Let $T$ have the exponential distribution with rate $\lambda = 1/2$. Let $R = \sqrt{T}$. Find the density of $R$.
I recognize this as the Rayleigh density. However, I am not sure how to derive it. Using the information given, I set up my problem as:
$f_T(t) = \lambda \cdot e^{-\lambda t}$, where $f_T(t)$ is the cdf of $T$
By substituting in the information, $\sqrt(R)$ is then $\sqrt(\frac{1}{2} \cdot e^{-\frac{1}{2} t})$. After this I do not know how to continue
The distribution function of $R=\sqrt{T}$ assuming that $T\sim \text{exp}\left(\frac12\right)$:
$$F_R(t)=P(\sqrt T<t)=P(T<t^2)=\color{blue}{1-e^{-\frac12t^2}}, \,\, 0\le t.$$
So, $$f_R(t)=\frac{d}{dt}F_T(t)=\color{blue}{te^{-\frac12t^2}}\,\text{ if } 0\le t, 0 \, \text{ otherwise}.$$
Yes, these are the cdf and the pdf of the Rayleigh distribution with $\sigma^2=1$.