hw1_solution

# hw1_solution - 131B HW#1 solution 3.7 Multivariate...

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131B HW#1 solution 3.7 Multivariate Distributions 1. (a) We have 1 0 1 0 1 0 f ( x 1 , x 2 , x 3 ) dx 1 dx 2 dx 3 = 3 c Since the value of this integral must be equal to 1, it follows that c = 1 / 3. (b) For 0 x 1 1 and 0 x 3 1, f 13 ( x 1 , x 3 ) = 1 0 f ( x 1 , x 2 , x 3 ) dx 2 = 1 3 ( x 1 + 1 + 3 x 3 ) . (c) The marginal joint p.d.f of X 1 and X 2 is f 12 ( x 1 , x 2 ) = 1 0 f ( x 1 , x 2 , x 3 ) dx 3 = 1 3 ( x 1 + 2 x 2 + 3 2 ) . The conditional p.d.f. of X 3 given that X 1 = x 1 and X 2 = x 2 is g 3 ( x 3 | x 1 , x 2 ) = f ( x 1 , x 2 , x 3 ) f 12 ( x 1 , x 2 ) = x 1 + 2 x 2 + 3 x 3 x 1 + 2 x 2 + 3 / 2 Therefore, P ( X 3 < 1 2 | X 1 = 1 4 , X 2 = 3 4 ) = 1 / 2 0 g 3 ( x 3 | x 1 = 1 4 , x 2 = 3 4 ) dx 3 = 1 / 2 0 ( 7 13 + 12 13 x 3 ) dx 3 = 5 13 . 8. For any given value x of X , the random variables Y 1 , . . . , Y n are i.i.d., each with the p.d.f. g ( y | x ). Therefore, the conditional joint p.d.f. of Y 1 , . . . , Y n given that X = x is h ( y 1 , . . . , y n | x ) = g ( y 1 | x ) · · · g ( y n | x ) = { 1 x n for 0 < y i < x ( i = 1 , . . . , n ) , 0 otherwise . 1

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This joint p.d.f. is positive if and only if each y i > 0 and x is greater than every y i . In other words, x must be greater than m = max { y 1 , . . . , y n } . (a) For y i > 0( i = 1 , . . . , n ), the marginal joint p.d.f. of Y 1 , . . . , Y n is g 0 ( y 1 , . . . , y n ) = -∞ f ( x ) h ( y 1 , . . . , y n | x ) dx = m 1 n ! exp( - x ) dx = 1 n ! exp( - m ) . (b) For y i > 0( i = 1 , . . . , n ), the conditional joint p.d.f. of X given that Y i = y i ( i = 1 , . . . , n ) is g 1 ( x | y 1 , . . . , y n ) = f ( x ) h ( y 1 , . . . , y n | x ) g 0 ( y 1 , . . . , y n ) = { exp( - ( x - m )) for x > m, 0 otherwise . 3.9 Functions of Two or More Random Variables 6. By Eq. (3.9.2) (with a change in notation), g ( z ) = -∞ f ( z - t, t ) dt for - ∞ < z < However, the integrand is positive only for 0 z - t t 1. Therefore, for 0 z 1, it is positive only for z/ 2 t z and we have g ( z ) = z z/ 2 2 zdt = z 2 . For 1 < z < 2, the integrand is positive only for z/ 2 t 1 and we have g ( z ) = 1 z/ 2 2 zdt = z (2 - z ) . 18. We need to transform ( X, Y ) to ( Z, W ), where Z = X/Y and W = Y . The joint p.d.f. of ( X, Y ) is f ( x, y ) = g 1 ( x | y ) f 2 ( y ) = { 3 x 2 f 2 ( y ) /y 3 if 0 < x < y, 0 otherwise . The inverse transformation is x = zw and y = w . The jacobian is J = det ( w z 0 1 ) = w. The joint p.d.f. of ( Z, W ) is g ( z, w ) = f ( zw, w ) w = 3 z 2 w 2 f 2 ( w ) w/w 3 = 3 z 2 f 2 ( w ) , for 0 < z < 1 . This is clearly factored in the appropriate way to show that Z and W are independent. Indeed, if we integrate g ( z, w ) over w , we obtain the marginal p.d.f. of Z , namely g 1 ( z ) = 3 z 2 , for 0 < z < 1.
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