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:

$$\phi (x) = \frac{1}{\sqrt{2 \pi}}\int e^{\frac{-t^{2}}{2}}dt$$


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?


1 Answer 1


Note that you have the pdf of the standard normal not quite right, or at least not consistently right. The constant in front should be $\frac{1}{\sqrt{2\pi}}$. So the constant in front of the joint density is $\frac{1}{2\pi}$.

Let random variable $R$ denote the distance to the origin. Then $R=\sqrt{X^2+Y^2}$. We want the cumulative distribution function (cdf) of $R$, so we want $P(R\le r)$. Clearly this is $0$ for $r \lt 0$. So from now on assume $r\ge 0$. We have $$P(R \le r)=\iint_{D_r} \frac{1}{2\pi}e^{-(x^2+y^2)/2}dx\,dy,$$ where $D_r$ is the disk of radius $r$, centre the origin. We need to evaluate this integral.

To do this, change to polar coordinates. (If that has been forgotten, look back to your several variable calculus course.) Because the letter $r$ is already taken, I will use $t$ for the polar coordinate distance to the origin. Let $x=t\cos\theta$, $y=t\sin\theta$. The integral becomes $$\int_0^{2\pi} \int_0^r \frac{1}{2\pi}e^{-t^2/2}t\,dt\, d\theta.$$ The integration is straightforward. The inner integral is $\frac{1}{2\pi}\left(1-e^{-r^2/2}\right)$, and for the outer integral you just multiply by $2\pi$, getting $1-e^{-r^2/2}$.

Now we have the cumulative distribution function of $R$. The rest follows without much trouble. For (c), differentiate. For (d), if the cdf is $F_R(r)$, we want $F_R(3)-F_R(2)$.

  • $\begingroup$ So I think I'm starting to grasp this... We're looking for W (or our target radius) such that W is less than w(or the radius of the larger disk), because, in this way, we can ensure that our shot hits inside the disk, correct? Therefore, the double integral serves to get the area of the disk(outer integral), and then the area in which the shot landed inside the disk(inner integral). Is this a correct assumption? $\endgroup$
    – gfppaste
    May 3, 2012 at 3:50
  • $\begingroup$ It takes a while, the stuff is not easy, you have to get fully comfortable with the notion of random variable. And then there are little technical tricks needed, like polar coordinates here. By the way, the answer to (d) would be the integral over the annulus (ring) going from $2$ to $3$, though I get to it a slightly different way later. The real point is that we want the cumulative distribution function of $W$. I called it $W$ not $R$, because wanted to save $r$ for polar coordinates. Maybe should not have. Remember, caps for random variables, always. $\endgroup$ May 3, 2012 at 3:54
  • $\begingroup$ Definitely should use $R$. $\endgroup$
    – Did
    May 3, 2012 at 5:24
  • $\begingroup$ @Didier: Was torn. Use $r$ for both? Poor hygiene. Use $R$ and $r$? But polar had $r$ first. However, have changed, maybe leaving a few typos behind. $\endgroup$ May 3, 2012 at 5:44

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