Ask Uncle Colin: A horrible CDF

Dear Uncle Colin

I have the set-up below - how do I find the cumulative density function for $d^2$?

CDF Puzzle

- Circles Don’t Fit

Hi, CDF, and thanks for your message!

Before I start, I’m going to make two simplifying assumptions that don’t change anything1 - that the radius of the outer circle $R=1$, and that point P lies on the positive $x$-axis.

This problem breaks down nicely if you think about the CDF as the probability a point in the outer circle lies inside a circle centred on P with a given radius, $d_0$. We’ll deal with the squariness shortly.

There are three ways the circles could be arranged:

  • The P-circle could be completely inside the O-circle, if $d_0 \le 1-p$.
  • The P-circle could intersect the O-circle, if $1-p < d_0 < 1+p$
  • The P-circle could be completely outside the O-circle, if $d_0 \ge 1-p$.

Case 1: $d_0 \le (1-p)$

If the P circle is inside the O circle, the probability of Q lying inside the P-circle is the ratio of the circles’ areas: $P(d < d_0) = d_0^2$.

Case 3: $d_0 \ge (1+p)$

If the P circle completely encloses the O circle, Q must lie inside it, so $P(d < d_0) = 1$.

Case 2: $(1-p) \lt d_0 \lt (1+p)$

This is the interesting situation. To work this out, this diagram is helpful: it shows how the (upper halves of the) two circles interact. In particular, the probability we want corresponds to the area in the intersection of the two semicircles - the triangle and the two regions immediately adjacent to it.

The area of these turns out to be the union of two sectors - one centred on O, with radius 1, and one centred on P, with radius $d_0$. The intersection of the two is the marked triangle - which tells us how to find the area!

We can use cosine rule to determine that:

  • $\cos(\theta) = \frac{1+p^2-d_0^2}{2p}$
  • $\cos(\phi) = \frac{p^2 + d_0^2 - 1}{2dp}$

The two sectors have areas:

  • $A_o = \frac{1}{2}\theta$
  • $A_p = \frac{1}{2}d_0^2 \phi$

And the triangle has area $\frac{1}{2}p \sin(\theta)$

Putting those together gives the area of the region as $\frac{1}{2}\left(\theta + d^2 \phi - p\sin(\theta) \right)$.

However, probability-wise, we want the relative area of this compared to the O semicircle, so we need to divide by $\piby 2$.

That gives $P(d < d_0) = \frac{1}{\pi}\left( \theta + d_0^2 \phi - p\sin(\theta)\right)$.

Here is the curve in desmos.

But what about the CDF of $d^2$?

We know the probability $P(d < d_0) = f(d_0)$ for any value of $d_0$. We want $P(d^2 < k)$ for any value of $k$, and since $d\ge 0$, that’s the same as $P(d < \sqrt{k})$.

In other words, $P(d^2 < k) = f(\sqrt{k})$.

Since $f(d_0) = \dots$

  • $\dots d_0^2$ for $0 < d_0 < 1-p$,
  • $\dots \frac{1}{\pi} \left( \theta + d_0^2 \phi - p\sin(\theta)\right)$ for $1-p < d_0 < 1+p$
  • $\dots 1$ for $1+p < d_0$

… we have $P(d^2 < k) = \dots$

  • $\dots k$ for $0 < k < (1-p)^2$
  • $\dots \frac{1}{\pi} \left( \theta + k \phi - p \sin(\theta)\right)$ for $(1-p)^2 < k < (1+p)^2$
  • $\dots 1$ for $(1+p)^2 < k$

… where $\theta = \cos^{-1}\left(\frac{ 1 + p^2 -k }{2p}\right)$ and $\phi=\cos^{-1}\left(\frac{k + p^2 - 1}{2p\sqrt{k}}\right)$.

At least, that’s what I think I get. Frankly, my eyes have gone wibbly.

Hope that helps!

- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. which is what we mean when we say “without loss of generality”, or “wlog” []

Share

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter