Factorial moments of the Binomial and related distributions

Factorial moments are defined as the expected value of the falling factorial of a random variable. In this post, we are going to try to compute the (scaled) Factorial moments of certain distributions without a direct computation.

The Factorial moment of a random variable $X$ is given by

$$\mathbb{E}[(X)_r]=\mathbb{E}[X(X-1)(X-2)\cdots (X-r+1)]$$

For our purposes, we focussing on the following definition which differs from the above only by a constant factor.

$$\displaystyle\mathbb{E}\left[\binom{X}{r}\right]=\mathbb{E}\left[\frac{(X)_r}{r!}\right]$$

Let's start with the famous Binomial distribution. The Binomial random variable $X$, with parameters $n$ and $p$, is the number of successes in a sequence of $n$ independent draws (with replacement), each with a success probability $p$.

Taking a cue from Problem of 7 of Strategic Practice 8 of Stat 110, $\binom{X}{k}$ denotes the set of draws where all the draws in the set result in a success. Creating an Indicator random variable for each set and using the linearity of expectations, we have

$$\displaystyle\mathbb{E}\left[\binom{X}{k}\right]=\binom{n}{k}\cdot p^k$$

We now move on to the Hypergeometric distribution (which is exactly the Stat 110 problem quoted above). Let $X$ denote the number of white balls from an Urn containing $N$ balls of which $K$ are white in a sample of $n$ draws. This is exactly the same problem given in Stat 110 quoted above.

$$\displaystyle\mathbb{E}\left[\binom{X}{k}\right]=\binom{n}{k}\cdot \frac{\binom{K}{k}}{\binom{N}{k}}$$

That is, we've considered an Indicator random variable for each $k$-set of draws and used the Linearity of Expectation. Note that $X$ is the number of successes where a 'success' is viewed as draw that gives a white ball.

Alternately, we can view a 'success' as a white ball that gets drawn. This way, we can solve the problem by considering an Indicator random variable for each $k$-set of white balls. Then,

$$\displaystyle\mathbb{E}\left[\binom{X}{k}\right]=\binom{K}{k}\cdot \frac{\binom{k}{k}\binom{N-k}{n-k}}{\binom{N}{n}}$$

Again, using the Linearity of Expectation, we restrict our attention to only a particular $k$-set of white balls. The probability that we are interested is that this particular set gets drawn when drawing $n$ balls. This is again a Hypergeometric distribution where only this particular set constitute 'successes' and rest are considered failures.

We now get to the Negative Binomial distribution with parameter $p$ and $j$ where $X$ denotes the number of failures we encounter before the $r$'th success. It can be verified with direct calculation that

$$\displaystyle\mathbb{E}\left[\binom{X}{k}\right]=\binom{k+r-1}{k}\left(\frac{q}{p}\right)^k$$

While the calculation is itself is not hard, I could not get a nice 'story' proof for the above. Rather we come back to this problem after the next case.

We now move to the Negative Hypergeometric distribution which will be most interesting aspect of this post. Here, again in the context of Urn and balls, $X$ denotes the number of black balls (failures) we get before the $r$'th white ball (success) from a population of $N$ balls in which $K$ are white.

Here again, $\binom{X}{k}$ denotes the $k$-set of draws that are black balls (failures) that gets drawn before the $r$'th white ball (success). Let $I_j,j \in \{1,2,\cdots,N-K\}$ be $1$ if the $j$th black ball (failure) gets drawn before the $r$'th white ball (success) and $0$ otherwise. Let the indicator random variable $I_S$ be the product of all the indicator random variable in the set $S$. Then,

$\displaystyle\mathbb{E}\left[\binom{X}{k}\right] =\sum_{S_k \subset \{1,2,\cdots,N-K\}}\mathbb{E}(I_{S_k})=\binom{N-K}{k}\mathbb{E}(I_1I_2\cdots I_k)=\binom{N-K}{k}\mathbb{P}(E)$

where $S_k$ denotes a $k$-element subset and $E$ denotes the event that failures $1$ to $k$ occur before the $r$'th success.

In other words, the probability that we want is the probability that we draw (without replacement) $k$ specific black balls from a bag containing $K$ white balls and $N-K$ black balls before the $r$'th white ball. This is a nice question by itself.

