[EM] Exact spatial model probabilities?

[Daniel Carrera]

On Fri, Jan 21, 2022 at 9:02 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> It's easy to approximately find such probabilities for spatial models as
> well by just creating an instance and randomly placing voters on it to
> determine what they vote, and hence in the limit of number of trials
> going to infinity, get the probability.
>
> But I was wondering, is it possible to do so exactly? Let's say that in
> a spatial model, first c candidates are placed uniformly at random in an
> unit d-cube. Then voters are also placed uniformly at random in this
> cube and they prefer candidates closer to them to candidates further away.
>
> Now I would suppose that the probability that a voter votes A first is
> equal to the volume of the Voronoi cell that belongs to A. (And the
> probability that a voter will vote A>B>C is the volume of the
> intersection of the closest-neighbor Voronoi cell for A with the
> second-closest-neighbor for B and the third-closest-neighbor for C.)
>
> Well, that eliminates one source of randomness -- assuming I can exactly
> calculate the vertices of these regions. But there's still the second
> source in the randomness of the candidates. Do you know of any calculus
> tricks to get the probabilities over every possible candidate position,
> or is this simply too hairy?
>


I have been thinking about this. I don't have a clever solution for you,
but I have some thoughts that could be a stepping stone toward a solution.
What you are trying to do is compute an integral in a highly dimensional
space. Let's say that for a given set of candidates you have a function f()
that outputs a real value. That might be a utility, or a probability, or
whatever. I just want it to be a real number and not binary so that we can
hope that f() is smooth in the space spanned by the candidates.

The set of candidates can be thought of as a point in a C*D dimensional
space where C is the number of candidates and D is the dimension of your
D-cube. So you have a vector:

x_jk = position of candidate j on issue k

x = [ x11, x12, ... , x1N, x21, ... , xCD ]

When you ask:

"Do you know of any calculus tricks to get the probabilities over every
possible candidate position, or is this simply too hairy?"

You are asking, "how do I integrate f()"? Your default plan of drawing
random candidates and computing f() for each is integration by Monte Carlo
sampling. Now, I don't know how to integrate f() analytically, but
realizing that it is an integral allows us to consider other forms of
numerical integration (hence my request that f() be a smooth function). One
such example is Laplace's method:

https://en.wikipedia.org/wiki/Laplace%27s_method

Let us imagine that f() looks like a hill. If you can compute the gradient
of f(), you can climb the hill and get to the peak. Up there, if you can
compute the Hessian of f(), you can approximate f() with the Gaussian
function with the same height and Hessian. This method is approximate, but
it is very fast because you do not have to sample across the entire space.
Laplace's method is often used in highly dimensional problems, where the
cost of random sampling is prohibitive.

How do you compute the derivative and Hessian? Most often this is used with
an Autodiff package. Alternatively, you can use finite differences but at
the cost of reduced accuracy. What if f() isn't one hill but many? You can
integrate each hill separately and add them up.

Anyway, I don't know whether Laplace's method is a realistic solution, but
if it isn't, perhaps there is a different form of numerical integration
tool that is available to you that is more efficient than Monte Carlo
integration. For example, importance sampling doesn't require derivatives.
There is a long list of numerical integrators that have been devised over
the ages.

Cheers,
-- 
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
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