6
$\begingroup$

Does there exist a Frequentist or ODE or Non-Bayesian solution to Gull's Lighthouse Problem which is correctly modeled with cauchy distribution?

See The Lighthouse Problem and Dave Harris' answer to this thread: Bayesian lighthouse location estimation.

$\endgroup$
8
$\begingroup$

Interesting problem indeed. Three non-Bayesian solutions follow.

Classical Physicist's view

Here's my physicist's solution to it. $$\alpha=median[x_k]=F^{-1}(1/2)$$ $$\beta=F^{-1}(3/4)-F^{-1}(1/4)$$ Here $F$ is the empirical CDF of Cauchy distribution, or you could also call it a quantile (percentile) function. enter image description here

Cauchy distribution (aka Breit-Wigner in physics) has no mean, but it is symmetric. So, the median is a decent estimate of the $\alpha$.

Since, it has no variance either, in physics when using this distribution the notion of width at half height is used to describe its dispersion. It happens so that the width at half height of PDF is $\beta$ and corresponds to the span between first and third quartiles (interquartile range).

MLE

Maximum likelihood estimation, of course, is more efficient. However, mine is very simple and intuitive.

OLS (doesn't work)

There's also a terrible regression solution. Look at the quantile function of the distribution: $$Q(p; \alpha,\beta) = \alpha + \beta\,\tan\left[\pi\left(p-\tfrac{1}{2}\right)\right]$$

It appears that, we can convert it into OLS problem (assuming $x_k$ are ordered!): $$x_k = \alpha + \beta\,\tan\left[\pi\left((k-1/2)/N-\tfrac{1}{2}\right)\right]$$ $$x_k = \alpha + \beta\,z_k,$$ where $z_k=\tan\left[\pi\left((k-1/2)/N-\tfrac{1}{2}\right)\right]$

OLS will give you immediately the estimates $\hat\alpha,\hat\beta$.

The problem is that, OLS assumes finite variance, and we don't have it with Cauchy. So, the output of OLS is garbage. Here's R code to experiment and see how my first method is pretty robust.

alpha <- 10 # unkonwn true values
beta  <- 30 # this is known for now
##################
set.seed(123)
N       <- 1024
theta_k <- runif(N,-pi/2,pi/2)
x_k     <- beta * tan(theta_k) + alpha
q123 = quantile(x_k,c(1/4,1/2,3/4),type=1)
print("alpha")
print(q123[2])
print("beta")
print((q123[3] - q123[1])/2)

y = sort(x_k)
x = tan(pi*((seq(1:N)-0.5)/N-1/2))
fit = lm(y~x)
print(fit)

Outputs:

alpha <- 10 # unkonwn true values
beta  <- 30 # this is known for now
##################
set.seed(123)
N       <- 1024
theta_k <- runif(N,-pi/2,pi/2)
x_k     <- beta * tan(theta_k) + alpha
q123 = quantile(x_k,c(1/4,1/2,3/4),type=1)
print("alpha")
print(q123[2])
print("beta")
print((q123[3] - q123[1])/2)

y = sort(x_k)
x = tan(pi*((seq(1:N)-0.5)/N-1/2))
fit = lm(y~x)
print(fit)
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The solution is obtained through quantile regression? $\endgroup$ – Numbers May 22 at 20:06
  • $\begingroup$ I wouldn't call it a quantile regression. That term is reserved for a different kind of technique, but OLS solution does use quantiles $\endgroup$ – Aksakal May 22 at 20:13
5
$\begingroup$

Since $x \sim Cauchy(\alpha, \beta)$ and you want to estimate the position of the lighthouse, wich is $(\alpha, \beta)$, your problem is simply to make inference on a Cauchy distribution. Here you will find what you need: https://en.wikipedia.org/wiki/Cauchy_distribution#Estimation_of_parameters

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.