Then RBLS? px=2op[责（∑2训-2NzA+N4 2wse(∑州-w} 1 h(x) g(∑x2(n),A) Sufficient statistics: whxiong@uestc.edu.cn
whxiong@uestc.edu.cn Then RBLS? = 1 (2¼A) N=2 exp ½ ¡ 1 2 µ 1 A Xx 2 [n] + NA ¶¾ | {z } g( Px2(n);A) ¢exp(Nx¹) | {z } h(x) Sufficient statistics: Xx 2 [n]
Then RBLS? x4利={a(∑时+NA-2wP)} w{(∑时-A} 1 h(x) g(∑x2(m),A) Sufficient statistics:>Completeness?Needs to check Assume its complete whxiong@uestc.edu.cn 12
whxiong@uestc.edu.cn Then RBLS? 12 p(x; A) = 1 (2¼A) N=2 exp ½ ¡ 1 2A ³Xx 2 [n] + NA ¡ 2Nx¹A 2 ´¾ = 1 (2¼A) N=2 exp ½ ¡ 1 2 µ 1 A Xx 2 [n] + NA ¶¾ | {z } g( Px2(n);A) ¢exp(Nx¹) | {z } h(x) Sufficient statistics: Xx 2 [n] Completeness ? Needs to check Assume its complete
Then RBLS? x4利={a(∑时+NA-2wP)} 防即{（∑+4小 1 h(x) g(∑x2(m),A) Sufficient statistics:]Completeness?Needs to check Assume its complete 1.Find g(x)such that Elg(x2n)]=A whxiong@uestc.edu.cn 13
whxiong@uestc.edu.cn Then RBLS? 13 p(x; A) = 1 (2¼A) N=2 exp ½ ¡ 1 2A ³Xx 2 [n] + NA ¡ 2Nx¹A 2 ´¾ = 1 (2¼A) N=2 exp ½ ¡ 1 2 µ 1 A Xx 2 [n] + NA ¶¾ | {z } g( Px2(n);A) ¢exp(Nx¹) | {z } h(x) Sufficient statistics: Xx 2 [n] Completeness ? Needs to check Assume its complete 1. Find g(x) such that E[g( Px 2 [n])] = A
Then RBLS? x4=27w即{a（∑r例+NA-2Na4的} 脚{（Σ+4} 1 h(x) g(∑x2(m),A) Sufficient statistics:]Completeness?Needs to check Assume its complete 1.Findg(x)such that E[g(∑x2[nl)】]=A 2.Or E(A2[n])for any unbiased estimator whxiong@uestc.edu.cn 14
whxiong@uestc.edu.cn Then RBLS? 14 p(x; A) = 1 (2¼A) N=2 exp ½ ¡ 1 2A ³Xx 2 [n] + NA ¡ 2Nx¹A 2 ´¾ = 1 (2¼A) N=2 exp ½ ¡ 1 2 µ 1 A Xx 2 [n] + NA ¶¾ | {z } g( Px2(n);A) ¢exp(Nx¹) | {z } h(x) Sufficient statistics: Xx 2 [n] Completeness ? Needs to check Assume its complete 1. Find g(x) such that E[g( Px 2 [n])] = A 2. Or E(A · j Xx 2 [n])for any unbiased estimator
How about these estimator a=∑2m whxiong@uestc.edu.cn 15
whxiong@uestc.edu.cn How about these estimator 15 A b 1 = 1 N Xx 2 [n]