I'm attemtping to solve this problem:
Suppose a shot is fired at a circular target. The vertical and the horizontal coordinates of the point of impact (taking the center of the target as origin) are independent random variables, each distributed according to the standardized normal distribution.
a. Write down the PDFs of the two coordinates.
b. Write down the joint PDF of the coordinates of the point of impact.
c. What is the PDF of the radius of the point of impact $r = \sqrt{x^2 + y^2}$ ? Show all steps you used to derive this expression.
Hint: you can use the CDF method for finding the PDF of a transformed RV.
d. What is the probability that the point of impact will land in the ring of radii 2-3?
MY attempt at a solution:
a) P.D.F. of x is: $\frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{2}}$ P.D.F. of y is: $\frac{1}{2\sqrt{\pi}}e^{-\frac{y^2}{2}}$
b) joint PDF is: PDF(x) * PDF(y) = $\frac{1}{2\sqrt{\pi}}e^{-\frac{y^2}{2}} * \frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{2}} = 2\pi^{-1}e^{\frac{-(x^2 + y^2)}{2}} $
c) this is where I'm getting hung up on...
So the C.D.F. of a std. normal R.V. $x$ is:
$\phi (x) = \frac{1}{\sqrt{2 \pi}}\int e^{\frac{-t^{2}}{2}}dt$
correct?
since $r = \sqrt{x^2 + y^2}$ is given, then is the CDF of $r$ just $\phi (r) = \frac{1}{\sqrt{2 \pi}}\int e^{\frac{-r^{2}}{2}}dr$ And then would the PDF of the radius just be the first derivative of this?