Final Exam Formula Sheet

Final Exam Formula - Formula sheet ACTSC 432/832 1 The Normal Equations Unbiaisedness equation E[Xn 1 = e 0 Cov(Xi Xn 1 = n X j =1 n X j =1 The

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Unformatted text preview: Formula sheet - ACTSC 432/832 1. The Normal Equations Unbiaisedness equation E [Xn+1 ] = e 0 + Cov (Xi ; Xn+1 ) = n X j =1 n X j =1 The other n equations e j E [Xj ] i = 1; 2; :::; n e j Cov (Xi ; Xj ) , 2. Bayesien Premium De…nition E [Xn+1 jX1 = x1 ; :::; Xn = xn ] = Alternative representation E [Xn+1 jX1 = x1 ; :::; Xn = xn ] = Z n+1 Z DX xn+1 fXn+1 jX1 ;:::;Xn (xn+1 jx1 ; :::; xn ) dxn+1 () D jX1 ;:::;Xn ( jx1 ; :::; xn ) d 3. Nonparametric Estimation (Buhlmann-Straub framework) unbiaised estimator of b unbiaised estimator of v r ni PP =X mij Xij r P Xi 2 v= b unbiaised estimator of a i=1 j =1 (ni 1) i=1 b= a i=1 m2 i=1 r P r P mi X i X 2 (r m2 i 1) v b m Continuous distributions X EXP( ) : Exponential distribution with parameters >0 1 density fX (x) = e x for x > 0 x c.d.f. FX (x) = 1 e mean E [X ] = variance V ar (X ) = 2 1 m.g.f. MX (t) = 1 t X GAM( ; ) : Gamma distribution with parameters > 0 and 1 for x > 0 >0 x density fX (x) = x e for x > 0 () mean E [X ] = variance V ar (X ) = m.g.f. MX (t) = 2 1 1 t >0 X PAR( ; ) : Pareto distribution with parameters > 0 and density fX (x) c.d.f. FX (x) mean E [X ] variance V ar (X ) = = = = (x + ) 1 1 ( +1 for x > 0 for x > 0 >1 for >2 x+ for 2 2 1) ( 2) Discrete distributions N POIS( ) : Poisson distribution with parameter ( > 0) k p.m.f. pN (k ) = e k! for k = 0; 1; ::: mean E [N ] = variance V ar (N ) = p.g.f. PN (s) = e (s 1) N BIN(m; q ) : Binomial distribution with parameters m (m = 1; 2; :::) and q (0 mk m q (1 q ) k mean E [N ] = mq variance V ar (N ) = mq (1 q ) m p.g.f. PN (s) = (1 q + qs) p.m.f. pN (k ) = q 1) k for k = 0; 1; :::; m ...
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This note was uploaded on 02/02/2010 for the course ACTSC 432 taught by Professor Davidlandriault during the Spring '09 term at Waterloo.

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Final Exam Formula - Formula sheet ACTSC 432/832 1 The Normal Equations Unbiaisedness equation E[Xn 1 = e 0 Cov(Xi Xn 1 = n X j =1 n X j =1 The

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