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Probability 2 - Notes 11 The bivariate and multivariate ...

Probability 2 - Notes 11 The bivariate and multivariate normal s(X,Y)have a bivariate normal distributionN( 1, 2, 21, 22, )if theirjoint isfX,Y(x,y) =12 1 2 (1 2)e 12(1 2)[(x 1 1)2 2 (x 1 1)(y 2 2)+(y 2 2)2](1)for allx,y. The parameters 1, 2may be any real numbers, 1>0, 2>0, and 1 is convenient to rewrite (1) in the formfX,Y(x,y) =ce 12Q(x,y),wherec=12 1 2 (1 2)andQ(x,y) = (1 2) 1[(x 1 1)2 2 (x 1 1)(y 2 2)+(y 2 2)2](2) marginal distributions ofN( 1, 2, 21, 22, )are normal with sXandYhaving density functionsfX(x) =1 2 1e (x 1)22 21,fY(y) =1 2 2e (y 2)22 expression (2) forQ(x,y)can be rearranged as follows:Q(x,y) =11 2[(x 1 1 y 2 2)2+(1 2)(y 2 2)2]=(x a)2(1 2) 21+(y 2)2 22,(3)wherea=a(y) = 1+ 1 2(y 2). HencefY(y) = fX,Y(x,y)dx=ce (y 2)22 22 e (x a)22(1 2) 21dx=1 2 2e (y 2)22 22,where the last step makes use of the formula e (x a)22 2dx= 2 with = 1 1 the formula forfX(x).

This is just the m.g.f. for the multivariate normal distribution with vector of means Am+b and variance-covariance matrix AVAT. Hence, from the uniqueness of the joint m.g.f, Y » N(Am+b;AVAT). Note that from (2) a subset of the Y0s is multivariate normal. NOTE. The results concerning the vector of means and variance-covariance matrix for linear

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Transcription of Probability 2 - Notes 11 The bivariate and multivariate ...

1 Probability 2 - Notes 11 The bivariate and multivariate normal s(X,Y)have a bivariate normal distributionN( 1, 2, 21, 22, )if theirjoint isfX,Y(x,y) =12 1 2 (1 2)e 12(1 2)[(x 1 1)2 2 (x 1 1)(y 2 2)+(y 2 2)2](1)for allx,y. The parameters 1, 2may be any real numbers, 1>0, 2>0, and 1 is convenient to rewrite (1) in the formfX,Y(x,y) =ce 12Q(x,y),wherec=12 1 2 (1 2)andQ(x,y) = (1 2) 1[(x 1 1)2 2 (x 1 1)(y 2 2)+(y 2 2)2](2) marginal distributions ofN( 1, 2, 21, 22, )are normal with sXandYhaving density functionsfX(x) =1 2 1e (x 1)22 21,fY(y) =1 2 2e (y 2)22 expression (2) forQ(x,y)can be rearranged as follows:Q(x,y) =11 2[(x 1 1 y 2 2)2+(1 2)(y 2 2)2]=(x a)2(1 2) 21+(y 2)2 22,(3)wherea=a(y) = 1+ 1 2(y 2). HencefY(y) = fX,Y(x,y)dx=ce (y 2)22 22 e (x a)22(1 2) 21dx=1 2 2e (y 2)22 22,where the last step makes use of the formula e (x a)22 2dx= 2 with = 1 1 the formula forfX(x).

2 N( 1, 21),Y N( 2, 22), we know the meaning of four parameters involved intothe definition of the normal distribution, namelyE(X) = 1,Var(X) = 21,E(Y) = 2,Var(X) = |(Y=y)is a normal verify this statement we substitute the necessary ingredientsinto the formula defining the relevant conditional density:fX|Y(x|y) =fX,Y(x,y)fY(y)=1 2 (1 2) 1e (x a(y))22 21(1 2).1In other words,X|(Y=y) N(a(y),(1 2) 21). (X|Y=y) =a(y)or, equivalently,E(X|Y) = 1+ 1 2(Y 2). In particular, we see thatE(X|Y)is a linear function (XY) = 1 2 + 1 (XY) =E[E(XY|Y)] =E[Y E(X|Y)] =E[Y( 1+ 1 2(Y 2)] = 1E(Y)+ 1 2[E(Y2) 2E(Y)] = 1 2+ 1 2[E(Y2) 22] = 1 2+ 1 2 Var(Y) = 1 2 + 1 (X,Y) = 1 2 . This follows from Corollary 4 and the formulaCov(X,Y) =E(XY) E(X)E(X).6. (X,Y) = . In words: is the correlation coefficient ofX,Y.)

3 This is now obvious from thedefinition (X,Y) =Cov(X,Y) Var(X)var(Y). thatXandYare independent iff =0. (We proved this in the lecture; it iseasily seen from either the joint ) is possible to show that the ofX,YisMX,Y(t1,t2) =e( 1t1+ 2t2)+12( 21t21+2 1 2t1t2+ 22t22)Many of the above statements follow from it. (To actually do this is a very useful exercise.)The multivariate normal vector and matrix study the joint normal distributions of more than s, it is convenient to use vectors and matrices. But let us first introduce these notations forthe case of two normal sX1,X2. We setX=(X1X2);x=(x1x2);t=(t1t2);m=( 1 2);V=( 21 1 2 1 2 22)Thenmis the vector of means andVis the variance-covariance matrix. Note that|V|= 21 22(1 2)andV 1=1(1 2)(1 21 1 2 1 21 22)HencefX(x) =1(2 )2/2|V|1/2e 12(x m)TV 1(x m)for allx.

4 AlsoMX(t) =etTm+ again use matrix and vector notation, but now there arenrandom variables so thatX,x,tandmare nown-vectors withithentriesXi,xi,tiand iandVis then nmatrix withiithentry 2iandi jthentry (fori6=j) i j. Note thatVis symmetric so thatVT= joint isfX(x) =1(2 )n/2|V|1/2e 12(x m)TV 1(x m)for allx. We say thatX N(m,V).We can find the joint quite (t) =E[e nj=1tjXj]=E[etTX] = .. 1(2 )n/2|V|1/2e 12((x m)TV 1(x m) 2tTx) do the equivalent of completing the square, we write(x m)TV 1(x m) 2tTx= (x m a)TV 1(x m a)+bfor a suitable choice of then-vectoraof constants and a constantb. ThenMX(t) =e b/2 .. 1(2 )n/2|V|1/2e 12(x m a)TV 1(x m a) just need to findaandb. Expanding we have((x m) a)TV 1((x m) a)+b= (x m)TV 1(x m) 2aTV 1(x m)+aTV 1a+b= (x m)TV 1(x m) 2aTV 1x+[2aTV 1m+aTV 1a+b]This has to equal(x m)TV 1(x m) 2tTxfor allx.

5 Hence we needaTV 1=tTandb= [2aTV 1m+aTV 1a]. Hencea=Vtandb= [2tTm+tTVt]. ThereforeMX(t) =e b/2=etTm+12tTVtResults obtained using the Any (non-empty) subset of multivariate normals is multivariate normal . Simply puttj=0foralljfor whichXjis not in the subset. For exampleMX1(t1) =MX1,..,Xn(t1,0,..,0) =et1 1+t21 21 N( 1, 21). A similar result holds forXi. This identifies the parameters iand 2iasthe mean and variance ofXi. AlsoMX1,X2(t1,t2) =MX1,..,Xn(t1,t2,0,..,0) =et1 1+t2 2+12(t21 21+2 12t1t2+ 22t22)HenceX1andX2have bivariate normal distribution with 12=Cov(X1,X2). A similar resultholds for the joint distribution ofXiandXjfori6=j. This identifiesVas the variance-covariancematrix forX1,.., a vector of independent random variables iffVis diagonal ( all off-diagonal entriesare zero so that i j=0fori6=j).

6 (1), if theX sare independent then i j=Cov(Xi,Xj) =0for alli6=j, so diagonal thentTVt= nj=1 2jt2jand henceMX(t) =etTm+12tTVt=n j=1(e jtj+12 2jt2j/2)=n j=1 MXj(tj)By the uniqueness of the joint ,X1,..,Xnare Linearly independent linear functions of multivariate normal random variables are multivari-ate normal random variables. IfY=AX+b, whereAis ann nnon-singular matrix andbisa (column)n-vector of constants, thenY N(Am+b,AVAT). the joint (t) =E[etTY] =E[etTAX+b] =etTbE[e(ATt)TX] =etTbMX(ATt)=etTbe(ATt)Tm+12(ATt)TV(ATt) =etT(Am+b)+12tT(AVAT)tThis is just the for the multivariate normal distribution with vector of meansAm+band variance-covariance matrixAVAT. Hence, from the uniqueness of the joint ,Y N(Am+b,AVAT).Note that from (2) a subset of theY sis multivariate results concerning the vector of means and variance-covariance matrix for linearfunctions of random variables hold regardless of the joint distribution ofX1.

7 , define the expectation of a vector of random variablesX,E[X]to be the vector of theexpectations and the expectation of a matrix of random variablesY,E[Y], to be the matrix ofthe expectations. Then the variance-covariance matrix ofXis justE[(X E[X])(X E[X])T].The following results are easily obtained:(i) LetAbe anm nmatrix of constants,Bbe anm kmatrix of constants andYbe ann kmatrix of random variables. ThenE[AY+B] =AE[Y]+ Thei jthentry ofE[AY+B]isE[ nr=1 AirYr j+Bi j] = nr=1 AirE[Yr j]+Bi j, which is thei jthentry ofAE[Y]+B. The result is then immediate.(ii) LetCbe ak mmatrix of constants andYbe ann kmatrix of random variables. ThenE[YC] =E[Y] Just transpose the equation. The result then follows from (i).Hence ifZ=AX+b, whereAis anm nmatrix of constants,bis anm-vector of constantsandXis ann-vector of random variables withE[X] = and variance-covariance matrixV, thenE[Z] =E[AX+b] =AE[X]+b=A +bAlso the variance-covariance matrix forYis justE[(Y E[Y])(Y E[Y])T] =E[A(X )(X )TAT] =AE[(X )(X )T]AT= thatE[X1] =1,E[X2] =0,Var(X1) =2,Var(X2) =4andCov(X1,X2) = +X2andY2=X1+aX2.

8 Find the means, variances and covariance and hence findaso thatY1andY2are in vector and matrix notation we haveE[Y] =Amand the variance-covariance matrixforYis justAVAT wherem=(10)V=(2 11 4)A=(1 11a)ThereforeAm=(1 11a)(10)=(11)AVAT=(1 11a)(2 11 4)(1 11a)=(83+5a3+5a2+2a+4a2)HenceY1andY2have means1and1, variances8and2+2a+4a2and covariance3+5a. Theyare therefore uncorrelated if3+5a=0, ifa=


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