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In an article I'm reading the probability of an observed point falling in bin i is written as:

$P_i=\frac{m_i}{\sum\limits_j m_j}$

where $m_i$ is the number of model points in bin i. Then, the cumulative probability of obtaining the entire data set $n_i$ is written as:

$P=\prod\limits_i \left( \frac{m_i}{\sum\limits_j m_j} \right)^{n_i}$

where $n_i$ if the observed value of bin i.

My question is: how was equation 2 obtained?

Edit: this is the article in question. I didn't add much information in the question because I wanted to keep it as simple as possible. Equations in the article are (14) and (15).

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  • $\begingroup$ I guess they are assuming independende and therefore the joint probability is the product of the probabilities of each observation. $\endgroup$
    – user10525
    May 15 '12 at 13:18
  • $\begingroup$ That would explain the product over all $i$, but how did that $n_i$ get up there? $\endgroup$
    – Gabriel
    May 15 '12 at 13:23
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    $\begingroup$ It would help if you let us know the reference in order to understand the context. At first glance, it seems like they are observing $n_i$ bins $i$ and then they take the product of these probabilities. That is how that $n_i$ climbed up there. $\endgroup$
    – user10525
    May 15 '12 at 13:25
  • $\begingroup$ Added the article in the question. $\endgroup$
    – Gabriel
    May 15 '12 at 13:35
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Assume that you have independent observations. If the first observation falls in bin $i$ then the probability of obtaining that observation is $P_i$.

If the second observation falls in bin $j$, the probability of obtaining those observations in that order is $P_i\cdot P_j$. If the third observation falls into bin $i$ the cumulative probability is now $P_i\cdot P_j\cdot P_i=P_i^2\cdot P_j$, and so on.

Continuing in this way, we find that the cumulative probability is $P_i^{n_i}\cdot P_j^{n_j}\cdots=\prod_i P_i^{n_i}$, which is the second formula.

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  • $\begingroup$ I'm choosing this as the answer because I find it a bit easier to understand (my statistics is pretty basic). Thanks! $\endgroup$
    – Gabriel
    May 15 '12 at 13:36
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It just used that each points falls independently into a bin. Remember that for independent events A and B $$ P(A \cap B)=P(A)P(B). $$

Each of the points falling into bin $i$ had a chance of $P_i$ to do so. Because of independence the probability of all the $n_i$ points falling into bin $i$ doing so, is just

$$ {P_i}^{n_i}=({\frac{m_i}{\sum_j m_j}})^{n_i}. $$ Repeat for all the other bins (the events there are independent) and you get the final result.

What I personally find more confusing is your first sentence.

The tag maximum-likelihood gives it much needed context, since otherwise it makes no sense to equate short-run frequencies and probabilities. I think the whole point of the argument is taking the short-run frequencies as estimates of the probabilities (i.e. long-run frequencies) maximizes the Likelihood of the observed data and is therefore the MLE for estimating the long-run frequencies.

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