2
$\begingroup$

I am reading a research paper and there is a random variable $R$ whose PDF is as below. (note that $\lambda$ is a constant which is a density of PPP)

$$f_{R}(r) = 2\lambda\pi re^{-\lambda\pi r^2} $$

I have to find the pdf of $\gamma$, i.e. $f_\gamma(x)$, such that: $\gamma = R^{\alpha}$.

I know that $\frac{d}{dx}\mathbb{P}[\gamma \leq x] = f_\gamma(x)$

First of all, I find the CDF of $R$ as: $\ F_R(r) = 1 - e^{-\lambda\pi r^2}$

I start as follows:

$$F_\gamma(x) = \mathbb{P}[\gamma \leq x]$$ $$F_\gamma(x) = \mathbb{P}[R^{\alpha} \leq x]$$ $$F_\gamma(x) = \mathbb{P}[R \leq x^{1/\alpha}] = F_R(x^{1/\alpha})$$ $$F_R(x^{1/\alpha}) = 1 - e^{\lambda\pi x^{2/\alpha}}$$ $$f_\gamma(x) = \frac{d}{dx} [1 - e^{\lambda\pi x^{2/\alpha}}]$$ $$f_\gamma(x) = \frac{2}{\alpha}\lambda\pi x^{2/\alpha-1}e^{\lambda\pi x^{2/\alpha}}$$

However, this is not in line with what authors have come across. Which is

$$ f_\gamma(x) = \frac{2\gamma}{\alpha}\lambda\pi x^{2\gamma/\alpha-1} e^{\lambda\pi x^{2\gamma/\alpha}}$$

(I have added the picture as well for the paper) enter image description here

$\endgroup$
4
  • $\begingroup$ It seems that the authors came up with the same solution with yours in the paper... $\endgroup$ Commented Apr 10, 2018 at 3:49
  • $\begingroup$ No, I am missing $\gamma$ in my solution. $\endgroup$
    – SJa
    Commented Apr 10, 2018 at 3:51
  • $\begingroup$ I was solving on paper with this variable and copied the same thing here. You can consider Y = $\gamma$ $\endgroup$
    – SJa
    Commented Apr 10, 2018 at 4:15
  • $\begingroup$ Yes but isn't the paper you referred to is also missing $\gamma$ on the term? $\endgroup$ Commented Apr 10, 2018 at 4:18

1 Answer 1

2
$\begingroup$

\begin{align} f_\gamma(x) & = \frac d {dx} \Pr(\gamma \le x) = \frac d {dx} \Pr(R^\alpha \le x) = \frac d {dx} \Pr(R\le x^{1/\alpha} ) \\[10pt] & = \frac d {dx} \left(1-e^{-\lambda \pi x^{2/\alpha}} \right) = \frac 2 \alpha \lambda \pi x^{(2/\alpha)-1} e^{-\lambda\pi x^{2/\alpha}} \end{align} The authors write $\delta = \dfrac 2 \alpha,$ so this becomes $$ \Large \delta\lambda\pi x^{\delta -1} e^{-\lambda\pi x^\delta}. $$ That is what the authors have.

Your question says the authors wrote something in which $\gamma$ appears on the right side of that last equality. That makes no sense, and it appears nowhere in the page you included in your question.

$\endgroup$
1
  • $\begingroup$ oh....stupid me.... I read the dot sign as the multiplication symbol $\endgroup$
    – SJa
    Commented Apr 10, 2018 at 5:07

You must log in to answer this question.

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