The following graph shows data from professional golfers on the proportion of successful putts as a function of distance from the hole. Unsurprisingly, the probability of making the shot declines as a function of distance:

The error bars associated with each point \(j\) in the above graph are simple classical standard deviations, \(\sqrt{\hat{p}_j(1-\hat{p}_j)/n_j}\), where \(\hat{p}_j=y_j/n_j\) is the success rate for putts taken at distance \(x_j\).

Can we model the probability of success in golf putting as a function of distance from the hole? Given usual statistical practice, the natural starting point would be logistic regression:

\[ y_j\sim\mbox{binomial}(n_j, \mbox{logit}^{-1}(a + bx_j)), \mbox{ for } j=1,\dots, J. \] In Stan, this is:

```
data {
int J;
array[J] int n;
vector[J] x;
array[J] int y;
}
parameters {
real a;
real b;
}
model {
y ~ binomial_logit(n, a + b * x);
}
```

The code in the above model block is (implicitly) vectorized, so that
it is mathematically equivalent to modeling each data point,
`y[i] ~ binomial_logit(n[i], a + b*x[i])`

. The vectorized
code is more compact (no need to write a loop, or to include the
subscripts) and faster (because of more efficient gradient
evaluations).

We fit the model to the data:

```
golf_data <- list(x=x, y=y, n=n, J=J)
fit_logistic <- stan("golf_logistic.stan", data=golf_data)
a_sim <- extract(fit_logistic)$a
b_sim <- extract(fit_logistic)$b
```

And here is the result:

```
Inference for Stan model: anon_model.
4 chains, each with iter=2000; warmup=1000; thin=1;
post-warmup draws per chain=1000, total post-warmup draws=4000.
mean se_mean sd 25% 50% 75% n_eff Rhat
a 2.23 0 0.06 2.20 2.23 2.27 1346 1
b -0.26 0 0.01 -0.26 -0.26 -0.25 1369 1
Samples were drawn using NUTS(diag_e) at Tue Sep 12 13:17:49 2023.
For each parameter, n_eff is a crude measure of effective sample size,
and Rhat is the potential scale reduction factor on split chains (at
convergence, Rhat=1).
```

Going through the columns of the above table: Stan has computed the posterior means \(\pm\) standard deviations of \(a\) and \(b\) to be 2.23 \(\pm\) 0.06 and -0.26 \(\pm\) 0.01, respectively. The Monte Carlo standard error of the mean of each of these parameters is 0 (to two decimal places), indicating that the simulations have run long enough to estimate the posterior means precisely. The posterior quantiles give a sense of the uncertainty in the parameters, with 50% posterior intervals of [2.20, 2.27] and [-0.26, -0.25] for \(a\) and \(b\), respectively. Finally, the values of \(\widehat{R}\) near 1 tell us that the simulations from Stanâ€™s four simulated chains have mixed well. (We have more sophisticated convergence diagnostics, and also we recommend checking the fit using simulated data, as discussed in the Bayesian Workflow Using Stan book, but checking that \(\widehat{R}\) is near 1 is a good start.)

The following graph shows the fit plotted along with the data:

The black line shows the fit corresponding to the posterior median estimates of the parameters \(a\) and \(b\); the green lines show 10 draws from the posterior distribution.

In this example, posterior uncertainties in the fits are small, and for simplicity we will just plot point estimates based on posterior median parameter estimates for the remaining models. Our focus here is on the sequence of models that we fit, not so much on uncertainty in particular model fits.

As an alternative to logistic regression, we shall build a model from first principles and fit it to the data.

The graph below shows a simplified sketch of a golf shot. The dotted line represents the angle within which the ball of radius \(r\) must be hit so that it falls within the hole of radius \(R\). This threshold angle is \(\sin^{-1}((R-r)/x)\). The graph, which is not to scale, is intended to illustrate the geometry of the ball needing to go into the hole.

The next step is to model human error. We assume that the golfer is attempting to hit the ball completely straight but that many small factors interfere with this goal, so that the actual angle follows a normal distribution centered at 0 with some standard deviation \(\sigma\).

The probability the ball goes in the hole is then the probability that the angle is less than the threshold; that is, \(\mbox{Pr}\left(|\mbox{angle}| < \sin^{-1}((R-r)/x)\right) = 2\Phi\left(\frac{\sin^{-1}((R-r)/x)}{\sigma}\right) - 1\), where \(\Phi\) is the cumulative normal distribution function. The only unknown parameter in this model is \(\sigma\), the standard deviation of the distribution of shot angles. Stan (and, for that matter, R) computes trigonometry using angles in radians, so at the end of our calculations we will need to multiply by \(180/\pi\) to convert to degrees, which are more interpretable by humans.

Our model then has two parts: \[\begin{align} y_j &\sim \mbox{binomial}(n_j, p_j)\\ p_j &= 2\Phi\left(\frac{\sin^{-1}((R-r)/x_j)}{\sigma}\right) - 1 , \mbox{ for } j=1,\dots, J. \end{align}\] Here is a graph showing the curve for some potential values of the parameter \(\sigma\).