| [ QuizWit ] in KIDS 글 쓴 이(By): sisyphe () 날 짜 (Date): 2001년 7월 3일 화요일 오전 05시 01분 22초 제 목(Title): Re: puking inverse 우선 E(Ci'Ci)를 구하면, Ci의 각 element를 Cip라 하면 E(sum_p^{Cip*Cip}) = sum_p^{ E(Cip*Cip) } (1) Cip는 각각 mean 0, var 1인 gaussian random var이므로 var(Cip) = E(Cip^2) - E(Cip)^2 로부터 1 = E(Cip^2) - 0, 그러므로 E(Cip^2) = 1. (2) (1)과 (2)로 부터 E(S_ii) = a*1 (j가 1에서 a까지일때). Ci와 Cj가 independent하므로 E(Ci'Cj)는 비슷한 요령으로 각각의 element를 Cip, Cjp라 하면 E(Cip*Cjp) = E(Cip)*E(Cjp) (각각이 independent하므로). 이것은 다시 E(Cip)=0, E(Cjp)=0 (mean 0이므로) 이므로 E(Cip*Cjp) = 0, 그러므로 E(Ci'Cj) = E(sum_p^{Cip*Cjp}) = sum_p^{E(Cip*Cjp)} = 0. 고로 E(S_ij) =0 (i!=j). 그러므로 S는 각 diagonal term의 Expectation이 a이고 나머지는 0이므로 E(S) = a*I. (I는 dentity matrix). 그러므로 E(S^(-1))=(1/a)*I. ======= The last argument is incorrect. :) But nice try. Your solution can be more elegantly written if you know the idea of tensorial contraction, which avoids the use of summation. the covariance structures of C_ij is the 2nd order tensor. Albert Einstein avoided using the summation notation in his derivation of general relativity. I am not sure if he is the first person to use the convention. It is usually called "Einstein summation" or "Einstein notation". Using Einstein notation, you write sum_p C_{ip}C_{jp} as C_{ip}C_{jp} with the understanding that when there is a paired indices, the summation notation is assumed to be present. Also you seem not knowing the most important result in Gaussian distribution. For any Gaussian variables X_1,X2, ... Xn with mean zeros, E(X_1*X_2*...*X_n) = E(X_1*X_2)E(X_3*X_4).... + E(X_1*X_3)E(X_2*X_4).. + the sum of the products of all possible paied combinations. This fact can be proved using the moment generating functions. Further, you can show that E(S^k) = a_k*I for any integer k. Why? You showed that E(S)= a_1*I. It is easy to realize that E(S^k) is symmetric matrix. E(S^k) = A + a_k*I where A is a symmetric matrix with the diagonal element zero. Then for any orthogonal matrix Q, it is easy to show that E(QS^kQ') = E(S^k). Then QAQ' + a_k*I = A + a_k*I Hence QAQ' = A for all orthogonal matrix Q. Therefore, A must be 0 and hence proving my claim. QED |