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二元正態分佈

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二元正態分佈是具有機率密度函式的統計分佈

P(x_1,x_2)=1/(2pisigma_1sigma_2sqrt(1-rho^2))exp[-z/(2(1-rho^2))],
(1)

其中

z=((x_1-mu_1)^2)/(sigma_1^2)-(2rho(x_1-mu_1)(x_2-mu_2))/(sigma_1sigma_2)+((x_2-mu_2)^2)/(sigma_2^2),
(2)

並且

rho=cor(x_1,x_2)=(V_(12))/(sigma_1sigma_2)
(3)

是 x_1 和 x_2 的相關性（Kenney 和 Keeping 1951, pp. 92 和 202-205; Whittaker 和 Robinson 1967, p. 329），V_(12) 是協方差。

二元正態分佈的機率密度函式實現為MultinormalDistribution[{mu1, mu2}, {{sigma11, sigma12}, {sigma12, sigma22}}] 在 Wolfram 語言 包中MultivariateStatistics` .

邊緣機率然後是

P(x_1)=int_(-infty)^inftyP(x_1,x_2)dx_2
(4)
=1/(sigma_1sqrt(2pi))e^(-(x_1-mu_1)^2/(2sigma_1^2))
(5)

並且

P(x_2)=int_(-infty)^inftyP(x_1,x_2)dx_1
(6)
=1/(sigma_2sqrt(2pi))e^(-(x_2-mu_2)^2/(2sigma_2^2))
(7)

（Kenney 和 Keeping 1951, p. 202）。

設 Z_1 和 Z_2 是兩個獨立的正態變數，具有均值 mu_i=0 和 sigma_i^2=1 對於 i=1, 2。然後，下面定義的變數 a_1 和 a_2 是具有單位方差和相關係數 rho 的二元正態分佈

a_1=sqrt((1+rho)/2)z_1+sqrt((1-rho)/2)z_2
(8)
a_2=sqrt((1+rho)/2)z_1-sqrt((1-rho)/2)z_2.
(9)

為了推導二元正態機率函式，設 X_1 和 X_2 是正態且獨立分佈的變數，具有均值 0 和方差 1，然後定義

Y_1=mu_1+sigma_(11)X_1+sigma_(12)X_2
(10)
Y_2=mu_2+sigma_(21)X_1+sigma_(22)X_2
(11)

（Kenney 和 Keeping 1951, p. 92）。變數 Y_1 和 Y_2 自身也呈正態分佈，具有均值 mu_1 和 mu_2，方差

sigma_1^2=sigma_(11)^2+sigma_(12)^2
(12)
sigma_2^2=sigma_(21)^2+sigma_(22)^2,
(13)

和協方差

V_(12)=sigma_(11)sigma_(21)+sigma_(12)sigma_(22).
(14)

協方差矩陣由下式定義

V_(ij)=[sigma_1^2 rhosigma_1sigma_2; rhosigma_1sigma_2 sigma_2^2],
(15)

其中

rho=(V_(12))/(sigma_1sigma_2)=(sigma_(11)sigma_(21)+sigma_(12)sigma_(22))/(sigma_1sigma_2).
(16)

現在，x_1 和 x_2 的聯合機率密度函式是

f(x_1,x_2)dx_1dx_2=1/(2pi)e^(-(x_1^2+x_2^2)/2)dx_1dx_2,
(17)

但從 (◇) 和 (◇) 中，我們得到

[y_1-mu_1; y_2-mu_2]=[sigma_(11) sigma_(12); sigma_(21) sigma_(22)][x_1; x_2].
(18)

只要

|sigma_(11) sigma_(12); sigma_(21) sigma_(22)|!=0,
(19)

就可以反轉得到

[x_1; x_2]=[sigma_(11) sigma_(12); sigma_(21) sigma_(22)]^(-1)[y_1-mu_1; y_2-mu_2]
(20)
=1/(sigma_(11)sigma_(22)-sigma_(12)sigma_(21))[sigma_(22) -sigma_(12); -sigma_(21) sigma_(11)][y_1-mu_1; y_2-mu_2].
(21)

因此，

x_1^2+x_2^2=([sigma_(22)(y_1-mu_1)-sigma_(12)(y_2-mu_2)]^2)/((sigma_(11)sigma_(22)-sigma_(12)sigma_(21))^2) +([-sigma_(21)(y_1-mu_1)+sigma_(11)(y_2-mu_2)]^2)/((sigma_(11)sigma_(22)-sigma_(12)sigma_(21))^2),
(22)

