[機統] 一題較難的連續隨機分佈!請高手指點!
讓 T1, ..., Tk
成為一組連續隨機變數(continuous random variables)
有marginal CDFs F1, ..., Fk
和(marginal densities)邊際密度 f1, ..., fk.
讓 x) = (2π)^(-1/2) × e^(-z^2/2)
表示標準常態密度 (standard normal density)
然後讓 Φ(x) =∫Φ(z) dz,(-∞,∞)
表示標準常態的CDF( standard normal CDF. )
(i) 證明Xj ~ N(0, 1).
假設Xj 可以被寫成
Xj = αjYj +√1 – √αj^2xZ, j = 1, ..., k,
where Y1, ..., Yk,Z ~ N(0, 1) are 獨立的(independent) and
0 <= α1, ..., αk<= 1 are constants.
(ii) 找出(joint distribution)聯合分配 of (X1, ...,Xk).
(iii)讓Σ 代表 the covariance matrix 與(X1, ...,Xk)相關. Find an expression
for Σ in terms of a1, ..., ak. What are the eigenvalues of Σ?
(iv) By conditioning on Z, 證明(unconditional) joint density(聯合密度) of
(X1, ...,Xk)能被寫成
Φ(x1, ..., xk; 0, Σ) =
∫Φ(x1 -√1 - √α1^2xZ)/α1-Φ(xk -√1 - √αk^2xZ)/αkΦ(z)x dz(-∞,∞).
(v) 運用 part (iv) 去找出an expression for (聯合密度)the joint density of
(T1, ..., Tk).
讓 Σ be as in這問題的 part (iii) .
The Gaussian copula:
C(u1, ..., uk) =Φ(Φ^-1(u1), ...,Φ^-1(uk); 0, Σ),
被稱為 the single-factor Gaussian copula.
他被稱為 a single-factor copula 由於 (1) implies that (X1, ..,Xk) follow a
single-factor model, with the common factor Z.
(vi) 用公式表示a three-factor Gaussian copula.為何a three-factor 比起the
single-factor model更受到選擇上的偏好? a three-factor model比較於 the
single-factor model 有哪些潛在的缺陷?
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