格林高级计量经济学第二版

fY ( y ) = f X ( x ) ∂x ∂y '
x = x( y ) fY ( y1 , y2 ,..., yn ) = f X ( x1 , x2 ,..., xn ) || J ( y ) || fY ( y1 , y2 ,..., yn ) = f X ( x1 , x2 ,..., xn .) || J || where J is the Jacobian matrix ∂x1 / ∂y1 / ∂x1 / ∂y2 ... ... ∂x ≡ J ≡ J ( y) ≡ ... ... ∂y ' ∂xn / ∂y1 / ∂xn / ∂y2 ... ∂x1 / ∂yn ... ... ... ... ... ∂xn / ∂yn
2010.3.1 Nanqiang2 208 16:30-18:10
Multivariate Distribution
Be very familiar with Distribution, Why? The foundation of econometrics. No distribution theory, no econometrics .
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Univariate Standard Normal Distribution Normal Distribution is universal distribution, why? Bernoulli, scaled symmetric random walk, brownian motion? Probability Density Function: PDF X ~ N ( 0,1)
Non-parametric estimation -- the central issue is to estimate
PDF;
Parameter estimation -- distribution theory plays major roles:
1. Derive the Linear Regression Equation 2. Maximum likelihood estimation 3. Hypothesis Testing 4. Asymptotic Distribution 5. Bayesian estimation Cumulative Distribution Function & Probability Density Function
Remark: image
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PDF of A Function of A Random Vector Theorem: Let f X ( x1 , x2 ,..., xn )
If Y ~ N ( µ , ∑) then
( y − µ )' ∑ −1 ( y − µ ) 1 1 −1 2 f ( y) = e || ∑ || 2π n
一般多元正态分布的密度函数
Remark : So the proportinal factor is Remark: compare with
−µ 1 1 − 2 yσ f ( y) = e 2 || σ || 2π
1 || ∑ ||
n 1
2
一般一元正态分布的密度函数 1 1 −2 f ( y) = e || σ 2 || 2π 1 1 −2 f ( y) = e || ∑ || 2π
fY ( y ) = f X ( x ) dx . dy
Remark: LHS x=x(y) “Proof”: ! By the definition of probability, within an interval ∆x : Probability = f X ( x ) ∆x Probability = fY ( y ) ∆y ⇒ f X ( x ) ∆ x = fY ( y ) ∆ y ∆x fY ( y ) = f X ( x ) ∆y ∆y dx fY ( y ) = f X ( x ) (taking limit) dy dx as a proportional factor! dy
PDF : A Function of A Random Variable Motivating example: Image that two grading systems - Chianese and American Prob { X ∈ [ 0,100]} = 1 and Prob {Y ∈ [0, 4]} = 1 How to change the PDF from Chinese to American system? Then the transformation is y=x/25 . Special case of scalar: suppose that X : f X ( x ) is known and Y is a function of X, then
X ~ N ( 0,1) ? The PDF of Y ~ N ( µ , σ 2 ) is : f ( y) = Proof: recall F ( y ) = ∫ (？ )dt
−∞ y
1 2πσ 2
e
1 y −µ − 2 σ
2
, need to find (?) cdf by using
be the joint probability density
function of X A ( X1 , X 2 ,..., X n ) , then the joint probability density function of Y is: X ,Y are randon vector, x y are vectors
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Construct a multivariate normal distribution2 Theorem: A linear transformation distribution is normal. of normal
x2 1 −1 e 2 is preferred. 2π
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2010.3.3 Nanqiang2 209 16:30-18:10
Univariate Normal Distribution How to derive the PDF of Y ~ N ( µ,σ 2 ) from
EX1: assume that X 1 , X 2 ,..., X n ~ iid N(0,1) then the PDF of X ≡ ( X 1 , X 2 ,..., X n ) is
x'x 1 −1 2 f X ( x1 , x2 ,..., xn ) = e 2π where iid stands for independently and identically distributed x ≡ ( x1 , x2 ,..., xn ) ' n
E ( X i ) ≡ µi E ( X ) =µ
then
Hale Waihona Puke Baidu
Second Moment of a Multivariate Distribution Denote: σ ij ≡ E ( X i − µi )( X j − µ j ) σ 11 σ 12 σ σ 22 ∑ ≡ 21 σ n1 σ n 2 σ 1n σ 23 σ nn
a1n x1 a2n x2 ann xn ( y1 , y2 ,..., yn ) .
n
1 1 − 2 y '( AA ')−1 y fY ( y1 , y2 ,..., yn ) = e || A−1 || (一般正态分布) 2π
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2010.3.9 Nanqiang2 209 16:30-17:10
First Moment of a Multivariate Distribution X1 X let X A 2 , Xn denote E ( X1 ) E ( X2 ) define E ( X ) ≡ E ( X ) n µ1 µ µ ≡ 2 µn
Alternatively:
derive
pdf through ∂ f ( y) = F ( y) ∂y
Suppuse σ =0.5 , then the density will be twice as stronger.
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Using matrix notation: ∑ A var ( X ) A E [ ( X − µ )( X − µ ) '] Remarks: 1) ∑ is symmetric 2) ∑ is nnd . quadratic form is non-negative, n.n.d. , d=definite. That is, for any vector a, a ' ∑ a ≥ 0 Proof: let w ≡ a1 X 1 + ... + an X n = a ' X （ no assumptions of normality）
EX21： x1 , x2 ,..., xn ~ iid N (0,1) , and
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To make life easy, now we abused notations sometimes without producing ambiguities: we use the same notation for regular variable and for random variable; the same notation for row vector and column vector! 5
− 1 x2 1 f ( x) = e 2 2π
Cumulative Distribution Function: CDF F ( x ) = Pr( X ≤ x ) =
−∞
∫
x
1 f ( t )dt = 2π
−∞
∫e
x
1 − t2 2
dt
Remark: to avoid using too many notations (X x t), simply using f ( x ) =
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y1 a11 a12 y a 2 = 21 a22 yn an1 an 2 Pls derive fY ( y1 , y2 ,..., yn ) , the PDF of Answer: