Question

Problem 2. (Conditional Distribution of MVN) Let Z1, Z2, Z3 be i.i.d. N(0,1) dis- tributed random variables, and set X1 = 21 – Z3 X2 = 2Z1 + Z2 – 223 X3 = -221 +3Z3 1) What distribution does X = (X1, X2, X3)T follow? Specific the parameters. 2) Find out P(X2 > 0|X1 + X3 = 0).

Answer 1

Solution : Given Z, iz Z, be lid N(011) f(2,22,23) = f(21) fle,) H(23) - ( t e bitz +43) So, fle) = 1 2 3 e 2 we know, if yn N (4,2) then cynnoleu CEC) where Cman matrix of constants. - Hene we have x. X2 - 2 1-2 - 121 1 -2 1 22. 0 3 1 Z3 lloon = 10 lool 0o 10-1 C = 12 1-2 1-2 0 31 (x, , +2, X3 ) follows multivariate normal with parameters 110-11 cu = (2 -2 1- 20 3 1 eze=1636) EC - 110-1 1 2 1 - 2 1-2 0 3 1 0 0 01 0 0 0 1 1 0 1-1 2 1 - 2 - 1 0 -1 2 - 2 1 2 1 -2 0 0 1-2 031-1-2 37 12 4 1-5 4 9 -10 -5 1 -10 13) P(x2.70 and X, + X₂ = 0) (6) P(X₂30 1x, + X₂ =0) - - P(x, + X3 = 0) We will find the joint density of (x2, X1 + X3) which can be written in the form (a! :) (2) which is multivariate normal with parameters :) )-1) and 0

19 lolol12 4 - 5 liol 1 4 9 -10 1-5-10 13lol lolol 14 -3 1 10 9 -6 1-10 8 - (0) 4.)) And ***3 = COI) (nn, (1.1(),C01.2x () So, x, + Xz follows normal distribution with mean (0) and woven vaniance (101) u 9 - 10 1-5 -10 13 = 1-3 -6 8) (6)=–3 + 8 = 5 P(x2 1 X, + X3 ) = P(x2, X, +X3) P(x, + X) 12,42 - (22, *,**) 31(X2, X+23)" (27) 2/2 e woman et (1942-012 +(3), El = 45-36 = 9 ) 25-26 (84) 124)

cohen nit23=0 (920) ET (*:-) = 5() (%) (2) 5 ( 522 622)() So, P (X2)X1 + X3=0) = - Jo, Ply,yox+850) - 956) av. w..5(37) 226-(197) ) oketat S

Sur (osv.]otend to
we the PDF for multivariate normal for
Yo Nη (μ, Σ) is f(X) = -ε (-μ)*Σ-1(y-μ) 2πΙΣ 1.
.

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1° dx = V27. we know 11)= VT

The derivation is as follows.