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Suppose that we have an urn that initially contains $n$ balls, partitioned into $k\geq 2$ color-classes with respect to some initial probability distribution $P=(p_1,\dots,p_k)$. At each discrete time step, our modified urn process distinguishes among two actions, i.e. we pick a ball from the urn uniformly at random, and perform one of the following:

  • throw the ball away,
  • return the ball to the urn along with an additional ball of the same color.

The choice of the mentioned actions is prescribed by a predefined sequence $S$ of characters $-$ and $+$, respectively.

So let $n\gg m$, fix some sequence $S\in\left\{ +,- \right\}^m$, and let $\alpha$ be the number of $+$ characters in $S$. Additionally, define $S'\in\left\{ +,- \right\}^m$ by pushing all the $-$ characters of $S$ to the left, i.e. $S'=(-^{m-\alpha}+^\alpha)$. Both $S$ and $S'$ may be used to define different urn-processes, which are denoted $S$ and $S'$-process, respectively.

Now fix an $S$ (or $S'$) process, and after $m$ time-steps measure the color distribution, denoted by $R=(r_1,\dots,r_k)$ (or $R'=(r'_1,\dots,r'_k)$), respectively. We expect both $R$ and $R'$ to deviate from $P$ to some extent. This question is about my intuition claiming that the deviation from the $S'$-process (i.e. $R'$) will be higher then the one in the $S$-process (i.e $R$). In other words, if one aims to maximize the change of color-ratios in the urn, one should do the removals first.

For simplicity, one may focus on the deviation of fraction of balls of a fixed color $i\in [ k ]$ only. Can one show that the r.v. $(p_i-r'_i)^2$ majorizes the r.v. $(p_i-r_i)^2$? In fact, I would be interested in any kind of statements confirming that the deviation of $R'$ from $P$ majorizes the one in the $S$-process.

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    $\begingroup$ Let $V_S=E_S[\sum(p_i-r_i)^2]$. Then $$V_{-+}=\frac{4}{n(n-1)}\sum_{i<j}p_ip_j,\ V_{+-}=\frac{4}{n(n+1)}\sum_{i<j}p_ip_j.$$ So indeed $V_{-+}>V_{+-}$. $\endgroup$
    – user44143
    Commented Sep 6, 2017 at 23:02
  • $\begingroup$ Could you say something in the sense $V_{-+}\geq V_S\geq V_{+-} $? Or would you mind expanding the above comment into a full answer? $\endgroup$ Commented Sep 18, 2017 at 13:11
  • $\begingroup$ I don't have an intuition for this. But you could derive similar formulas for {--+, -+-, +--} and check the patterns there. $\endgroup$
    – user44143
    Commented Sep 18, 2017 at 13:59

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