Finding the probably density function of $Z=\sqrt{X^2+Y^2}$ where Y~N(0,1) and X~N(0,1) X and Y are normal random variables that are independant. Finding the probably density function of  $Z=\sqrt{X^2+Y^2}$ where Y~N(0,1) and X~N(0,1).
Attempt:
Let $z \in R$. If $z \lt 0$ then $P(Z\le z)=0$ since $Z=\sqrt{X^2+Y^2}\ge 0$
Let $z\ge0$, then:
$$F_z(z)= P(Z\le z) = P(\sqrt{X^2+Y^2} \le z)= P(\sqrt{X^2+Y^2} \le z)     $$
This is where I'm stuck. I need to use $\phi=\frac12\left[1+erf\frac{x}{\sqrt2}\right] $ but I'm not sure how to continue with the inequality $P(\sqrt{X^2+Y^2} \le z) $ in order to be able to plug it into $\phi$. 
 A: The easy way to do this is to note  for $c \geq 0$, $P(\sqrt{X^2+Y^2} \leq c) = P( X^2+Y^2 \leq c^2) = \iint_{\sqrt{x^2+y^2} \leq c} \frac{1}{2 \pi} e^{-(x^2+y^2)/2} dx dy = \iint_{0 \leq r \leq c, -\pi \leq \theta \leq \pi} \frac{r}{2 \pi} e^{-r^2/2} dr d\theta$ by a transformation to polar coordinates. This integral is easy to evaluate via substitution of $u=r^2$. 
The result will be a rayleigh distribution. 
A: Let $X, Y \overset{ind}{\sim}\mathcal{N}(0, 1)$. Then it is well-known that $X^2 + Y^2 = Z^2 \sim \chi^2_2$. Thus, $Z \sim \left(\chi^2_2\right)^{1/2}$. This is a monotonic transformation of the $\chi^2_2$ random variable.
Let $W \sim \chi^2_2$, so that $Z \sim W^{1/2}$. Then, as you can see here,
$$f_{W}(w) = \dfrac{x^{2/2-1}e^{-x/2}}{2^{2/2}\Gamma\left(\dfrac{2}{2} \right)} = \dfrac{1}{2}e^{-x/2}\text{.}$$
(This is particularly interesting... a $\chi^2_2$ distribution is just an exponential distribution.) Now if $Z \sim W^{1/2}$, $Z^2 \sim W$; hence,
$$f_{Z}(z) = f_{W}(z^2)\left|\dfrac{\text{d}}{\text{d}z}[z^2] \right| = \dfrac{2}{2}ze^{-z^2/2} = ze^{-z^2/2}\text{, } z > 0\text{.}$$
