Suppose X1 X2 has the joint pdf Let Y1 3X1 and Y2 2X2 X1

Suppose (X_1, X_2) has the joint p.d.f. Let Y_1 = 3X_1 and Y_2 = 2X_2 - X_1. Sketch the support of (Y_1, Y_2) on the plane. Find the joint p.d.f. of (Y_1, Y_2). Are X_1 and X_2 independent? Warning: carefully check the shape of the support. Independence means that the joint p. d. f. equals the product of the two marginals.

Solution

i) I am trying again and again graph is not uploading up. I\'l try it.

ii)

Y1=3X1 => X1=Y1/2
Y2=2X2-X1 => X2 = (Y2 + X1 )/2 = Y1/4 + Y2/2

Both infers:

J =
|1/2 0 |
|1/4 1/2|
= 1/4

Since this is a one to one transformation , so the joint pdf of y1 and y2 is
f_y1,y2(y1,y2) = h(g^-1(y1,y1))|J|
= 2 exp (-y1-y2)*1/4
= exp (-y1-y2)/2 Answer

iii)

f_x1(X1) = integral ( 2e^(-x1-x2)) dx2

= integral ( 2e^-x1 e^-x2)) dx2
= 2e^-x1 integral e^-x2 dx2
= -2e^(-x1-x2)

f_x2(X2) = integral ( 2e^(-x1-x2)) dx1
= integral ( 2e^-x1 e^-x2)) dx1
= 2e^-x2 integral e^-x1 dx1
= -2e^(-x1-x2)

f_x1(X1)*_fx2(X2) = -2e^(-x1-x2) * -2e^(-x1-x2) = 4e^2(-x1-x2)
which is not equal to pdf ,
Hence,
Not independent.