516hw12 - Section 5.4 1. a) r , 1 2 X2 1 r1, 1 1 dx if 0...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 5.4 1. a) r , 1 2 X2 1 r1, 1 1 dx if 0 < z < 1 2 0 z 0 b) f X1 + X 2 ( z ) = f X1 ( x) f X 2 ( z - x)dx = 1 1 x<2 2 dx = 0< x<1;zx< z 2 dx = 0 < x <1;0 < z - - 2< dx 2 0 1 1 1 if 1 < z < 2 1 dx if 2 < z < 3 2 z-2 Xz if 0 < z < 1 2 1 = if 1 < z < 2 2 3- z if 2 < z < 3 2 z c) FX1 + X 2 ( z ) = - f X1 + X 2 ( x )dx 0 if z < 0 z z2 if 0 < z < 1 0 4 1 z x 1 1 z -1 = dx + dx if 1 < z < 2 =+ if 1 < z < 2 12 4 2 0 2 1 2 z x (3 - z ) 2 1 3- x 1 if 2 < z < 3 dx + dx + dx if 2 < z < 3 - 12 2 2 4 2 0 1 if z > 3 1 if z > 3 2. a) 0.875 x dx if 0 < z < 1 2 0 if z < 0 1 X2 1/2 0 r1, 1 b) By symmetry, this probability would be . c) Use the formula for the density of S3 derived in this chapter. t2 3 P ( S3 ( 1.1) = dt + t 2 + 3t - = 0.2213 - dt 2 2 0 1 Z d) P (1.0 ( S3 1.001) f S3 (1) 0.001 = 0.0005 1 1.1 5. a) It is clear that the range of Z = 10 X + 10Y is [10, 70]. Take any point z in that range. Then there is a unique integer x between 1 and 6 and a unique number in the interval [0, 1) such that z = 10 x + 10 y . So, P ( Z ( z ) = P (10 X + 10Y 10 x + 10 y ) = P( X + Y x + y ) = P( X = 1) + ... + P ( X = x - 1) + P ( X = x, 0 < Y < y ) x - 1 y x + y - 1 z - 10 = + = = 6 6 6 60 This gives the uniform distribution on [10, 70]. b) 0.4833 6. Use induction and the fact that if X 1 follows Gamma( r , ) and X 2 follows Gamma( s, ) then X 1 + X 2 follows Gamma( r + s, ). To show: For any non-negative integer n, if X 1 ~ Gamma( r1 , ),..., X n ~ Gamma( rn , ) and are independent, then X 1 + ... + X n ~ Gamma(r1 + ... + rn , ) . The result holds for n = 2. Assume the result holds for n = m, then will show that the result holds for n = m + 1. Note that this proves the assertion for all n. For n = m + 1, we have , X 1 ~ Gamma( r1 , ),..., X m ~ Gamma( rm , ), X m +1 ~ Gamma( rm +1 , ) and are independent. This means that Y = X 1 + ... + X m ~ Gamma(r = r1 + ... + rm , ) , by the assumption that the result holds for n = m. Hence, X 1 + ... + X m + X m +1 = Y + X m +1 ~ Gamma(r + rm +1 , ) Gamma( r1 + ... + rm + rm +1 , ) 9. FX ( x) = P ( X x ) = P(UV x) = area shaded region x = 1.x + du = x - x log x total area u x 1 Hence the density is: dFX ( x) = - log x dx 1 0 x 1 E 5.4.9. a) T = U , X = UV U = T , V = X / T . Hence, 0 X T 0 X X T 1. b) U = T , V = T / X . 1, 0 X /T 1 , i.e., X U T Jacobian: V T U X V X 1 = X - 2 T 0 , 1 determinant = 1/ T . Hence the joint density is: T 1 t for 0 x t 1 fT , X (t , x) = fU ,V (t , x / t ) = X 1 t 1 T ,X for 0 X x t 1 . c) For 0 < x < 1, f X ( x) = P 13. FZ ( z ) =S ( Z = Now, P ( X Y - - f 1 (t , x)dt = = - log x . dt t x z ) = P( X Y + z) y + z )dy z ) = P( X - Y f ( y ) P( X y z) + = 0 1- e y + z )dy = - y P ( X e 0 if y + z - ( y+ z ) 0 if y + z > 0 Hence, if z e 0, then 1 FZ ( z ) = - y ( 1 - e - ( y + z ) ) dy = - y dy -e - z -2 y dy = 1 - e - z e e e 2 0 0 0 And if z < 0, then FZ ( z ) = -z e ( 1 - e - y - ( y+ z ) ) dy = e -z - y dy -e - z -2 y dy = e z - e - z e -z 1 2 z 1 z e = e 2 2 So differentiating we get the density to be: E 5.4.13. a) T = X + Y , Z = X - Y X = - | z| e . 2 T +Z T -Z ,Y = . Hence the possible values 2 2 are T + Z > 0 and T - Z > 0 , i.e., | Z |< T . b) X = T +Z T -Z ,Y = 2 2 T 1 Y = 2 Z 1 Y 2 1 2 So the determinant of the jacobian is -1/2. . 1 - 2 X T X Hence Jacobian = Z X Hence the joint density is, t 1 + z t - z fT , Z (t , z ) = f X ,Y , for 0 | z<| t 2 2 2 1 for 0 X| z |< t = 2 e - t 2 c) The marginal of Z is: fZ (z) = 1 2 - t - | z| fT ,Z (t , z)dt = | e dt = 2 e . 2 - z| 1 ...
View Full Document

This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at Berkeley.

Ask a homework question - tutors are online