Question

Let X1 and X2 be uniformaly distributed over the interval zero to one, (a) What is the joint probability density function of Y1 = X1+X2 and Y2 = X2-X1? (b) Are Y1 and Y2 independent? Explain.

Answer 1

Given X and Y are uniformly distributed between 0 and 1, so

f(3) = 1:0 <I< 1,0; otherwise.if we consider X1 and Y1 jointly uni- formly distributed thanf(11,12) = 1; 0 < (11,12) <10; otherwise

So we use Jacobian to find the PDF for  
(Y,Y)
from the PDF of
(x1, X₂)
. The solution is as follows
let y = x1 + x2 and Y2 2 X2-XI, ow, these two funtions are one-to-one and continuons and denivative (pantial) enest and also continuon. So we can use Jacobian to find the. paf for Y, and Y2 7 = x1 + x2 - () 2 2 X2-41 - (11) (1) + (11) = x2 2 11tY2 (i) - (ii) 2x1 = 1,-Y2 = x = Yi-Y2 2 x 2 = +42 2 2 2 Now, pt I NA 2x1 DY 272 2x2 ut 2421 olx, Li o < Y-Y₂ <1 7 04 Y, -Y ₂ C2 – (W) 2 - (ing o <X2 <1 =) ol 11 +72 4 = 0 < 7,4Y2 42 So, region defined by th) and (1y & given by

ang 592 9,-4,22 0) ( 9+4 = 2 ,- - 2 h (a,02) 9 ₂ (ag, 2) 242 hi (4, 9) = x1 halb,,Y₂) = x2 fly, I₂) = f (h. 19, 2), 5 (81, 8)) 151. dy, dyz = f (sity, S ) 11 dy, dyz So, - 1 tl sy, dyz = . هله ( y (3) = 1 h 13, 42) dyz Now from me diagram for & c(0,1), we have any <% ZY for y, E (1,2) we have 9 -2 LY 2C-9, +2 S. 14,16) oly, Li I hl992) dyz -4, , sody? I -942 h (9, 1921dy 9-2 124, 22

fy, (3,) = 1 1 (4) dy Oy ocy, ci y + 2 V (1/2) dyz, 849,22 9,2 ocycl { [921 of 5,4 1 - +2 lay, L2 2192), -2 2 x+24) ocy, ci cy, la g 2-y, 09, icy, c2 So 30, s. 16) , u , ocy, 21 2-9, 929, 12 Nom from diagram for IE(0,1), will have, IL 9, < 2-72 92 € (-1,0) we have, - 42 <Y, 2+2 ty2 (4) 2 ( h 13, 142) dy, 2 -42 OLY₂21 fy2(92) = 1 s hly, 2) dy 2 +42 -149240 I h (319) dy,

2-42 04 3 day, occi +7, (3) Ocy Y2 2+42 LY 20 2-42 OLY₂ 21 - 3125, 12, ocyan yz 2742 - 24,20 { [y]2492 -129240 02 - 12-92-4] , 02424/ + 244 +2] – 49221 = ) ! - Y2 , 0492<! 1+ , - <4.40 So, hp 12) f (81) f(Y2) 1+4 -12920 &, muy are not indepeedas. fy, (8,y, oy,<! si f(82) y oyo olyan I-4₂ 2-91 129.22

h(y1, y2) = 5 + fy(y)fY,(92),so Y and Y2 are not independent, if they were independent then it should satisfy h(y1, y2) = fy, (y)fy, (y2).