# HW2solcor - Chapter 3 11 The densities are p Î x = exp Î·x...

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Unformatted text preview: Chapter 3. 11. The densities are p Î· ( x ) = exp( Î·x- A ( Î· )) sin( x )1 (0 ,Ï€ ) ( x ) , where A ( Î· ) = log Z R exp( Î·x ) sin( x )1 (0 ,Ï€ ) ( x ) dx. By integration by parts, Z Ï€ exp( Î·x ) sin( x ) dx = 1 Î· exp( Î·x ) sin( x ) Ï€- 1 Î· 2 exp( Î·x ) cos( x ) Ï€- 1 Î· 2 Z Ï€ exp( Î·x ) sin( x ) dx = 1 + exp( Î·Ï€ ) Î· 2- 1 Î· 2 Z Ï€ exp( Î·x ) sin( x ) dx, so that we obtain Z Ï€ exp( Î·x ) sin( x ) dx = 1 + exp( Î·Ï€ ) 1 + Î· 2 > for any Î· âˆˆ R . Therefore, the natural parameter space Î˜ is R , and the densities are p Î· ( x ) = 1 + Î· 2 1 + exp( Î·Ï€ ) exp( Î·x ) sin( x )1 (0 ,Ï€ ) ( x ) . 15. From results about a radius of convergence, we have A ( Î· ) = ln âˆž X x =1 e- Î·x x 2 < âˆž for Î· > . Thus, Î˜ = R + . For Î¸ âˆˆ Î˜, by Theorem 3.1, âˆž X x =1 e- Î·x x 2 = âˆž X x =1 âˆ‚ 2 âˆ‚Î· 2 e- Î·x = âˆ‚ 2 âˆ‚Î· 2 âˆž X x =1 e- Î·x = âˆ‚ 2 âˆ‚Î· 2 e- Î· 1- e- Î· = âˆ‚ 2 âˆ‚Î· 2 1 e Î·- 1 = âˆ‚ âˆ‚Î·- e Î· ( e Î·- 1) 2 = e Î· ( e Î· + 1) ( e Î·- 1) 3 . 1 Therefore, we obtain A ( Î· ) = ln e Î· ( e Î· + 1) ( e Î·- 1) 3 = Î· + ln( e Î· + 1)- 3 ln( e Î·- 1) and p Î· ( x ) = ( e Î·- 1) 3 e Î· ( e Î· + 1) e- Î·x x 2 . From the argument about the cumulant generating function, we have E [ X ] = E [- T ( X )] =- âˆ‚A ( Î· ) âˆ‚Î· =- 1- e Î· e Î· + 1 + 3 e Î· e Î·- 1 = e 2 Î· + 4 e Î· + 1 e 2 Î·- 1 Var( X ) = âˆ‚ 2 A ( Î· ) âˆ‚Î· 2 = âˆ‚ âˆ‚Î· 1 + 1 1 + e- Î·- 3 1- e- Î· = e- Î· (1 + e- Î· ) 2 + 3 e- Î· (1- e- Î· ) 2 = 4 e- Î· ( e- 2 Î· + e- Î· + 1) ( e- 2 Î·- 1) 2 19. a) Suppose that | h | < K for some K > 0. Also, since h is differentiable, for a > 0, there exists a Î´ > 0 such that | h ( x ) | x = | h ( x )- h (0) | x < | h (0) | + for any 0 < x < Î´. 2 From the above inequalities, we obtain that Z âˆž | h (1 /x 2 ) | dx = Z âˆž 1 / âˆš Î´ + Z 1 / âˆš Î´ ! | h (1 /x 2 ) | dx < Z âˆž 1 / âˆš Î´ ( | h (0) | + ) 1 x 2 dx + Z 1 / âˆš Î´ Kdx = âˆš Î´ ( | h (0) | + ) + K 1 âˆš Î´ < âˆž . b) By DCT which is justified by a), lim n â†’âˆž n E h (1 / ( n 2 Z 2 ) = lim n â†’âˆž Z âˆž-âˆž n âˆš 2 Ï€n 2 h (1 /x 2 ) exp- x 2 2 n 2 dx = r 2 Ï€ lim n â†’âˆž Z âˆž h (1 /x 2 ) exp- x 2 2 n 2 dx = r 2 Ï€ Z âˆž h (1 /x 2 ) lim n â†’âˆž exp- x 2 2 n 2 dx = r 2 Ï€ Z âˆž h (1 /x 2 ) dx. 20. a) f ( x ) = lim n â†’âˆž f n ( x ) = 0. b) Let g = âˆ‘ âˆž n =1 | f n | , which satisfies | f n | â‰¤ g,n = 1 , 2 ,... . Suppose âˆƒ g * ( m ) < g ( m ) = c m for some point m âˆˆ N . Then, g * doesnâ€™t satisfy | f n | â‰¤ g * ,n = 1 , 2 ,... , in particular g * ( m ) < | f m ( m ) | = c m . Thus, g is the smallest function bounding...
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## This note was uploaded on 04/14/2010 for the course STATS 610 taught by Professor Moulib during the Fall '09 term at University of Michigan.

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

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