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I have some probabilities: $P_1, P_2,...,P_n$. Sum of them equals $S_n$.

if I delete some $i$-th probability $P_i$, this equation is now valid: $S_n$=$S_{n-1}$+$P_i$

Now I have n-1 probabilities left, I want to apply some update operation on them, such that, sum of updated elements $S'_{n-1}$ must be equal to $S_n$.

Update Operation: split deleted $P_i$ probability into multiple $n-1$ pieces (they may not be equal, in most cases): $x_1, x_2,..., x_{n-1}$; $\sum_{k=1}^{n-1}x_k=P_i$. Add these pieces to our list: $P_1+x_1, P_2+x_2,...,P_{n-1}+x_{n-1}$ and the equation $S'_{n-1}=\sum_{k-1}^{n-1}{(P_k+x_k)}=S_n$ will become valid.

But when applying update operation on given probabilities, they should remain this property (this is something I can't explain well, but I'll try my best): smallest probability should be the smallest again and highest should be the highest again and so on for ALL elements. In other words: to remain order and 'weight'.

Question: What is the algorithm to choose such $x$ values to remain that property (mentioned above) in a best possible manner and that $S'_{n-1}=S_n$ become valid?

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  • $\begingroup$ Can you give a numerical example showing a before-after transition? $\endgroup$ – quasi Mar 10 '18 at 12:08
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Assuming $S_n\le 1$, a simple way is to just keep the ratios the same.

Thus, for $k=1,..,n-1$, you can define $P_k'$ by $$P_k'=\left(\frac{P_k}{S_n-P_n}\right)S_n$$

The new sum is still $S_n$, and all pairwise ratios are the same, so the order remains the same.

With this definition, you don't even need $x_k$; you can just compute $P_k'$ directly, but if you want $x_k$ for some reason, then using the above transition formula, you would get \begin{align*} x_k &= P_k'-P_k\\[4pt] &=\left(\frac{P_k}{S_n-P_n}\right)S_n-P_k\\[4pt] &=P_k\left(\frac{P_n}{S_n-P_n}\right)\\[4pt] \end{align*}

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  • $\begingroup$ Thanks for answer, but I think your solution might give us some $P>1$ numbers, but they are probabilities... $\endgroup$ – onlydmc123 Mar 10 '18 at 11:56
  • $\begingroup$ I was assuming $S_n \le 1$. If so, there's no issue. $\endgroup$ – quasi Mar 10 '18 at 12:11

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