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(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.

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

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