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

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

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