The key is to realise that we are concerned only about the $k$ specific black balls. This means we can completely ignore the remaining $N-K-k$ black balls and focus only on the $K$ white balls and the specific $k$ black balls.

The probability that we want is then the probability of getting $k$ black balls before the $r$'th white ball from an Urn containing $K+k$ balls of which $K$ are white. But that is exactly Negative Hypergeometric. Therefore,

$\displaystyle \mathbb{E}\left[\binom{X}{k}\right]=\binom{N-K}{k}\binom{k+r-1}{k}\frac{\binom{K+k-k-r}{K-r}}{\binom{K+k}{K}}=\binom{k+r-1}{k}\frac{\binom{N-K}{k}}{\binom{K+k}{k}}$

Alternately, the required probability can be seen as the probability of drawing $k$ black balls and $j-1$ white balls from an Urn containing $K+k$ balls of which $K$ are white. Then,

$\displaystyle \mathbb{E}\left[\binom{X}{k}\right]=\binom{N-K}{k}\frac{\binom{k}{k}\binom{K}{r-1}}{\binom{k+K}{k+r-1}}=\binom{N-K}{k}\frac{\binom{K}{r-1}}{\binom{k+K}{k+r-1}}$

Now we get back to the Negative Binomial case. Though I can't get a nice story proof for it, we can note that the first expression of the Negative Hypergeometric can be alternatively written as

$\displaystyle \mathbb{E}\left[\binom{X}{k}\right]=\binom{k+r-1}{k}\frac{\binom{N}{K+k}}{\binom{N}{K}}$

Using the asymptotic expression we derived in an earlier post, we can see that it matches with the result that we got from a direct calculation.

Hope you enjoyed the discussion. See ya in the next post.

Until then
Yours Aye
Me

The Inverse Sum Theorem

This post talks about the 'Inverse Sum theorem' which is at the heart of the video for my SoME2 submission.

Consider two right angled triangles $OAC$ and $OBC$ having a common side $OC=1$ and right angled at $O$. Let $F$ and $E$ be the foot of altitude of the hypotenuse of the respective triangles. Let the line connecting $E$ and $F$ meet the $x$-axis at $D$. We claim that

$$\frac{1}{OD}=\frac{1}{OA}+\frac{1}{OB}$$

We give three proofs of this result with Coordinate geometry.

Co-ordinate Geometry

Let $\angle OCA=\alpha$. Because $OC=1$, we can immediately see that $OA=\tan \alpha$, $CA=\sec \alpha$ and $OF=\sin \alpha$. As $\angle AOF=\alpha$, the co-ordinates of point $F$ can be seen to be $F \equiv (\sin \alpha \cos \alpha, \sin^2 \alpha)$.

Similarly, if we have $\angle OCB=\beta$, then we have $E \equiv (\sin\beta \cos\beta, \sin^2\beta)$

Then the equation of line $EF$ is

$$\frac{y-\sin^2\beta}{\sin^2\alpha-\sin^2\beta}=\frac{x-\sin\beta\cos\beta}{\sin\alpha\cos\alpha-\sin\beta\cos\beta}$$

$D$ is the point on where this line meets the $x$-axis. Therefore, using $y=0$ in the above equation and solving for $x$ gives us $OD$. After some algebraic manipulations, this gives us 

$$x=\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}$$

Therefore,

$$\displaystyle\frac{1}{OD}=\frac{1}{\tan\alpha}+\frac{1}{\tan\beta}=\frac{1}{OA}+\frac{1}{OB}$$

Trigonometry

Using the generalized Angled Bisector theorem in triangle $OEB$,

$$\frac{OD}{DB}=\frac{OE\sin\angle OED}{BE\sin\angle BED}$$

Using the law of Sines in triangle $OEB$, we get $OE\sin\angle EOB=BE\sin\angle EBO$. Therefore,

$$\frac{OD}{DB}=\frac{\sin\angle EBO}{\sin\angle EOB}\frac{\sin\angle OED}{\sin\angle BED}$$

But $\angle EOB$ and $\angle EBO$ are complementary. So are $\angle BED$ and $\angle OED$. Therefore,

$$\frac{OD}{DB}=\frac{\tan\angle EBO}{\tan\angle BED}=\frac{OC}{OB}\frac{1}{\tan\angle BED}$$

If we now draw a circle with $OC$ as diameter, we know that $E$ and $F$ must lie on this circle because of Thales theorem. Also, $OEF$ and $OCF$ must be equal because they are angles subtended by the chord $OF$. Therefore,

$$\angle BED=90^\circ - \angle OEF=90^\circ-\angle OCF=\angle OAC$$

Therefore,

$$\frac{OD}{DB}=\frac{OC}{OB}\frac{OA}{OC}\implies \frac{OD}{OB-OD}=\frac{OA}{OB}$$

Solving for $OD$, we get the desired result.

