# Derive the PDF of Y from the Rayleigh distribution

(i) With the CDF of $$Y=e^r$$ and $$r\sim N(0,1)$$ ( $$\Phi_Y$$ and $$\Phi_r$$), I want to find a relation to obtain the PDF for $$Y=e^r$$. I believe the PDF will be something like the Log-Normal distribution.

Firstly, suppose $$u$$, $$v\sim N(0,\frac{1}{2})$$ are independent samples, and setting $$z=u+iv$$ we have $$z\sim N_\mathbb{c}(0,1)$$ . The PDF of $$r\equiv|z|=(u^2+v^2)^{1/2}$$ is

$$\phi_r(r)=2re^{-r^2},$$

Know as the Rayleigh distribution.

Let's Consider the notation $$\Phi'_Y=\phi_Y$$ and $$\Phi'_r=\phi_r$$ for the derivative of the CDF; where $$\Phi_r$$ is the CDF and $$\phi_r$$ is the PDF of the Rayleigh distribution. Note that

$$\Phi_Y(y)\equiv P(Y \leq y)=P(e^r\leq y)=P(r\leq \ln(y))=P(r\leq y)=\Phi_r(\ln(y))$$

To get the PDF $$\phi_Y$$ in terms of the PDF of the Rayleigh distribution $$\phi_r$$, just differentiate $$\Phi_Y(y)=\Phi_r(\ln(y))$$.

$$\Phi'_Y(y)=\phi_Y(y)=\Phi'_r(\ln(y))=\phi_r(\ln(y))\frac{1}{y}$$

$$\phi_Y(y)=\phi_r(\ln(y))\frac{1}{y}$$

And them we get

$$\phi_Y(y)=\frac{2\ln(y)}{y}e^{-\ln(y)^2}$$

I don't know if my results are correct or if it is possible to obtain what I want in the way I am trying to do. So any insight or idea is very welcome.

• What is meant by $N_c(0,1)$? – Math1000 Feb 16 at 20:42
• That $z$ is a random variable from a complex normal distribution with mean zero and variance 1. – Marcos Benício Feb 16 at 20:46