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This seems like a simple question , but I cant seem to figure it out. Let $I\subseteq [n]$ be a fixed subset. As we run through subsets $A\subseteq [n] $, exactly half of $A\cap I$ will be even sized.

I came up with it while trying to prove that $\sum_{A\subseteq[n]} (-1)^{\vert A\cap I\vert} = 0$, this identity came up in showing that $\mathbb{E}[w_A(x).w_A(y)] = 1$ if $x=y$, $0$ otherwise. Here $w_A :\{-1,1\}^n\rightarrow\{-1,1\}$ is Walsh function defined as $w_A(x) = \prod_{i\in A}x_i$ and $A$ is a uniform random subset of $[n]$.

Note that if $I=[n]$, then this is true and we have both algebraic and combinatorial proof of that.

Can anyone please help with this one? Both algebraic and combinatorial justification are welcome.

RobPratt
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Sudipta Roy
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1 Answers1

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Obviously, $I$ needs to be nonempty. Using your proof for $I=[m]$, then for $I\subset [n]$, $$\sum_{A\subseteq [n]} \mathbb{1}\{|A\cap I|\ \mbox{is even}\}=\sum_{B\subseteq I^c} \sum_{C\subseteq I} \mathbb{1}\{|(B\cup C)\cap I|\ \mbox{is even}\}=\sum_{B\subseteq I^c} \sum_{C\subseteq I} \mathbb{1}\{|C|\ \mbox{is even}\}$$ $$=\sum_{B\subseteq I^c} \frac{1}{2} 2^{|I|}=\frac{1}{2}2^{|I|}\sum_{B\subseteq I^c} 1=\frac{1}{2}2^{|I|} 2^{|I^c|}=\frac{1}{2}2^n.$$

The Other Terry
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