Poisson Binomial Distributed Score weighted by entropy / rareness
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I have a game whose score (at chance performance) is a poisson binomial distribution (i.e. a sum of x independent Bernoulli trials, not necessarily equally distributed). It's possible in this scenario, that two scores are the same, yet one's likelihood is lower. E.g. In the case of two Bernoullis, p = 0.6, and p = 0.1. A person can score 1 (out of 2), in two ways one with a likelihood of 0.4 * 0.1, and another with a likelihood of 0.6 * 0.9.
Is it possible to create a score that takes into account the entropy of the variable, and therefore fairly weights rarer events higher than common events? (..and conversely penalises rare non-events)
In the example above, the first combination is more difficult to achieve by chance, so should be weighted higher.
probability entropy
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
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add a comment |
$begingroup$
I have a game whose score (at chance performance) is a poisson binomial distribution (i.e. a sum of x independent Bernoulli trials, not necessarily equally distributed). It's possible in this scenario, that two scores are the same, yet one's likelihood is lower. E.g. In the case of two Bernoullis, p = 0.6, and p = 0.1. A person can score 1 (out of 2), in two ways one with a likelihood of 0.4 * 0.1, and another with a likelihood of 0.6 * 0.9.
Is it possible to create a score that takes into account the entropy of the variable, and therefore fairly weights rarer events higher than common events? (..and conversely penalises rare non-events)
In the example above, the first combination is more difficult to achieve by chance, so should be weighted higher.
probability entropy
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
$endgroup$
add a comment |
$begingroup$
I have a game whose score (at chance performance) is a poisson binomial distribution (i.e. a sum of x independent Bernoulli trials, not necessarily equally distributed). It's possible in this scenario, that two scores are the same, yet one's likelihood is lower. E.g. In the case of two Bernoullis, p = 0.6, and p = 0.1. A person can score 1 (out of 2), in two ways one with a likelihood of 0.4 * 0.1, and another with a likelihood of 0.6 * 0.9.
Is it possible to create a score that takes into account the entropy of the variable, and therefore fairly weights rarer events higher than common events? (..and conversely penalises rare non-events)
In the example above, the first combination is more difficult to achieve by chance, so should be weighted higher.
probability entropy
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
$endgroup$
I have a game whose score (at chance performance) is a poisson binomial distribution (i.e. a sum of x independent Bernoulli trials, not necessarily equally distributed). It's possible in this scenario, that two scores are the same, yet one's likelihood is lower. E.g. In the case of two Bernoullis, p = 0.6, and p = 0.1. A person can score 1 (out of 2), in two ways one with a likelihood of 0.4 * 0.1, and another with a likelihood of 0.6 * 0.9.
Is it possible to create a score that takes into account the entropy of the variable, and therefore fairly weights rarer events higher than common events? (..and conversely penalises rare non-events)
In the example above, the first combination is more difficult to achieve by chance, so should be weighted higher.
probability entropy
probability entropy
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
asked Dec 13 '18 at 14:28
james_alvarezjames_alvarez
1266
1266
add a comment |
add a comment |
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