# Independent, Normally Distributed R.V.

Working on this:

A shot is fired at a circular target. The vertical and the horizontal coordinates of the point of impact (with the origin sitting at the target’s center) are independent and normally distributed with $\nu(0, 1)$. Show that the distance of the point of impact from the center is distributed with PDF $$p(r) = re^{-r^2/2}, r \geq 0.$$ Find the median of this distribution.

So I'm guessing this would be graphed on an X and Y axis. I can intuit that I need to take the integral of the PDF from the lower bound to $m$ (or from $m$ to the upper bound), but I don't know what the normal distribution with $\nu$(0, 1) mean.

Also, how would I show that the point of impact has the desired PDF?

Thank you.

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 They mean normal mean $0$ variance $1$. For showing density is $re^{r^2/2}$, it is handy to have some $2$ variable calculus, polar coordinates. For finding median, only need to find $m$ such that $\int_0^m re^{-r^2/2}\,dr=1/2$. The integration is easy. – André Nicolas May 1 '12 at 0:32

Let $r\ge 0$; put $R = \sqrt{X^2 + Y^2}$, where $X$ and $Y$ are the coordinates of the shot. Then $$P(R\le r) = {1\over 2\pi} \mathop{\int\!\!\int}_{B_r(0)} \exp\{(x^2 + y^2)/2\}\,dx\,dy.$$ Change to polars to get $$P(R\le r) = {1\over 2\pi}\int_0^r \int_0^{2\pi} \exp(r^2/2)r\,d\theta\, dr =\int_0^r r\exp\{r^2/2\}\,dr$$ Differentiate and you are there.
Here's an intuitive approach: On the circle of radius $r$, the value of the joint density is $ce^{-(x^2+y^2} = ce^{-r^2}$. The size of that region where the density has that value is the circumference of the circle, which is $(\text{constant}\cdot r)$ (not the $2$-dimensional size, of course; that is $0$). So the probability in that region is $\text{constant}\cdot re^{-r^2}$.