Circle Inversion

Using the diagram from above, we see that $OD^2=DF\cdot DE$ because of the tangent-secant theorem. If we imagine a circle $c$ with center $D$ and $OD$ as radius, then the above relation tells us that $E$ and $F$ are inverses of each other w.r.t circle $c$.

Here comes the nice part. Consider two points $U$ and $V$ which are reflections of each other w.r.t a line $l$. Let's say $U$, $V$ and $l$ are all reflected w.r.t a line $m$ to give $U'$, $V'$ and $l'$. Now it should be easy to see that $U'$ and $V'$ are reflections of each other w.r.t line $l'$. The amazing thing here is that this relation holds in case of Circle inversion as well.

Now imagine a circle $c'$ with $C$ as center and $CO$ as radius. Point $O$ is invariant as under an inversion w.r.t $c'$. Points $F$ and $A$ become inverses of each other under $c'$ whereas $E$ and $B$ become inverses. Circle $c$ we described above is invariant as it is orthogonal to Circle $c'$.

We know that $E$ and $F$ are inverses w.r.t $c$. Therefore, $A$ (inverse of $F$ w.r.t $c'$) and $B$ (inverse of $E$ w.r.t $c'$) must be inverses of each other w.r.t $c$ (which is its own inverse under an inversion in $c'$).

The cross ratio $\lambda$ of any four points $A$, $B$, $C$ and $D$ is defined as

$$\lambda=\frac{AB}{AC}\frac{CD}{BD}$$

If one of these points is at infinity, then the terms containing that point disappear from the definition. It is amazing that Cross ratio is invariant under Circle inversion.

The Cross ratio of $O$, $D$, $A$ and $B$ is then

$$\lambda=\frac{OD}{OA}\frac{AB}{DB}$$

The inverses of these points under an inversion w.r.t $c$ become $O$, $\infty$, $B$ and $A$. Therefore, the cross ratio after inversion is

$$\lambda=\frac{BA}{OB}$$

Because cross ratio is invariant under inversion. we have,

$$\frac{OD}{OA}\frac{AB}{DB}=\frac{BA}{OB} \implies \frac{OD}{DB}=\frac{OA}{OB}$$

which is the same relation we had in the previous case.

Interestingly, we also see that $OD^2=DA\cdot DB$ because $A$ and $B$ are inverses of each other w.r.t a circle with radius $OD$.

Until then

Yours Aye

Me

Expected values with Bertrand's Paradox

Bertrand's paradox is a neat problem that shows what happens when we take the words 'at random' a bit too casually. In this post, we will be looking at some problems in the same setup and how the different methods yield answers for the 'seemingly' same question.

Bertrand's paradox asks for the probability of a 'random' chord in a unit circle being longer than the side of an inscribed triangle. We are interested in the expected length of such a chord. We now see how the expected length varies if we choose the chord according to the three methods in the paradox.

(1) Circumference method

In this method, we choose two points uniformly randomly on the circumference of the circle and connect those points to create the chord. After the first point is chosen, the second point is uniformly distributed in the circumference.

With this method, if we draw a tangent at the first point, then the angle between the chord and the tangent is uniform in the range $(0,\pi)$. If we let that angle be $\theta$, then the length of the chord is simply $2\sin\theta$. Therefore,

$\displaystyle\mathbb{E}(L)=\int_0^{\pi/2}2\sin\theta\,\frac{d\theta}{\pi/2}=\frac{4}{\pi}\approx 1.2732$

(2) Radial method

In this method, we first choose a random radius. We then choose a point uniformly on this radius and draw a chord perpendicular to this line. If we let $x$ be the length of the point from the origin, then the length of the chord is $2\sqrt{1-x^2}$.

