0
$\begingroup$

Let $X_1$ and $X_2$ be two independent random variables, each with pdf $f(x) = e^{-x}$, $0

I made the transform and afterwards used the Jacobian and got $1/2$ however the answer to the question is $1/2 e^{-y_2}$. My question is where is the $e^{-y_2}$ coming from? I understand I had to change $e^{-x}$ in terms of $y$. I just don't understand where $-y_2$ is coming from as the transform...not sure if I'm making much sense thanks in advance for the help!

  • 0
    You need to use LaTeX since part of your question (the definition of $Y_1=X_1-X_2$ and $Y_2=X_1+X_2$) is invisible – 2017-02-23
  • 0
    Start with the accurate formula for the joint distribution or f $X_1$ and $X_2$. And you'll find where $y_2$ is. – 2017-02-23

1 Answers 1

0

Because $Y_1=X_1-X_2$ and $Y_2=X_1+X_2$, then $Y_1+Y_2=2X_1$ and $Y_2-Y_1=2X_2$

Thusly, and since $X_1,X_2$ are independent, we have:

$$\begin{align}f_{Y_1,Y_2}(y_1,y_2) &= \left\lVert\frac{\partial(\frac{y_1+y_2}2,\frac{y_2-y_1}2)}{\partial(y_1,y_2)}\right\rVert \cdot f_{X_1,X_2}(\frac{y_1+y_2}2,\frac{y_2-y_1}2)\\[1ex] &= \tfrac 12\cdot f_{X_1}(\frac{y_1+y_2}2)\cdot f_{X_2}(\frac{y_2-y_1}2) \\[1ex]\end{align}$$

Having correctly obtained the Jacobian's absolute determinant you should have reached this point .   The rest is just substitution and algebra.

  • 0
    I am still not getting the correct answer. fx1 is (y1+y2)/2 I get that and fx2 (y2-y1)/2 that's easy to however this e^-y2 to me is still coming out of left field. – 2017-02-23
  • 0
    She never said anything about substituing while we did this in class she said to look at the bounds in order to realize that e^-y2 is what you need however...idk maybe I'm misunderstanding the formula or her notes but here is what she has y2 = X1 + X2 implies 00 implies y1>-y2 and lastly X2=(y2-y1)/2>0 implies Y1 – 2017-02-23
  • 0
    First comment: $f_{X_1}(x)~=~f_{X_2}(x)~=~f(x)~=~e^{-x}\mathbf 1_{0\lt x \lt \infty}$.   So $f_{X_1}((y_1+y_2)/2) ~=~ e^{-(y_1+y_2)/2}\;\mathbf 1_{0\lt (y_1+y_2)/2\lt \infty}$ and similarly for $f_{X_2}((y_2-y_1)/2)$. – 2017-02-23
  • 0
    Second comment: Yes, that is how you determine the support of the joint distribution function: by *substituting* the transformation into the supports for $X_1, X_2$. $$f_{Y_1,Y_2}(y_1, y_2) = \tfrac 12 e^{-y_2}\;\mathbf 1_{0< y_2<\infty, -y_2< y_1< y_2}$$ – 2017-02-23
  • 0
    OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO wow...I'm an idiot..thank you so much..haha that was...really dumb of me. – 2017-02-23