展開 (22) 的分子得到

sigma_(22)^2(y_1-mu_1)^2-2sigma_(12)sigma_(22)(y_1-mu_1)(y_2-mu_2)+sigma_(12)^2(y_2-mu_2)^2+sigma_(21)^2(y_1-mu_1)^2-2sigma_(11)sigma_(21)(y_1-mu_1)(y_2-mu_2)+sigma_(11)^2(y_2-mu_2)^2,
(23)

所以

(x_1^2+x_2^2)(sigma_(11)sigma_(22)-sigma_(12)sigma_(21))^2 =(y_1-mu_1)^2(sigma_(21)^2+sigma_(22)^2)-2(y_1-mu_1)(y_2-mu_2)(sigma_(11)sigma_(21)+sigma_(12)sigma_(22))+(y_2-mu_2)^2(sigma_(11)^2+sigma_(12)^2) =sigma_2^2(y_1-mu_1)^2-2(y_1-mu_1)(y_2-mu_2)(rhosigma_1sigma_2)+sigma_1^2(y_2-mu_2)^2 =sigma_1^2sigma_2^2[((y_1-mu_1)^2)/(sigma_1^2)-(2rho(y_1-mu_1)(y_2-mu_2))/(sigma_1sigma_2)+((y_2-mu_2)^2)/(sigma_2^2)].
(24)

現在，(◇) 的分母是

sigma_(11)^2sigma_(21)^2+sigma_(11)^2sigma_(22)^2+sigma_(12)^2sigma_(21)^2+sigma_(12)^2sigma_(22)^2-sigma_(11)^2sigma_(21)^2 -2sigma_(11)sigma_(12)sigma_(21)sigma_(22)-sigma_(12)^2sigma_(22)^2=(sigma_(11)sigma_(22)-sigma_(12)sigma_(21))^2,
(25)

所以

1/(1-rho^2)=1/(1-(V_(12)^2)/(sigma_1^2sigma_2^2))
(26)
=(sigma_1^2sigma_2^2)/(sigma_1^2sigma_2^2-V_(12)^2)
(27)
=(sigma_1^2sigma_2^2)/((sigma_(11)^2+sigma_(12)^2)(sigma_(21)^2+sigma_(22)^2)-(sigma_(11)sigma_(21)+sigma_(12)sigma_(22))^2).
(28)

可以簡單地寫成

1/(1-rho^2)=(sigma_1^2sigma_2^2)/((sigma_(11)sigma_(22)-sigma_(12)sigma_(21))^2),
(29)

並且

x_1^2+x_2^2=1/(1-rho^2)[((y_1-mu_1)^2)/(sigma_1^2)-(2rho(y_1-mu_1)(y_2-mu_2))/(sigma_1sigma_2)+((y_2-mu_2)^2)/(sigma_2^2)].
(30)

求解 x_1 和 x_2 並定義

rho^'=(sigma_1sigma_2sqrt(1-rho^2))/(sigma_(11)sigma_(22)-sigma_(12)sigma_(21))
(31)

得到

x_1=(sigma_(22)(y_1-mu_1)-sigma_(12)(y_2-mu_2))/(rho^')
(32)
x_2=(-sigma_(21)(y_1-mu_1)+sigma_(11)(y_2-mu_2))/(rho^').
(33)

但 雅可比行列式是

J((x_1,x_2)/(y_1,y_2))=|(partialx_1)/(partialy_1) (partialx_1)/(partialy_2); (partialx_2)/(partialy_1) (partialx_2)/(partialy_2)|=|(sigma_(22))/(rho^') -(sigma_(12))/(rho^'); -(sigma_(21))/(rho^') (sigma_(11))/(rho^')|
(34)
=1/(rho^('2))(sigma_(11)sigma_(22)-sigma_(12)sigma_(21))
(35)
=1/(rho^')=1/(sigma_1sigma_2sqrt(1-rho^2)),
(36)

所以

dx_1dx_2=(dy_1dy_2)/(sigma_1sigma_2sqrt(1-rho^2))
(37)

並且

1/(2pi)e^(-(x_1^2+x_2^2)/2)dx_1dx_2=1/(2pisigma_1sigma_2sqrt(1-rho^2))exp[-z/(2(1-rho^2))]dy_1dy_2,
(38)

其中

z=((y_1-mu_1)^2)/(sigma_1^2)-(2rho(y_1-mu_1)(y_2-mu_2))/(sigma_1sigma_2)+((y_2-mu_2)^2)/(sigma_2^2).
(39)

