Answer to Question #320994 in Statistics and Probability for Amoh

Question #320994

 Let X1, X2 have the joint probability density function f(x1, x2) = 2e −(x1+x2) , 0 < x1, x2 < ∞ Let Y1 = X1, Y2 = X2 − X1.

(i) Using the change of variable technique, find the joint probability density function of Y1, Y2

(ii) Find the conditional distribution of Y2 given Y1


1
Expert's answer
2022-04-06T11:43:35-0400

"f\\left( x_1,x_2 \\right) =2e^{-\\left( x_1+x_2 \\right)},0<x_1<x_2<\\infty \\\\i:\\\\\\left( Y_1,Y_2 \\right) =\\left( X_1,X_2-X_1 \\right) \\\\X_1=Y_1\\\\X_2=Y_1+Y_2\\\\\\left| \\frac{\\partial \\left( X_1,X_2 \\right)}{\\partial \\left( Y_1,Y_2 \\right)} \\right|=\\left| \\begin{matrix}\t1&\t\t0\\\\\t1&\t\t1\\\\\\end{matrix} \\right|=1\\\\f_{\\left( Y_1,Y_2 \\right)}\\left( y_1,y_2 \\right) =f_{X_1,X_2}\\left( y_1,y_1+y_2 \\right) \\cdot 1=2e^{-\\left( y_1+\\left( y_1+y_2 \\right) \\right)}I\\left\\{ 0<y_1<y_1+y_2<\\infty \\right\\} =\\\\=2e^{-2y_1}e^{-y_2},y_1>0,y_2>0\\\\ii:\\\\Since\\\\f_{Y_1,Y_2}\\left( y_1,y_2 \\right) =2e^{-2y_1}I\\left\\{ y_1>0 \\right\\} \\cdot e^{-y_2}I\\left\\{ y_2>0 \\right\\} =f_{Y_1}\\left( y_1 \\right) f_{Y_2}\\left( y_2 \\right) ,\\\\the\\,\\,conditional\\,\\,distribution\\,\\,is\\,\\,the\\,\\,distribution\\,\\,of\\,\\,Y_2:\\\\f_{Y_2|Y_1}\\left( y_2|y_1 \\right) =f_{Y_2}\\left( y_2 \\right) =e^{-y_2}I\\left\\{ y_2>0 \\right\\}"


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