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We are given a vector $n$-dimensional random vector $\mathbf{X}$ whose components are the Bernoulli random variables $X_1, X_2, \ldots X_n$, such that the probability $\mathbb{P}(X_1=X_2=\ldots=X_n=0)$ is null.


Question: For $n>1$, is there any joint distribution of $\mathbf{X}$’s components and a vector $\mathbf{v}\in [0,1]^n$ such that $$\frac{\mathbb{E}\langle\mathbf{X},\mathbf{v}\rangle}{\mathbb{E}\,{\sum_{\!i=1}^{\!n}\!X_i}}>\frac{1}{n}\mathbb{E}\left[\frac{\langle\mathbf{X},\mathbf{v}\rangle}{{\sum_{\!i=1}^{\!n}\!X_i}}\right]\ ?$$

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  • $\begingroup$ You did not specify the correlation structure. Equicorrelation? $\endgroup$ Commented Mar 10, 2023 at 14:56
  • $\begingroup$ @kjetilbhalvorsen , the Bernoulli random variables correspond to edges incident to a given vertex $u$ in a given (simple) graph $G$. They can be included or not in a uniformly generated random spanning tree $T$ of $G$. The inclusion in $T$ of each edge $\{u,v\}$ incident to $u$ in $G$ is a Bernoulli random variable with parameter equal the effective resistance in $G$ between $u$ and $v$, which is known given $G$. Note that for each realization of $\mathbf{X}$, at least one component of $\mathbf{X}$ (one node $v$ adjacent to $u$ in $G$) has outcome equal to $1$ ($\{u,v\}\in T$). $\endgroup$ Commented Mar 10, 2023 at 16:03
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    $\begingroup$ Please add this clarifications as an edit to the post! Comments here are often seen by only a few people, and we want posts to be self-sufficient $\endgroup$ Commented Mar 10, 2023 at 16:09
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    $\begingroup$ Thank you @JamesMartin . Yes, done (and added how large can be such absolute difference). $\endgroup$ Commented Mar 10, 2023 at 23:56
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    $\begingroup$ The dependence on $a$ is just linear, right? Replacing $\mathbf{v}$ by $\mathbf{v}/a$ gives you a vector in $[-1,1]^d$ and multiplies the quantity you're interested in by $1/a$. $\endgroup$ Commented Mar 11, 2023 at 16:05

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$\newcommand\X{\mathbf X}\newcommand\v{\mathbf v}$

Indeed, $1\le\sum_{i=1}^n X_i\le n$ and hence $$n\sum_{i=1}^n X_i\ge n=En\ge E\sum_{i=1}^n X_i,$$ so that $$\frac1n\,E\frac{\langle\X,\v\rangle}{\sum_{i=1}^n X_i} =E\frac{\langle\X,\v\rangle}{n\sum_{i=1}^n X_i}\le E\frac{\langle\X,\v\rangle}{E\sum_{i=1}^n X_i} =\frac{E\langle\X,\v\rangle}{E\sum_{i=1}^n X_i},$$ as claimed. $\quad\Box$

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