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In the process of computing Shapley values, I observed an interesting combinatorial constant. I am not exactly sure where such behavior comes. And here is the conjecture.

Notations

For any finite non-empty sequence of number $Q \subset \mathbb{R}$, I can construct a vector $C = (c_0, c_1, c_2, \cdots, c_{|Q|})$ using such a mapping $C(Q)$: $$ \sum_{i=0}^{|Q|} c_i x^ i = \prod_{q_i \in Q} (1+q_ix) $$

Given a sequence $A = a_1, a_2, \cdots, a_m $, denote $A_{\neg i}$ as $A \setminus \{a_i\}$. $|A| = m$ denotes size/length of $A$ is $m$. With size $m \in \mathbb{N}$, denote a constant vector $N(m)$ as $(1/{m-1 \choose 0}, 1/{m-1 \choose 1}, \cdots, 1/{m-1 \choose m-1 } )$. And $\langle\cdot, \cdot \rangle$ is the inner product operation.

Conjecture

For any finite non-empty sequence $A \subset \mathbb{R}$ with $|A| = m$, i conjecture if $0 \in A$ then:

$$ \sum_{i \in A} (1 - i) \langle C(A_{\neg i}), N(m) \rangle = m $$

I'm able to obtain consistent results from simulations, but I don't know how to prove it. Any help on a proof/dis-proof would be appreciated!

Example

For $A = 1, 2, 3$, the size $m$ is clearly 3.

$A_{\neg 1} = 2, 3$, $A_{\neg 2} = 1, 3$, $A_{\neg 3} = 1, 2$

$C(A_{\neg 1}) = (1, 5, 6)$, $C(A_{\neg 2}) = (1, 4, 3)$, $C(A_{\neg 3}) = (1, 3, 2)$.

$N(3) = (1, 0.5, 1)$

$$ (1-1)* \langle (1, 5, 6), (1, 0.5, 1) \rangle \\ + (1-2)* \langle (1, 4, 3), (1, 0.5, 1) \rangle \\ + (1-3)* \langle (1, 3, 2), (1, 0.5, 1) \rangle \\ = -15 $$ which is not $m$.

On the other hand: For $A=0, 2, 3$, the size $m$ is still 3.

$A_{\neg 0} = 2, 3$, $A_{\neg 2} = 0, 3$, $A_{\neg 3} = 0, 2$

$C(A_{\neg 0}) = (1, 5, 6)$, $C(A_{\neg 2}) = (1, 3, 0)$, $C(A_{\neg 3}) = (1, 2, 0)$.

$N(3) = (1, 0.5, 1)$

$$ (1-0)* \langle (1, 5, 6), (1, 0.5, 1) \rangle \\ + (1-2)* \langle (1, 3, 0), (1, 0.5, 1) \rangle \\ + (1-3)* \langle (1, 2, 0), (1, 0.5, 1) \rangle \\ = 3 $$ which is $m$.

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  • $\begingroup$ So, if I'm correctly understanding your question, you are claiming that if $a_1, a_2, \ldots, a_m$ are $m$ numbers, and if $k$ is a nonnegative integer, then $\sum\limits_{i=1}^k \dfrac{e_k\left(a_1, a_2, \ldots, \widehat{a_i}, \ldots, a_m\right)}{\dbinom{m-1}{k}} - \sum\limits_{i=1}^k a_i \dfrac{e_{k-1}\left(a_1, a_2, \ldots, \widehat{a_i}, \ldots, a_m\right)}{\dbinom{m-1}{k-1}}$ equals $0$ when $k$ is positive and equals $m$ if $k = 0$. Here, $e_k$ means the $k$-th elementary symmetric polynomial (which is $0$ if $k$ is negative). This should be easy: compute each monomial's coefficient. $\endgroup$ Nov 24, 2021 at 23:07
  • $\begingroup$ And easy it is: The coefficient of a given monomial $a_{i_1} a_{i_2} \cdots a_{i_k}$ (with $i_1, i_2, \ldots, i_k$ being distinct) is $\dfrac{m-k}{\dbinom{m-1}{k}} - \dfrac{k}{\dbinom{m-1}{k-1}} = 0$. All other monomials have coefficient $0$ since they appear in neither of the two sums. $\endgroup$ Nov 24, 2021 at 23:10
  • $\begingroup$ not completely sure I follow the newly constructed term, but $ 0 \in A$ is the condition, basically at least one of $a_1, \cdots, a_m$ is 0. $\endgroup$
    – yupbank
    Nov 24, 2021 at 23:11
  • $\begingroup$ That condition isn't necessary. $\endgroup$ Nov 24, 2021 at 23:11
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    $\begingroup$ @yupbank: I will after tomorrow's FPSAC deadline. $\endgroup$ Nov 25, 2021 at 5:19

1 Answer 1

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We can rewrite $$\sum_{i=0}^{|Q|} c_i x^ i = \prod_{q_i \in Q} (1+q_ix)$$ as $$c_i = \sum_{\substack{S \subseteq Q \\ |S| = i}} \prod_{q \in S} q$$

Then $$\sum_{a \in A} (1 - a) \langle C(A_{\neg a}), N(m) \rangle$$ can be split into two sums: $$\left(\sum_{a \in A} \langle C(A_{\neg a}), N(m) \rangle \right) - \left(\sum_{a \in A} a \langle C(A_{\neg a}), N(m) \rangle \right)$$ Call the first sum $L$ (for lower) and the second sum $U$ (for upper). Note that $L$ only contains terms with products of $0$ to $m-1$ elements of $A$, and $U$ only contains terms with products of $1$ to $m$ elements of $A$.

Consider a subset $S \subset A$ of size $0 < k < m$. It occurs in $L$ once for every $a \in A \setminus S$ and each time with weight $\binom{m-1}{k}^{-1}$ from the dot product, for a total weight of $(m-k)\binom{m-1}{k}^{-1} = \frac{(m-k)!k!}{(m-1)!}$; it occurs in $U$ once for every $a \in S$ and each time with weight $\binom{m-1}{k-1}^{-1}$ from the dot product, for a total weight of $k\binom{m-1}{k-1}^{-1} = \frac{(m-k)!k!}{(m-1)!}$; and so all of these terms cancel.

Therefore the only terms which survive are the term from $L$ corresponding to the subset of size $0$ and the term from $U$ corresponding to the subset of size $m$. The former occurs with weight $\frac{(m-0)!0!}{(m-1)!} = m$; the latter with weight $\frac{(m-m)!m!}{(m-1)!} = m$ (but negated because of the subtraction).

Therefore the sum evaluates to $$m\left(1 - \prod_{a \in A}a\right)$$

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  • $\begingroup$ I'm sure Darij won't mind me answering, even though the argument is essentially the same as his. $\endgroup$ Nov 25, 2021 at 14:37

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