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I have obtained a pdf for

                  G= variable1 -variable2 

The pdf for this comes out to be a triangle, which is given by

                  fG(g)=g+1              for   0

Now , I want to find out the pdf for G^2,So I have to find out

                       Y=G^2 

which can be calculated using this formula

$g(y)=G'(y)=\frac{1}{2\sqrt{y}}(f(\sqrt{y})+f(-\sqrt{y}))$

https://math.stackexchange.com/a/263943/53984 But the main problem is when I put g+1 in the above formula for the specific range from 0 to 1 .The pdf is somehow not coming correct.Please help me out in this problem.

2 Answers 2

1

If you have $g(y) = \frac{1}{2\sqrt{y}}(f(\sqrt{y})+f(-\sqrt{y}))$ for $0 \le y \le 1$ then you can use your earlier expression for $f(x)$ to produce

$g(y) = \frac{1}{2\sqrt{y}}((1-\sqrt{y})+(1-\sqrt{y})) = \frac{1-\sqrt{y}}{\sqrt{y}}= \frac{1}{\sqrt{y}}-1.$

  • 1
    When $0 \le y \le 1$ you have $-1 \le -\sqrt{y}\le 0$. You know that the original density for $-1 \le x \le 0$ is $f(x)=1+x$ so letting $x=-\sqrt{y}$ that becomes $f(-\sqrt{y})= 1 +(-\sqrt{y})$.2012-12-23
1

You had reached the stage $g(y)=G'(y)=\frac{1}{2\sqrt{y}}(f_G(\sqrt{y})+f_G(-\sqrt{y})).$ So all that needs to be done is to evaluate (i) $f_G(\sqrt{y})$ and (ii) $f_G(-\sqrt{y})$.

For (i), $\sqrt{y}$ is non-negative, so we look up the formula for $f_G(g)$ when $g\ge 0$, and substitute $\sqrt{y}$ wherever we see $g$. We get $1-\sqrt{y}$.

For (ii), note that $-\sqrt{y}$ is $\le 0$. So we look up the formula for $f_G(g0$ in that case. For $-1\le g\le 0$ we have $f_G(b)=g+1$. Substitute $-\sqrt{y}$ for $g$. We get $-\sqrt{y}+1$. Finally, add and simplify.

Remark: We can use the geometry to avoid the unpleasantness of negative values of $G$. The density function of $G$ is symmetric. Let $0\le y\le 1$. We have $Y\le y$ if and only if $|G|\le \sqrt{y}$. By symmetry, $\Pr(|G|\le \sqrt{y})=2\Pr(0\le G\le \sqrt{y}).$ Now we can work purely with non-negative values of $G$.