(Probability/Statistics) Transformation of Bivariate Random Variable

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Homework Statement



Let [itex]X_1, X_2[/itex] have the joint pdf [itex]h(x_1, x_2) = 8x_1x_2, 0<x_1<x_2<1 [/itex], zero elsewhere. Find the joint pdf of [itex]Y_1=X_1/X_2[/itex] and [itex]Y_2=X_2[/itex].

Homework Equations



[tex]p_Y(y_1,y_2)=p_X[w_1(y_1,y_2),w_2(y_1,y_2)][/tex] where [itex]w_i[/itex] is the inverse of [itex]y_1=u_1(x_1,x_2)[/itex]

The Attempt at a Solution


We can get [itex]X_1=Y_1Y_2[/itex] and [itex]X_2=Y_2[/itex]. Naively plugging in [itex]y[/itex], we can get [itex]8y_1y_2^2[/itex]. However this isn't right according to the back of the book.

I thought it might have to do with finding the marginal distributions of [itex]x_1, x_2[/itex], but that doesn't seem to lead me anywhere either. Any thoughts welcome!
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Let [itex]X_1, X_2[/itex] have the joint pdf [itex]h(x_1, x_2) = 8x_1x_2, 0<x_1<x_2<1 [/itex], zero elsewhere. Find the joint pdf of [itex]Y_1=X_1/X_2[/itex] and [itex]Y_2=X_2[/itex].

Homework Equations



[tex]p_Y(y_1,y_2)=p_X[w_1(y_1,y_2),w_2(y_1,y_2)][/tex] where [itex]w_i[/itex] is the inverse of [itex]y_1=u_1(x_1,x_2)[/itex]

The Attempt at a Solution


We can get [itex]X_1=Y_1Y_2[/itex] and [itex]X_2=Y_2[/itex]. Naively plugging in [itex]y[/itex], we can get [itex]8y_1y_2^2[/itex]. However this isn't right according to the back of the book.

I thought it might have to do with finding the marginal distributions of [itex]x_1, x_2[/itex], but that doesn't seem to lead me anywhere either. Any thoughts welcome!
You forgot the Jacobian, necessary to transform dx1*dx2 into h(y1,y2)*dy1*dy2
 
  • #3
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Thanks! That was it.
 

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