證畢。

二元正態分佈的特徵函式由下式給出

phi(t_1,t_2)=int_(-infty)^inftyint_(-infty)^inftye^(i(t_1x_1+t_2x_2))P(x_1,x_2)dx_1dx_2
(40)
=Nint_(-infty)^inftyint_(-infty)^inftye^(i(t_1x_1+t_2x_2))exp[-z/(2(1-rho^2))]dx_1dx_2,
(41)

其中

z=[((x_1-mu_1)^2)/(sigma_1^2)-(2rho(x_1-mu_1)(x_2-mu_2))/(sigma_1sigma_2)+((x_2-mu_2)^2)/(sigma_2^2)]
(42)

並且

N=1/(2pisigma_1sigma_2sqrt(1-rho^2)).
(43)

現在讓

u=x_1-mu_1
(44)
w=x_2-mu_2.
(45)

然後

phi(t_1,t_2)=N^'int_(-infty)^infty(e^(it_2w)exp[-1/(2(1-rho^2))(w^2)/(sigma_2^2)])int_(-infty)^inftye^ve^(t_1u)dudw,
(46)

其中

v=-1/(2(1-rho^2))1/(sigma_1^2)[u^2-(2rhosigma_1w)/(sigma_2)u]
(47)
N^'=(e^(i(t_1mu_1+t_2mu_2)))/(2pisigma_1sigma_2sqrt(1-rho^2)).
(48)

在內積分中完成平方

int_(-infty)^inftyexp{-1/(2(1-rho^2))1/(sigma_1^2)[u^2-(2rhosigma_1w)/(sigma_2)u]}e^(t_1u)du =int_(-infty)^inftyexp{-1/(2sigma_1^2(1-rho^2))[u-(rho_1sigma_1w)/(sigma^2)]^2}{1/(2sigma_1^2(1-rho^2))((rho_1sigma_1w)/(sigma_2))^2}e^(it_1u)du.
(49)

重新排列以將取決於 w 的指數項移到外積分之外，令

v=u-rho(sigma_1w)/(sigma_2),
(50)

並寫作

e^(it_1u)=cos(t_1u)+isin(t_1u)
(51)

得到

phi(t_1,t_2)=N^'int_(-infty)^inftye^(it_2w)exp[-1/(2sigma_2^2(1-rho^2))w^2]exp[(rho^2)/(2sigma_2^2(1-rho^2))w^2]int_(-infty)^inftyexp[-1/(2sigma_2^2(1-rho^2))v^2]{cos[t_1(v+(rhosigma_1w)/(sigma_2))]+isin[t_1(v+(rhosigma_1w)/(sigma_2))]}dvdw.
(52)

展開括號內的項得到

[cos(t_1v)cos((rhosigma_1wt_1)/(sigma_2))-sin(t_1v)sin((rhosigma_1w)/(sigma_2t_1))]+i[sin(t_1v)cos((rhosigma_1w)/(sigma_2t_1))+cos(t_1v)sin((rhosigma_1wt_1)/(sigma_2))] =[cos((rhosigma_1wt_1)/(sigma_2))+isin((rhosigma_1wt_1)/(sigma_2))][cos(t_1v)+isin(t_1v)]=exp((irhosigma_1w)/(sigma_2)t_1)[cos(t_1v)+isin(t_1v)].
(53)

但是 e^(-ax^2)sin(bx) 是奇函式，因此正弦項上的積分消失，我們剩下

phi(t_1,t_2)=N^'int_(-infty)^inftye^(it_2w)exp[-(w^2)/(2sigma_2^2)]exp[(rho^2w^2)/(2sigma_2^2(1-rho^2))]exp[(irhosigma_1wt_1)/(sigma_2)]dwint_(-infty)^inftyexp[-(v^2)/(2sigma_1^2(1-rho^2))]cos(t_1v)dv =N^'int_(-infty)^inftyexp[iw(t_2+t_1(rho(sigma_1)/(sigma_2)))]exp[-(w^2)/(2sigma_2^2)]dwint_(-infty)^inftyexp[-(v^2)/(2sigma_1^2(1-rho^2))]cos(t_1v)dv.
(54)

現在評估高斯積分

int_(-infty)^inftye^(ikx)e^(-ax^2)dx=int_(-infty)^inftye^(-ax^2)cos(kx)dx
(55)
=sqrt(pi/a)e^(-k^2/4a)
(56)