Because $x$ is uniform in $(0,1)$, we have

$\displaystyle \mathbb{E}(L)=\int_0^12\sqrt{1-x^2}dx=\frac{\pi}{2}\approx1.5708$

But here's something interesting. If we use $x=\cos\theta$ in the above integral, the integral becomes,

$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}2\sin\theta\cdot\sin\theta\,d\theta$

From what we saw in the 'Circumference method', we know that the $2\sin\theta$ denotes the length of the chord. Hence, whatever remains after that must be the PDF of the variable $\theta$.

Therefore, the PDF, $f(\theta)$, for this case is $\displaystyle f(\theta)=\sin\theta$

(3) Midpoint method

Here, a point is chosen randomly in the circle and the unique chord which has this point as the midpoint is drawn. The expected length is slightly tricky here. The trick is to identify the parameter of interest here is the distance of the point from the origin.

If we let $r$ be that distance, then the length of the chord is $2\sqrt{1-r^2}$. But $r$ is not uniform. It is well known that the density of function of $r$ is $2r$ for a unit circle. We then have

$\displaystyle \mathbb{E}(L)=\int_0^12\sqrt{1-r^2}\cdot 2r\,dr=\frac{4}{3}\approx 1.3334$

If we use the substitution $r=\cos\theta$ in the above integral, we can see that the PDF of $\theta$ in this case is $f(\theta)=\sin2\theta$

(4) Random point and line at random angle method

The video also specifies another method of creating a chord which involves choosing a point randomly in the circle and drawing a line at a random angle through this point. To make things clear, let the $r$ be the distance between the point and other centre of the circle.

Now we draw a line perpendicular to the radial line containing this point. The angle, $t$, that the random line (that we draw to create the chord) makes with this perpendicular line is uniformly distributed between $(0,\pi)$.

It can shown then the distance of the random chord and the centre of the circle is then $r\cos t$. Therefore,

$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}\int_0^12\sqrt{1-(r\cos t)^2}\cdot 2r\,dr\frac{dt}{\pi/2}=\frac{16}{3\pi}\approx 1.69765$

The above integral was solved using Wolfram Alpha.

Here we use the substitutions $u=r\cos t$ and $v=r\sin t$. Then we know that $dudv=rdrdt$. The limits of $r$ and $t$ covers the first quadrant of the unit circle and so should the limits of $u$ and $v$. Therefore,

$\displaystyle \mathbb{E}(L)=\int_0^{1}\int_0^{\sqrt{1-u^2}}2\sqrt{1-u^2}\cdot 2\,\frac{dvdu}{\pi/2}=\int_0^{1}2\sqrt{1-u^2}\cdot 2\sqrt{1-u^2}\,\frac{du}{\pi/2}$

If we now use $u=\cos\theta$, then we see that

$\displaystyle f(\theta)=\frac{2\sin^2\theta}{\pi/2}$

(5) Random point on a radial line and a line at random angle method

Just for the sake of needlessly complicating things, we look at another method where we choose a point randomly in a randomly chosen radial line and draw a at a random angle through this point (the number of randoms in this sentence!). This is almost exactly like the previous case.

$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}\int_0^12\sqrt{1-(x\cos t)^2}\cdot \,dx\frac{dt}{\pi/2}=\frac{2}{\pi}(2G+1)\approx 1.80286$

where $G$ is the Catalan constant. Note that this integral can be expressed as the integral of the Elliptic integral of the second kind. With that and some known results, this can also be calculated manually.

Again using the subtitution $u=x\cos t$ and $v=x\sin t$ with the Jacobian $dudv=xdxdt$, we see 

$\displaystyle \mathbb{E}(L)=\int_0^{1}\int_0^{\sqrt{1-u^2}}2\sqrt{1-u^2}\cdot \,\frac{dvdu}{x\cdot\pi/2}$

But $x=\sqrt{u^2+v^2}$. Using this, the fact that the integral of the secant function is the inverse hyperbolic sine of the tangent function and the substitution $u=\cos t$, we can see that the PDF in this case as

$\displaystyle f(\theta)=\frac{\sin\theta\sinh^{-1}(\tan\theta)}{\pi/2}$

That sums up the Expected values. But we can do more. There was a recent Numberphile video on similar lines. The question discussed was to find the probability that two points selected at random in a circle is lies on different sides of a 'random' chord. Here's where the PDFs we've found so far will be really useful. We'll see those in the next post.

