# sport idea – Identification of a inquisitive duty Answer

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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 $$sum_{j. Let $$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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