以獲得特徵函式的顯式形式，

phi(t_1,t_2)=(e^(i(t_1mu_1+t_2mu_2)))/(2pisigma_1sigma_2sqrt(1-rho^2)){sigma_2sqrt(2pi)exp[-1/4(t_2+rho(sigma_1)/(sigma_2)t_1)^22sigma_2^2]}{sigma_1sqrt(2pi(1-rho^2))exp[-1/4t_1^22sigma_1^2(1-rho^2)]} =e^(i(t_1mu_1+t_2mu_2))exp{-1/2[t_2^2sigma_2^2+2rhosigma_1sigma_2t_1t_2+rho^2sigma_1^2t_1^2+(1-rho^2)sigma_1^2t_1^2]} =exp[i(t_1mu_1+t_2mu_2)-1/2(sigma_1^2t_1^2+2rhosigma_1sigma_2t_1t_2+sigma_2^2t_2^2)].
(57)

在奇異情況下，

|sigma_(11) sigma_(12); sigma_(21) sigma_(22)|=0
(58)

（Kenney 和 Keeping 1951, p. 94），由此得出

sigma_(11)sigma_(22)=sigma_(12)sigma_(21)
(59)
y_1=mu_1+sigma_(11)x_1+sigma_(12)x_2
(60)
y_2=mu_2+(sigma_(12)sigma_(21))/(sigma_(11))x_2
(61)
=mu_2+(sigma_(11)sigma_(21)x_1+sigma_(12)sigma_(21)x_2)/(sigma_(11))
(62)
=mu_2+(sigma_(21))/(sigma_(11))(sigma_(11)x_1+sigma_(12)x_2),
(63)

所以

y_1=mu_1+x_3
(64)
y_2=mu_2+(sigma_(21))/(sigma_(11))x_3,
(65)

其中

x_3=y_1-mu_1
(66)
=(sigma_(11))/(sigma_(21))(y_2-mu_2).
(67)

標準化的二元正態分佈取 sigma_1=sigma_2=1 和 mu_1=mu_2=0。在這種特殊情況下，象限機率透過分析給出

P(x_1<=0,x_2<=0)=P(x_1>=0,x_2>=0)
(68)
=int_(-infty)^0int_(-infty)^0P(x_1,x_2)dx_1dx_2
(69)
=1/4+(sin^(-1)rho)/(2pi)
(70)

（Rose 和 Smith 1996; Stuart 和 Ord 1998; Rose 和 Smith 2002, p. 231）。類似地，

P(x_1<=0,x_2>=0)=P(x_1>=0,x_2<=0)
(71)
=int_(-infty)^0int_0^inftyP(x_1,x_2)dx_1dx_2
(72)
=(cos^(-1)rho)/(2pi).
(73)

另請參閱

Box-Muller 變換, 二元正態分佈的相關係數, 多元正態分佈, 正態分佈, Price 定理, 三元正態分佈

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參考文獻

Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 936-937, 1972.Holst, E. "The Bivariate Normal Distribution." http://www.ami.dk/research/bivariate/.Kenney, J. F. and Keeping, E. S. Mathematics of Statistics, Pt. 2, 2nd ed. Princeton, NJ: Van Nostrand, 1951.Kotz, S.; Balakrishnan, N.; and Johnson, N. L. "Bivariate and Trivariate Normal Distributions." Ch. 46 in Continuous Multivariate Distributions, Vol. 1: Models and Applications, 2nd ed. New York: Wiley, pp. 251-348, 2000.Rose, C. and Smith, M. D. "The Multivariate Normal Distribution." Mathematica J. 6, 32-37, 1996.Rose, C. and Smith, M. D. "The Bivariate Normal." §6.4 A in Mathematical Statistics with Mathematica. New York: Springer-Verlag, pp. 216-226, 2002.Spiegel, M. R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, p. 118, 1992.Stuart, A.; and Ord, J. K. Kendall's Advanced Theory of Statistics, Vol. 1: Distribution Theory, 6th ed. New York: Oxford University Press, 1998.Whittaker, E. T. and Robinson, G. "Determination of the Constants in a Normal Frequency Distribution with Two Variables" and "The Frequencies of the Variables Taken Singly." §161-162 in The Calculus of Observations: A Treatise on Numerical Mathematics, 4th ed. New York: Dover, pp. 324-328, 1967.

在 上被引用

二元正態分佈

請引用為

Weisstein, Eric W. “二元正態分佈。” 來自 Web 資源。https://mathworld.tw/BivariateNormalDistribution.html

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