Assume $(X_1,X_2)$ has joint density $f_X$. Determine the joint density of $X_1+X_2,X_1-X_2$, then using this find the density of $X_1+X_2$
I have a theorem which says I can determine the joint density is $f_Y(y)=f_X(h^{-1}(y))\vert\det J_{h^{-1}}(y)\vert$ if $h$ satisfies these properties:
$S\subseteq \mathbb{R}^n$ and $P(X\in S)=1$
- $h\to h(S)$ is one to one.
- $h$ has continuous partial derivatives
- the jacobian matrix $J_{h}(x)$ has non zero determinant in $S$.
In this case my $h(X_1,X_2)=(X_1+X_2,X_1-X_2)$
For 1. if $(a+b,a-b)=(c+d,c-d)$ I can conclude $(a=c+d-b)$ and $a=c-d+b$ and then using that get $b=d$ and $a=c$ so its injective.
For 2. I computed the partials as $\partial f_1/\partial x_1=\partial f_1/\partial x_2=\partial f_2\partial x_1=1$ and $\partial f_2/\partial x_2=-1$ So if these are all constant functions they are all continuous.
And for 3. I get this jacobian, $\begin{pmatrix}1&1\\1&-1\\\end{pmatrix}$ which has determinant $2$, so I satisfy all my conditions.
then $f_Y(y)=f_X(h^{-1}(y))\vert\det J_{h^{-1}}(y)\vert$
I assume $\vert det J_{h^{-1}}(y)\vert =1/2$ I'm not sure what $f_X(h^{-1}(y))$ is exactly.
For the last part I know finding the marginal density from a joint density like $f_{X,Y}(x,y)$ you integrate with respect to $y$ on the boundary of $y$, which here I believe means $f_{X_1+X_2}(x)=\int_{-\infty}^\infty f_Y(y) d(x_1-x_2)$? Should I be doing some kind of change of variables?
Best Answer
You are doing the transformation $$ \Bbb R^2 \ni y := \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} 1 &1 \\ 1 &-1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} =: h(x) $$ and so the inverse transformation is $$ \Bbb R^2 \ni x := \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 &1 \\ 1 &-1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} =: h^{-1}(y) $$
Hence $$ f_Y(y_1, y_2) := f_Y(y) = f_X(h^{-1}(y)) \ |\det J_{h^{-1}}(y)| = \frac{1}{2}f_X\big(\frac{1}{2}(y_1 + y_2), \frac{1}{2} (y_1 - y_2)\big) \quad y \in \Bbb R^2 $$ Finally recall that $Y_1 = X_1 + X_4$. So you have to integrate out the other variable $y_2$ from the above pdf: $$ f_{Y_1}(y_1) = \int_{-\infty}^\infty \frac{1}{2}f_X\big(\frac{1}{2}(y_1 + y_2), \frac{1}{2} (y_1 - y_2)\big) \ dy_2 $$