2
$\begingroup$

Let $\vec{X}=(X_1,X_2)^T$ (no info if dependent) have a given probability density function $f_{\vec{X}}(x_1,x_2)=\begin{cases} 6x_1, \hspace{2mm} x_1+x_2<1, \hspace{2mm} x_1,x_2>0,\\ 0 \hspace{12mm} \text{elsewhere} \end{cases} $

and $\vec{Y}=(X_1+X_2,X_1-X_2)^T$. Find $f_\vec{Y}(y_1,y_2)$

I found the marginals: $f_{x_1}(x_1)=\int_0^{1-x_1}6x_1dx_2 = 6x_1-6x_1^2$ and $f_{x_2}(x_2)=\int_0^{1-x_2}6x_1dx_1 = 3(1-2x_2-x_2^2)$

Should I first find $F_{\vec{Y}}(x_1,x_2)=P(X_1+X_2\leq x_1, X_1-X_2\leq x_2)$? And then it can be integrated into a PDF?

So

$$ P(X_1+X_2\leq x_1, X_1-X_2\leq x_2)= P(X_1\leq x_1-X_2, X_1\leq x_2+X_2) $$

but this is where I got stuck. I feel like I should try using CDF definition to write it as an integral:

$$ P(X_1\leq x_1-X_2, X_1\leq x_2+X_2) = \int_0^{x_1-X_2}\int_0^{x_1+X_2}f_{\vec{X}}(u,v)dudv = 3\int_0^{x_1-X_2}(x_1+X_2)^2dv = 3(x_1+X_2)^2(x_1-X_2) $$

But this seems very wrong. Even if the method suits, I think I should somewhere fix $X_2=x_2$?

I also thought of using the Bayes' conditional probability formula, but I cannot see how this would help since we don't know $P(X_1\leq x_1-X_2 | X_1\leq x_2+X_2)$

Thanks!

$\endgroup$

1 Answer 1

2
$\begingroup$

If you are familiar with the Jacobian Transformation method, you can directly find pdf.

$Y_1 = X_1 + X_2, ~Y_2 = X_1 - X_2$ $$\implies X_1 = \frac{Y_1 + Y_2}{2},~ X_2 = \frac{Y_1 - Y_2}{2}$$

$|J| = \frac 12$

$ \displaystyle f_{Y_1 Y_2} (y_1, y_2) = |J|~f_{X_1X_2}(\frac{y_1 + y_2}{2}, \frac{y_1-y_2}{2}) = \frac 32 (y_1 + y_2)$

For support of the joint density function, note that $0 \lt x_1 + x_2 \lt 1$ and $x_1, x_2 \gt 0$,
$~~~~~~~~~~~~~~~~~~\text { }$ So we must have $0 \lt |y_2| \lt y_1 \lt 1$.

If you are first finding CDF,

Given $Y = (X_1 + X_2, X_1 - X_2)$,
$F_{Y_1Y_2}(y_1, y_2) = \displaystyle P(X_1 + X_2 \lt y_1, X_1 - X_2 \lt y_2)$

a) For $y_2 \lt 0$,

$ \displaystyle F_{Y_1Y_2}(y_1, y_2) = \int_0^{(y_1 + y_2)/2} \int_{x_1 - y_2}^{y_1 - x_1} f(x_1, x_2) ~ dx_2 ~dx_1$

$$ = \frac 14 (y_1 + y_2)^3$$

b) For $y_2 \gt 0$,

$ \displaystyle F_{Y_1Y_2}(y_1, y_2) = 1 - \int_0^{(y_1 - y_2)/2} \int_{x_2 + y_2}^{y_1 - x_2} f(x_1, x_2) ~ dx_1 ~dx_2$

$$ = 1 - \frac 34 (y_1 - y_2)^2 (y_1 + y_2)$$

Differentiating wrt $y_1$ and $y_2$, you get the same pdf as we obtained using Jacobian Transformation method.

$\endgroup$
1
  • 1
    $\begingroup$ Wow, I like the application of the Jacobian theorem! Very nice, thank you! $\endgroup$
    – statistic
    Jan 26, 2022 at 14:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .