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sport idea – Identification of a inquisitive duty

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During computation of some Shapley values (particulars beneath), I encountered the next duty:

$$

fleft(sum_{okay geq 0} 2^{-p_k}privilege) = sum_{okay geq 0} frac{1}{(p_k+1)binom{p_k}{okay}},

$$

the place $p_0 > 0$ and $p_{okay+1} > p_k$ for all $okay$. In different phrases, the enter to $f$ is the binary growth of a actual quantity within the meander $[0,1]$, and the $p_k$ coincide to the positions of $1$s within the binary growth.

For instance, $f(2^{-t}) = 1/(t+1)$, so $f(1/2) = 1/2$, $f(1/4) = 1/3$ and so forth. More difficult examples are $f(5/8) = f(2^{-1} + 2^{-3}) = 1/2 + 1/(4cdot 3) = 7/12$ and

$$ f(2/3) = fleft(sum_{okay geq 0}2^{-(2k+1)}privilege) = sum_{okay geq 0} frac{1}{(2k+2)binom{2k+1}{okay}} = frac{pi}{sqrt{27}}.$$

The duty $f$ is a steady rising duty satisfying $f(0) = 0$, $f(1) = 1$, and $f(1-t) = 1-f(t)$ for $t in [0,1]$. It has upright asymptotes at dyadic factors.

Here is a plot of $f$:

Is the duty $f$ identified?

Here is the place $f$ got here from. Let $n geq 1$ breathe an integer and let $t in [0,1]$. For a permutation $pi$ of the numbers ${ 2^{-m} : 0 leq m leq n-1 }$ satisfying $pi^{-1}(1) = i$, we are saying that $pi$ is *pivotal* if $sum_{j<i} pi(j) < t$. Let $f_n(t)$ breathe the chance {that a} random $pi$ is pivotal. Then $f(t) = lim_{n rightarrow infty} f_n(t)$.

For instance, take $n = 4$. The permutation $1/8,1/2,1,1/4$ is pivotal for $t in (5/8,1]$. For all $n geq 2$ we now have $f_n(1/2) = 1/2$, since $pi$ is pivotal iff $1$ seems earlier than $1/2$ in $pi$. The common system for $f$ is derived in the same route.

We go away it to the reader to design out how $f_n$ measures some Shapley worth. The capabilities $f_n$ are step capabilities with steps of size $1/2^{n-1}$. They are left-continuous, and are equal to $f$ on the breakpoints.

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