Hope you enjoyed this.

Until then

Yours Aye

Me

Candles and Cake problem - Probabilities with Bertrand's Paradox

The video I referred to in the earlier post only shows a Monte Carlo solution but we can find the exact values quite easily using the density functions.

Because the two candles (or points) are assumed to be uniformly randomly distributed in the circle (or the cake), an important parameter that keeps repeating in all the cases below is the area of a circular segment which can given in terms of the central angle $t$ as $(t-\sin t)/2$. The probability that a point randomly chosen in a unit circle lies in this segment is then $(t-\sin t)/2/\pi$.

Note that the density we found in the earlier refers are for the random variable $\theta$, the angle the chord makes with the tangent. The angle subtended by the chord is twice this value.

Let $E$ denote the event that two randomly chosen points on the circle lie on the opposite side of the chord chosen according to the following methods.

(1) Circumference method

Like before, if we let $\theta$ be the angle between the tangent and the chord, the central angle becomes $2\theta$. Therefore, the probability that a point randomly chosen in the circle lies in the segment created by this chord is the ratio of the segment's area to that of the circle. Therefore,

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\,\frac{d\theta}{\pi/2}=\frac{1}{3}-\frac{5}{4\pi^2}\approx 0.20668$

Even though we can solve this easily, Wolfram Alpha does the job perfectly.

(2) Radial line method

We know from our earlier post that the density function of the tangent angle in this case is just $\sin\theta$. Therefore the required probaility

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin\theta\,d\theta=\frac{128}{45\pi^2}\approx 0.28820$

This can also be simplified manually, but again Wolfram Alpha minimizes our effort and shows

Note that this is $\pi$ times the expected distance between two points chosen at random inside a circle. Coincidence?

(3) Midpoint method

Again, using the density from the previous post, the probability in this case is,

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin2\theta\,d\theta=\frac{1}{8}+\frac{2}{3\pi^2}\approx 0.19255$

Simple even without Wolfram Alpha.

(4) Random point and line at random angle method

Like the cases above we use the density function for this case that we already found in the previous post.

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot2\sin^2\theta\,\frac{d\theta}{\pi/2}=\frac{1}{3}\approx 0.33334$

Solved with Wolfram Alpha.

(5) Random point on a radial line and a line at random angle method

Diving right into the probability with the density function at our disposal,

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin\theta\sinh^{-1}(\tan\theta)\,\frac{d\theta}{\pi/2}\approx 0.386408$

This time Wolfram Alpha only gives us a numerical approximation. But with Mathematica, we can confirm that the exact value is

$\displaystyle \mathbb{P}(E)=\frac{427+60\pi^2-480\log 2}{180\pi^2}$

For the sake of completion, let's finish this post with one more method that was used in the video. Here, the cut is created by choosing two random points on the circumference while the two candles were chosen randomly on a randomly chosen radius.

The first point on the circumference can be chosen to be the 'south pole' of the circle. After the second point is chosen on the circumference, let the chord subtend an angle of $2t$ at the centre. Using the symmetry of the problem, we can limit the range of $t$ to be in $(0,\pi/2)$.

We now calculate the probability that the first candle lies to the right of the chord and the second on the left. Let the radii selected for the first candle make an angle $x$ with the radius perpendicular to the chord. We get a a non-zero probability only if $-t \leq x \leq t$. When $x$ lies in this range, the probability of the first candle landing to the right of the cut is $1-\cos t/\cos x$ and $0$ otherwise.

Similarly, let $y$ be the angle between the radius chosen for the second candle and the radius perpendicular to the cut. When $-t \leq y \leq t$, the probability of the second candle ending up to the left of the cut is $\cos t/\cos y$. Else, the probability is $1$.

Therefore probability in this case depends on the following two integrals.

