Finding pdf of $Y=\sqrt{2X/\lambda}$ when $X\sim \text{Exp}(\lambda)$ Suppose that $X\sim \text{Exp}(\lambda)$ and  $Y=\sqrt{2X/\lambda}$
Find probability density function of $Y$ and hence find $E(Y)$.
The answer I got for my pdf was $ye^{-y^{2}/2}$ for positive $y$ values but I'm not sure if this is correct. 
 A: For $\mathbb{E}[Y]$, you don't need the density. Note that $Y$ is supported on $[0,\infty)$, hence all you need is to compute:
$$
\mathbb{E}[Y] = \int_0^\infty \mathbb{P}(\sqrt{2X/\lambda}>t) \; dt = \int_0^\infty \mathbb{P}(X\geq \lambda t^2/2)\; dt = \int_0^\infty e^{-\lambda^2 t^2/2}\; dt.
$$
using the fact that for any $k$, $\mathbb{P}(X\geq k)=e^{-\lambda k}$, if $X\sim {\rm Exp}(\lambda)$. Now, we compute this integral, using a trick that the integrand resembles the PDF of the Gaussian distribution with mean zero, and an appropriate variance. 
Consider $Z\sim N(0,1/\lambda^2)$. Then, $f_Z(z)= \frac{1}{\sqrt{2\pi/\lambda^2}}\exp (-z^2\lambda^2/2)$. Hence,
$$
\int_{-\infty}^\infty \frac{\lambda}{\sqrt{2\pi}} \exp (-z^2\lambda^2/2)\;dz = 1.
$$
Therefore,
$$
\int_0^\infty e^{-\lambda^2 z^2/2}\;dz = \frac{1}{2}\int_{-\infty}^\infty \exp(-z^2\lambda^2/2)\; dz = \frac{\sqrt{\pi}}{\lambda\sqrt{2}}.
$$
A: Sometimes when people write $X\sim\operatorname{Exp}(\lambda)$ they mean $\Pr(X>x) = e^{-\lambda x}$ for $x\ge0,$ and sometimes they mean $\Pr(X>x) = e^{-x/\lambda}$ for $x\ge0.$ I'm going to assume you intended the latter since that is consistent with your having no $\text{“}\lambda\text{''}$ in the bottom line.
\begin{align}
f_Y(y) & = \frac d {dx} F_Y(y) = \frac d {dx} \Pr(Y\le y) \\[8pt]
& = \frac d {dy} \Pr\left( \sqrt{\frac{2X}\lambda} \le y \right) \\[8pt]
& = \frac d {dy} \Pr\left( X \le \frac{\lambda y^2} 2 \right) \\[8pt]
& = \frac d {dy}\left( 1 - e^{-y^2/2} \right) \\[8pt]
& = y e^{-y^2/2} \quad \text{for } y\ge0.
\end{align}
