I need help with these two parts of this question, as I'm sure that I did it correctly (the first part), but the second part is alarmingly wrong, since I need to find the integral of the PDF (with limits not being $\infty$ either).
Would appreciate any help, and if anyone can see if I did it correctly (my attempt is at the bottom)


Let $X$ and $Y$ be two random variables and suppose $X_1 = X+Y$ and $X_2 = X-Y$. Given that the joint density function of $(X_1,X_2)$ is given by
$$f_{X_1,X_2}(x_1,x_2)= \frac{1}{2\sqrt{3}\pi}e^{-\frac{1}{2}\left[\frac{(x_1 -4)^2}{3} + (x_2 - 2)^2\right]}, \qquad x_1\in\mathbb{R},x_2\in\mathbb{R}.$$
(i) Compute the joint density function $f_{X,Y}(x,y)$.
(ii) Compute the marginal density function $f_X(x)$.


Given that $X_1 = X+Y, X_2 = X-Y$, then the joint density function of $(X,Y)$ is $$f_{X,Y}(x,y) = |\det(J)|f_{X_1,Y_1}(x_1,x_2) \qquad x_1\in\mathbb{R},x_2\in\mathbb{R}.$$

$$J=\begin{pmatrix}\frac{\partial x_1}{\partial x} & \frac{\partial x_1}{\partial y} \\ \frac{\partial x_2}{\partial x} & \frac{\partial x_2}{\partial x}\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}. $$

Hence, $$f_{X,Y}(x,y) = 2f_{X_1,X_2}(x_1,x_2)= \frac{1}{\sqrt{3}\pi}e^{-\frac{1}{2}\left[\frac{(x+y -4)^2}{3} + (x-y - 2)^2\right]}, \qquad x\in ?,y \in ?.$$
I think the tricky part is to find the bounds for $x,y$.

I think it is: $x\geq 0, -x \leq y \leq x$. But I'm not sure why actually, I tried looking at the mapping: $(x_1,x_2)\mapsto (x,y)$.

(ii) This will be $$f_X(x) = \int_{\text{all} \ y} \frac{1}{\sqrt{3}\pi}e^{-\frac{1}{2}\left[\frac{(x+y -4)^2}{3} + (x-y - 2)^2\right]}dy$$

but the bounds of $y$ aren't infinity for me to use any sort of $PDF$ property....


The bounds are infinity.

$(X_1,X_2)$ ranges over the entire plane. The variable transformation is just a coordinate change where $X$ and $Y$ are coordinates on an axis rotated by 45 degrees. To see this, notice the "X-axis" is given by $Y=0,$ which means $X_2=X_1,$ i.e. the 45 degree line in the $X_1-X_2$ plane. Plot a few more points and you'll see. (However note $(X,Y) = (1,0)$ is not distance $1$ from the origin... the coordinates are also stretched.)

Thus $(X,Y)$ also ranges over the entire plane.

  • $\begingroup$ Thank you, is my working out good until then? And I think I could have noticed that $X_1,X_2$ are independent of each other at first right (to find the marginal densities of $X$)? $\endgroup$ – Natash1 Jun 15 '17 at 10:46
  • 1
    $\begingroup$ @natash1 looks fine though haven't checked in detail. Yes there is a shortcut to doing that integral for finding marginal of $X$. You know it's normal so just compute its mean and variance from what you know about $X_1$ and $X_2$. I'd suggest still doing the integral for practice completing the square and as a check. $\endgroup$ – spaceisdarkgreen Jun 15 '17 at 14:23

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