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One-dimensional golf is a function $g$ on $\mathbb R$ such that $g(x)= 1+\min_\mu E[g(x+N(\mu,c\mu^2))]$ if $|x|>1$ and 0 if $|x|\le 1.$
Here $N$ is the normal distribution, whose mean $\mu$ you can choose in each step, but the deviation will depend on it, and on your talent $c$.

What is $g$? Is there an explicit formula for it? Roughly how does it behave?

I set the dependence of the deviation as $\sigma^2=c\mu^2$, but other models might be more realistic.
Also note that I assumed the terrain to be flat.
The problem of course can be similarly defined in $\mathbb R^2$, in which case choosing the direction is obvious (if the terrain is flat), so the problem would reduce to the one-dimensional case, with a possibly different distribution function.
I wonder if this model is realistic to any extent, and how well it would fit on data from professional golf tournaments.
One would of course need to use different $c$'s, or even different distributions for different players.

My question has been inspired by this question by Nate River .

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    $\begingroup$ You will probably enjoy Gelman's and Nolan's foray into (2-d) golf. They wrote up the first part here stat.columbia.edu/~gelman/research/published/golf.pdf and added data here statmodeling.stat.columbia.edu/2019/03/21/… with further links there. $\endgroup$
    – Dror Speiser
    Commented Jan 22 at 8:04
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    $\begingroup$ Haha, calling this process golf is so apt. Is there any direct relation to the planar survival problem I have in the linked post? $\endgroup$
    – Nate River
    Commented Jan 22 at 9:43
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    $\begingroup$ @Nate I don't see any direct relation. It's more like the opposite of your question: we pick the direction, but we have no control over the step size. $\endgroup$
    – domotorp
    Commented Jan 22 at 11:40
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    $\begingroup$ @James Sorry, $f$ is the same as $g$, and the expectation sign was also missing - I fixed both. $\endgroup$
    – domotorp
    Commented Jan 22 at 11:43
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    $\begingroup$ @James Martin Yes your interpretation is correct. With $X$ as you’ve defined, set $g(x) = \inf \mathbb E^x[ \tau]$, where $\tau := \inf\{n \geq 0 \, | \, |X_n| \leq 1\}$, $\mathbb E^x$ denotes the expectation operator under a probability measure such that $X_0 = x$, and the infimum is taken over all sequential choices of $\mu$ at each step. Then $g$ satisfies the given recursion by the dynamic programming principle argument. $\endgroup$
    – Nate River
    Commented Jan 22 at 11:57

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