exp- x 2 2 n 2 dx = r 2 π Z ∞ h (1 /x 2 ) dx. 20. a) f ( x ) = lim n →∞ f n ( x ) = 0. b) Let g = ∑ ∞ n =1 | f n | , which satisfies | f n | ≤ g,n = 1 , 2 ,... . Suppose ∃ g * ( m ) < g ( m ) = c m for some point m ∈ N . Then, g * doesn’t satisfy | f n | ≤ g * ,n = 1 , 2 ,... , in particular g * ( m ) < | f m ( m ) | = c m . Thus, g is the smallest function bounding...
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HW2solcor - Chapter 3 11 The densities are p x = exp ηx...

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