joint density function of two independent random variables

Question

Suppose that $𝑋_1$ and $𝑋_2$ are independent and follow a uniform distribution over $[0, 1]$. Let $𝑌_1 = 𝑋_1 + 𝑋_2$, and $𝑌_2 = 𝑋_2 − 𝑋_1$.

a) Find the joint pdf $𝑓_{𝑌_1,𝑌_2} (𝑦_1, 𝑦_2)$ of $𝑌_1$ and $𝑌_2$.

b) Sketch the region 𝐷 = {(𝑦1, 𝑦2)} for $𝑓_{𝑌_1,𝑌_2}(𝑦_1, 𝑦_2) > 0$.

I just find that $$f(Y_1) = \begin{cases} y_1 & \text{for $0 < y_1 < 1$} \\ 2-y_1 & \text{for $1 \le y_1 < 2$} \\ 0 & \text{otherwise.} \end{cases}$$

$$f(Y_2) = \begin{cases} y_2+1 & \text{for $-1 < y_2 < 0$} \\ 1-y_2 & \text{for $0 \le y_2 < 1$} \\ 0 & \text{otherwise.} \end{cases}$$

how about the next steps?

Answer 1

$$f_{Y_1,Y_2}(y_1,y_2)=\frac{1}{2}\mathbf{1}_{D}(y_1,y_2)$$ where $D$ is the square delimited by the points $(0,0),(1,1),(0,2),(-1,-1)$, which is the image of the unit square by the linear transformation $\phi(x_1,x_2)=\begin{pmatrix}x_1+x_2\\ x_2-x1\end{pmatrix}$.

To see this, let $h$ a non-negative measurable function. Then you can compute $E(h(Y_1,Y_2))$ using the substitution $\phi$. You'll see that the $\frac{1}{2}$ is the absolute value of the jacobian determinant of $\phi^{-1}$. You should obtain $$E[h(Y_1,Y_2)]=\frac{1}{2}\int\int h(y_1,y_2)\mathbf{1}_D(y_1,y_2)dy_1dy_2$$ which proves the result.

Answer 2

Because $~Y_1=X_1+X_2~,~ Y_2=X_2-X_1$

Then $~~~~~X_1=\dfrac{Y_1-Y_2}2~,~ X_2=\dfrac{Y_2+Y_1}2$

Therefore the Jacobian change of variable transformation is: $$\begin{align}f_{Y_1, Y_2}(y_1, y_2) ~=~& f_{X_1}\Big(\frac{y_1-y_2}2\Big)~f_{X_2}\Big(\frac{y_1+y_2}2\Big)\left\lVert\dfrac{\partial (\frac{y_1-y_2}2,\frac{y_1+y_2}2)}{\partial (y_1,y_2)}\right\rVert\\ ~=~& \boxed? \cdot\mathbf 1_{(y_1,y_2)\in\mathcal D}\end{align}$$

The support domain $\mathcal D$ is the transformation of the support of $X_1,X_2$, which is the square $\Box(0,0)(0,1)(1,1)(1,0)$.

$$\begin{align}\mathcal D~=~&\Diamond(0+0,0-0)(0+1,1-0)(1+1,1-1)(1+0,0-1)\\~=~&\Diamond(0,0)(1,1)(2,0)(1,{-1})\\~=~&\{(y_1,y_2)\in \boxed{?}\}\end{align}$$

Can you complete?

Answer 3

Hint: The joint density of $(X_1, X_2)$ is easy to compute (product of marginal densities). Write the random vector $(Y_1,Y_2) = A (X_1,X_2)$ where $A$ is the matrix

$$ \begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{array} $$

Do you know how to compute the "transformed" joint density?