$\displaystyle \mathcal{I}_1=\int_0^{\pi/2} \int_{-t}^t \int_{-t}^t \frac{\cos t}{\cos y}\left(1-\frac{\cos t}{\cos x}\right)\,dy\,dx\,dt$

$\displaystyle \mathcal{I}_2=\int_0^{\pi/2} \int_{-t}^t \int_{t}^{2\pi-t} \left(1-\frac{\cos t}{\cos x}\right)\,dy\,dx\,dt$

These are very hard to evaluate even with Mathematica. Numerically evaluating these shows,

$\displaystyle \mathbb{P}(E)=2\cdot\frac{\mathcal{I}_1+\mathcal{I}_2}{\pi/2\cdot 2\pi \cdot 2\pi}\approx 0.161612$

Hope you enjoyed this.

Until then

Yours Aye

Me

The Problem of points - Without Replacement

Consider an Urn with $A$ green balls and $B$ red balls. Two players A and B play a game where they draw balls from this Urn one at a time (without replacement). If the drawn ball is green, Player A gets a point. Else, Player B gets a point.

If Player A needs $a$ points to win the game whereas Player B needs $b$ points, the question that we address in this post concerns the probability that Player A wins the match.

As the title says, this is just the Problem of points but without replacement. We've already encountered the problem in our blog in the post titled A note on Partial fractions where we saw how Pascal and Fermat independently solved the problem with different reasoning.

Interestingly, Pascal reasoning involves the (CDF of) Neg. Binomial distribution whereas Fermat's idea involves the (CDF of) Binomial distribution. The central insight in both their ideas is in realising that we'll need a maximum of $a+b-1$ games to conclude the match.

Let's consider the original Problem for points for a moment. If we let $p$ and $q$ represent the probability of A and B winning a single game, then using Pascal's idea,

$\displaystyle \mathbb{P}(\text{A wins the match})=\sum_{k=0}^{b-1}\binom{k+a-1}{k}p^aq^k$

where the individual terms are probabilities from the Neg. Binomial distribution, $NB(a,p)$.

Using Fermat's idea,

$\displaystyle \mathbb{P}(\text{B wins the match})=\sum_{k=b}^{a+b-1}\binom{a+b-1}{k}q^kp^{a+b-1-k}$

where the individual terms are probabilities from the Binomial distribution, $B(a+b-1,p)$.

Now using our discussion about the Neg. Hypergeometric distribution in the previous case and Pascal's idea, we could solve the Without-Replacement case of the Problem of points. Or we could go for simplicity and use the Hypergeometric distribution and Fermat's idea.

Either way, let $P(A,a,B,b)$ be the probability Player A winning the game in the setup described at the start of the post and $M$ be a very large number. It should be clear the classical case can be derived out of this setup as $P(pM,a,qM,b)$ with $p+q=1$.

Obviously, when $A=B$ and $a=b$, either player has an equal chance of winning because of symmetry. Similarly, when $A=B$ and $a<b$, Player A has an advantage. Same goes when $a=b$ and $A>B$.

But consider $p(50,25,40,20)$. Player A has 10 more balls than Player B but needs 5 more points to win the game. It isn't immediately clear whether this is advantageous to Player A or not? In this case, the probability of Player A winning the match is only (approx.) 49.07%.

Naturally, we might think as we break symmetry, the match skews towards one way. Perhaps the most surprising result about the Without-Replacement case of Problem of points is the one family of parameters that form an exception.

Consider $p(21, 11, 19, 10)$. Player A has two more balls than Player B but needs one extra point to win the match. Surprisingly, this match is evenly poised. Not just this, for positive integer $n$,

$\displaystyle p(2n+1,n+1,2n-1,n)=\frac{1}{2}$

Clearly, the parameters that define the game lacks symmetry. Still the game is evenly poised between the two players. I could neither come up with a reason nor a simple explanation of why this should be the case? If you can, please post it in the comments below.

Clear["Global`*"];
pp[A_, a_, B_, b_] := (A / (A + B)) Binomial[A - 1, a - 1] Sum[Binomial[B, j] / Binomial[A + B - 1, j + a - 1], {j, 0, b - 1}];
p[A_, a_, B_, b_] := Sum[Binomial[j + a - 1, j] Binomial[A - a + B - j, B - j], {j, 0, b - 1}] / Binomial[A + B, B];
k = 2;
ListPlot[Table[p[2n + k, n + k, 2n - k, n], {n, 100}]]
p[50, 25, 40, 20]
N[%]
p[21, 11, 19, 10]
N[%]
p[26, 13, 22, 12]
N[%]

Until then
Yours Aye
Me

Expected distance between two points inside a Sphere

This post picks up where we left earlier. As discussed before, we want to find the expected distance between two points randomly selected in an $n$-sphere.

