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Let $\Omega$ be a set of $W$ possible outcomes of an experiment with probability assignments $p_i$ and thus $\sum_{i=1}^{W}p_i=1$. Now, let's divide $\Omega$ into $K$ non-intersecting subsets each containing $W_i$ elements, $i=1,2,\dots,K$ (so $\sum_{k=1}^{K}W_k=W$, $1\le K\le W$). Let us define the following probabilities: $$ \pi_1\equiv \sum_{i\in W_1}p_i,\\ \pi_2\equiv \sum_{i\in W_2}p_i,\\ \dots,\\ \pi_K\equiv \sum_{i\in W_K}p_i $$ It holds $\sum_{k=1}^K\pi_k=1$. If we say: "$\{p_i|\pi_k\}$ are conditional probabilities", how is that defined?

I know that a conditional probability for events $A$ and $B$ can be deduced as $P(A|B)=\frac{P(A\cap B)}{P(B)}$ but in this case we do not have events but probabilities, so I would guess $\{p_i|\pi_k\}=\frac{p_i\cdot\pi_k}{\pi_k}=p_i$ which is obviously wrong. At least I hope that $P(B)=\pi_k$ and if so, what is equivalent to $P(A\cap B)$?

It should also hold $\sum_{i\in W_k}(p_i|\pi_k)=1$, ($k=1,2,\dots,K$).

A more general problem is understanding how this conditional probability would work in Tsallis entropy defined as: $$ S_q(\{p_i\})\equiv k\frac{1-\sum_{i=1}^{W}p_i^q}{q-1} $$ So what actually is $S_q(\{p_i|\pi_k\})$?

This is based on book Tsallis, Constantino (2009). Introduction to nonextensive statistical mechanics : approaching a complex world (Online-Ausg. ed.). New York: Springer. Specifically page 47.

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  • $\begingroup$ You notation is unclear. Is $W_k$ the count of outcomes in the $k$-th subset of the partition, or is it the set of indices for the outcomes in that subset? $\endgroup$ – Graham Kemp Oct 21 '15 at 14:54
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You notation is unclear. Is $W_k$ the count of outcomes in the $k$-th subset of the partition, or is it the set of indices for the outcomes in that subset?

Let $\Omega$ be the set of $W$ atomic outcomes of an experiment. Let $\omega_i$ represent the $i$-th indexed outcome of $\Omega$, and $p_i$ be its probability assignment. Let us partition $\Omega$ into $K$ disjoint subsets, such that $D_k$ is the $k$-th indexed partition, and $W_k$ be its size.   Then the probability assignment of $D_k$ is: $$\pi_k = \sum_{i:\omega_i\in D_k} p_i$$

Then using you notation that $\{p_i\mid \pi_k\}$ is the conditional probability that outcome $\omega_i$ occurs given that one of the outcomes in event $W_k$, we have:

$$\{p_i\mid \pi_k\} = \dfrac{p_i \cdot [\omega_i\in D_k]}{\pi_k}$$

Where $[\omega_i\in D_k]$ is the Iverson bracket notation for the indicator function that outcome $\omega_i$ is in partition $D_k$.

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