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Given a list of $i$ probability distributions $f_j(x)$ and a list of $i$ probabilities $p_j$ and an interval I.

A probability distribution $f_z(x)$ is chosen from the list with probability p_z and a random number R from the interval I is chosen with this distribution.

Given $R$,$I$ and the two lists of distributions and associated probabilities, how to find the probability that the distribution with an index of $n$ was chosen?

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Your accounts are merged. – Zev Chonoles Nov 16 '11 at 3:49
@Zev Thanks, sorry for the hassle – TROLLKILLER Nov 16 '11 at 3:50
up vote 4 down vote accepted

This seems like standard Bayesian stuff: one knows that $f_j(x)\mathrm{d}x$ is the probability to choose (a number around) $x$ conditionally on $j$ and that $p_j$ is the probability to chose $j$, so Bayes formula indicates that the probability $p(j|x)$ that $j$ was chosen conditionally on $x$ is proportional to the probability of $x$ and $j$, which is $p_jf_j(x)\mathrm{d}x$. Normalizing this, one gets $$ p(j|x)=p_jf_j(x)/g(x),\qquad g(x)=\sum_kp_kf_k(x). $$

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