Let the two points be $P$ and $Q$, and $O$ be the centre. The first step is to apply the technique of projective Reduction with $O$ as the scaling point. This shows that the expected length between two points inside a $n$-sphere is $2n/(2n+1)$ times the expected length between a point on the surface of $n$-sphere and another point inside the $n$-sphere.

WLOG, let $P$ be the point on the surface and let's orient the sphere so that point $P$ becomes the south pole. The next step is apply another Projective Reduction with the south pole as the scaling point. This step gives another factor of $n/(n+1)$ for the expected length. Note that while the first reduction preserves the uniformity of the points on the respective 'surfaces', it is not the case with the second reduction. This is what makes the problem tricky.

To find the density of $Q$, consider a random chord from point $P$ (the south pole) along with the axis on which $P$ lies. Let $\theta$ be the angle between the chord and the axis.

Rotating the chord along this axis cuts out a (for a lack of a better word) spherical-cone which contains a cone with its apex at the south pole and a spherical cap at the top (Imaging the other end of the chord to be on the opposite side of $P$). With simple geometry, it is also apparent that the angle between $OQ$ and the axis is $2\theta$.

If we let $V_{\text{sp. cone}}(\theta, r)$ be the volume of the spherical cone from a $n$-sphere of radius $r$ and apex angle $\theta$, then we have,

$\begin{align} \displaystyle V_{\text{sph. cone}}(\theta, r) &= V_{\text{cone}} + V_{\text{sph. cap}}\\&=  \frac{1}{n}V_{n-1}(r \sin 2\theta)(r+r\cos2\theta)+\int_{0}^{2\theta}V_{n-1}(r \sin t) \,d(-r\cos t)\\ &= r^n\frac{v_{n-1}}{n}\sin^{n-1}2\theta \cdot 2  \cos^2\theta + r^nv_{n-1}\int_0^{2\theta}\sin^nt\,dt \\ &=\frac{2^nr^nv_{n-1}}{n}\sin^{n-1}\theta\cos^{n+1}\theta + + r^nv_{n-1}\int_0^{2\theta}\sin^nt\,dt \end{align}$

We have used the fact that the area of an $n$-dimensional cone is $1/n$ times the volume of the circumscribing 'cylinder'. For the volume of the cap, we have used that the volume is the sum of the areas of $(n-1)$ dimensional 'circles' with infinitesimal thickness (see here for more). 

Now, $\frac{\partial V_{\text{sph. cone}}}{\partial r \partial \theta}$ gives volume between $V_{\text{sph. cone}}(\theta + d\theta, r+dr)$ and $V_{\text{sph. cone}}(\theta, r)$. When $Q$ lies in this infinitesimal volume, the distance between $P$ and $Q$ is just $2\cos\theta$.

Because we are only interested in distribution of $Q$ on the surface of the $n$-sphere, we can integrate out the $r$. Therefore, the density is given by,

$\displaystyle f(\theta)=\frac{1}{v_n}\int_{0}^1\frac{\partial V_{\text{sph. cone}}}{\partial r \partial \theta}\,dr$

We now get,

$\displaystyle f(\theta)=\frac{2^nv_{n-1}}{nv_n}((n-1)\cos^{n+2}\theta\sin^{n-2}\theta-(n+1)\sin^n\theta\cos^n\theta)+\frac{v_{n-1}}{v_n}\cdot 2\sin^n2\theta$

where we have used Leibniz differential under integral sign for the last term. Further simplification then gives,

$\displaystyle f(\theta)=\frac{2^nv_{n-1}}{nv_n}(n-1)\sin^{n-2}\theta\cos^n\theta$

This nice simplification allows us to express the expected length $e_n$ in terms of Beta function.

