If $X \sim \text{Exp}(1)$, show that $\sqrt X$ follows a Rayleigh distribution. I have a task to show that if $$X  \sim \text{Exp}(1)$$ then $$\sqrt X \sim \mathrm{Rayleigh}\left(\frac{1}{\sqrt 2}\right)$$
How do I start from here?
For $X\sim$ exponential, I know that the pdf is defined as
$$f_X(x) = \lambda e^{-\lambda x}$$
and for Rayleigh
$$f_X(x) = \frac{x}{\lambda ^2} e^{\frac{-(\lambda x)^2}{2}}$$


How do I start?

Notice that $Y = \sqrt X$ is one to one over the possible values of $X$. Hence you can you use the one to one transformation; $X = Y^2$, and so
$$f_Y(y) = \frac{f_X(y^2)}{\left|\frac{dy}{dx}\right|_{x = y^2}}.$$
Otherwise, you can use the cdf method,
$$P(Y\leq y) = P(\sqrt X\leq y) = P(X\leq y^2).$$
