Transformation of an exponential distribution $R^2 \sim \exp(\lambda)$

Question

The distribution is $R^2\sim \exp(\lambda)$ and I want to find the pdf(R). Here is what I did.

Let $$Y=R^2$$ Then $$P(R \le r)=P(\pm \sqrt{Y} \le r)=P(-r\le\sqrt{Y}\le r)$$

At this point, the probability would cancel out. I am not sure if I am on the right track. Any tips would be greatly appreciated!

Answer 1

For $R \ge 0$

$$P(R \le r)=P(\sqrt{Y} \le r)=P(Y \le r^2) = 1 - \exp(-\lambda r^2)$$

See Rayleigh distribution.

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In general,

$$1 - \exp(-\lambda y) = P(Y \le y) = P(R^2 \le y)=P(-\sqrt{y} \le R \le \sqrt{y}) = F_R(\sqrt{y}) - F_R(-\sqrt{y})$$

If R has a pdf, then

$$\int_{-\infty}^{\sqrt{y}} f_R(s) ds - \int_{-\infty}^{\sqrt{-y}} f_R(s) ds$$

Answer 2

Let $R\ge0$. Therefore $$\Pr(R<r){=\Pr(R^2<r^2)\\=1-e^{-\lambda r^2}\\=F_R(r)\to \\f_R(x)=2x\lambda e^{-\lambda x^2}}.$$

Answer 3

Comment: Using R to sample $n=100,000$ observations from $\mathsf{Exp}(rate = .5)$ in order to illustrate @MostafaAyaz's (+1) Rayleigh density function [red curve].

y = rexp(10^5, .5);  r = sqrt(y)
hist(r, prob=T, col="skyblue2", main="Simulated Rayleigh Dist'n; Rate=.5")
curve(2*x*.5*exp(-.5*x^2), add=T, lwd=2, col="red")

Note: The Rayleigh distribution has practical uses. One of them occurs in the 'Box-Muller transformation' to generate standard two normal random variables from two uniform random variables. If interested, you can look at Wikipedia pages on 'Rayleigh distribution' and 'Box-Muller transformation' for more information.