$\displaystyle e_n=\int_0^{\pi/2}2\cos\theta\cdot f(\theta)\,d\theta=\frac{2^n}{n}\frac{v_{n-1}}{v_n}(n-1)B(n/2-1/2, n/2+1)$

Finally, we see that the expected length $L$ between two points randomly selected inside an $n$-sphere is,

$\displaystyle \mathbb{E}(L)=\frac{2n}{2n+1}\frac{n}{n+1}\frac{2^n(n-1)}{n}\frac{B(n/2-1/2,n/2+1)}{B(n/2+1/2,1/2)}$

Interestingly, at $n\to \infty$, irrespective of whether we choose points on the surface or the inside, the expected length goes to $\sqrt{2}$. I can't clearly see why but if you have an idea, please do share it with us.

UPDATE 3 Jul 2023: Generalising this, we can see that

$\displaystyle \mathbb{E}(L^k)=\frac{2n}{2n+k}\frac{n}{n+k}\frac{2^{n+k-1}(n-1)}{n}\frac{B(n/2-1/2,n/2+k/2+1/2)}{B(n/2+1/2,1/2)}$

Using this, we can see, using WA, that

$\displaystyle \lim_{n \to \infty}\mathbb{E}(L^k)=\sqrt{2}^k$

Also, using the limit definition of the logarithm function,

$\displaystyle \mathbb{E}(\ln L)=-\frac{1}{n}+\frac{n}{2}\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k(n+k)}$

This shows us the geometric mean of distance between two points chosen uniformly randomly from a unit circle is $e^{-1/4}$ and that of the sphere is $2e^{-3/4}$

Until then

Yours Aye

Me

Expected distance between two points on the Surface of a sphere

Problems in Geometric Probability are always fascinating. In this post, we going to see two of the well known and challenging problems in this topic.

The first problem is about the expected distance between two points selected at random from a surface of an $n$-sphere. To clarify, I'm considering a circle to be a $2$-sphere, a sphere to be a $3$-sphere and so on.

This problem is relatively easy as can seen from this stackexchange post. Let $S_n(R)$ and $V_n(R)$ denote the surface area and the volume of an $n$-sphere with radius $R$. We know that,

$S_n(R)=s_nR^{n-1}$ and $V_n(R)=v_nR^n$

where $s_n=\dfrac{2\pi^{n/2}}{\Gamma(n/2)}$ and $v_n=\dfrac{2}{n}\dfrac{\pi^{n/2}}{\Gamma(n/2)}$

For the problem under consideration, we can choose one of the points to be on the south pole of the $n$-sphere. Consider one of the axis that connects this point to the origin. A hyperplane perpendicular to this axis cuts the surface of the $n$-sphere to give a $(n-1)$-spherical surface. 

If we consider multiple such hyperplanes parallel to one another along the axis, the $n$-spherical surface is sliced into multiple rings. The picture below shows this in case of a $3$-sphere (taken from this Stack Exchange post).

Because second point is uniformly distributed in the $n$-spherical surface, it is equally likely to be present in any of these rings.

Let $\theta$ be the angle subtended between the radius that contains the second point and radius containing the first point. Then the distance between the points is $2\sin(\theta/2)$.

If we consider these rings to be infinitesimally small, then the density function of the second point lying in any of these rings is equal to the area of the ring. But this is easy. This is just the surface area of the  sliced $(n-1)$-hypersphere times $rd\theta$ (obviously $r=1$ for the unit sphere; used here for illustrative purposes). Therefore,

$\displaystyle \mathbb{E}(L)=\frac{1}{s_n}\int_0^\pi 2\sin(\theta/2)\cdot S_{n-1}(r\sin\theta)\cdot r\,d\theta=\frac{s_{n-1}}{s_n}\int_0^\pi 2\sin(\theta/2)\cdot \sin^{n-2}\theta \,d\theta$

Using Wolfram Alpha, we then see that

$\displaystyle \mathbb{E}(L)=\frac{\Gamma(n/2)}{\sqrt{\pi}\Gamma((n-1)/2)}\cdot \frac{2\sqrt{\pi}\Gamma(n-1)}{\Gamma(n-1/2)}=2\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\cdot \frac{\Gamma(n-1)}{\Gamma(n-1/2)}$

Now, the second problem asks for the expected distance between a point randomly selected on the $n$-spherical surface and a point selected on the $n$-spherical volume. This problem is more involved and requires a little more trickery. We'll see this in the next post.

Until then

Yours Aye

Me