Another Nice Application of Difference Equations

Here's a nice problem I encountered in a course on applied stochastic processes.

Problem:

Show that the set of all pairs of positive integers can be placed into one-one correspondence with the positive integers by giving an explicit one-one mapping between the two sets.

Solution:

This can be done expediently using the theory of partial difference equations. A standard diagonalization can be characterized by the following relations:
1. $$f(1,1)=1$$
2. $$f(x,y) = f(x-1,y)+x+y$$
3. $$f(x,y) = f(x,y-1)+x+y-1$$
These can be written in difference notation as:
1. $$f(1,1)=1$$
2. $$\frac{\Delta f}{\Delta x} = x+y$$
3. $$\frac{\Delta f}{\Delta y} = x+y-1$$
Equation 2 gives $$\frac{{\Delta}^2 f}{{\Delta x}{\Delta y}} = 1$$ and equation 3 gives $$\frac{{\Delta}^2 f}{{\Delta y}{\Delta x}} = 1,$$ so this is an exact partial difference equation. Summing using equation 2, $f(x,y) = x(x+1)/2 + xy + g(y).$ Differencing and setting this equal to equation 3, $x+\frac{\Delta g}{\Delta y} = x+y-1$ and so $$g(y) = y(y+1)/2 - y + C$$ . Finally $$f(1,1)=1$$ implies that $$C=-1$$ , and so $f(x,y) = x(x+1)/2 + xy + y(y+1)/2 - y - 1.$

A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

Mixed Models in R - Bigger, Faster, Stronger

When you start doing more advanced sports analytics you'll eventually starting working with what are known as hierarchical, nested or mixed effects models. These are models that contain both fixed and random effects. There are multiple ways of defining fixed vs random random effects, but one way I find particularly useful is that random effects are being "predicted" rather than "estimated", and this in turn involves some "shrinkage" towards the mean.

Here's some R code for NCAA ice hockey power rankings using a nested Poisson model (which can be found in my hockey GitHub repository):
model <- gs ~ year+field+d_div+o_div+game_length+(1|offense)+(1|defense)+(1|game_id) fit <- glmer(model, data=g, verbose=TRUE, family=poisson(link=log) ) The fixed effects are year, field (home/away/neutral), d_div (NCAA division of the defense), o_div (NCAA division of the offense) and game_length (number of overtime periods); off…

What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows \[ \frac{\partial W}{\partial P}(P=1) = …