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Example for divergent probability distribution ?

by Jano L.
Tags: distribution, divergent, probability
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Jano L.
#1
Nov13-12, 05:38 AM
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Sometimes it is said that the probability distribution which does not add up to 1 still can be used to find relative probabilities.

For example, consider probability distribution [itex]p_n = 1/n[/itex] for all natural numbers. Does it make sense to say [itex]n = 1[/itex] is twice as probable as [itex]n=2[/itex], even if total probability does not add up to 1? Or does probabilistic description necessarily require that total probability is 1?
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mfb
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Nov13-12, 10:23 AM
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If the sum of all values is finite (!), you can normalize the values to get a meaningful probability distribution. ##p_n=\frac{1}{n^2}## would allow that, for example, with normalization factor ##\frac{6}{\pi^2}##.

If you do not care about absolute probabilities, you can ignore that prefactor and work with relative probabilities.
Jano L.
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Nov13-12, 11:41 AM
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Thank you. I was thinking mainly of the case when the distribution cannot be normalized. Can I work with relative probabilities given by [itex]p_n = 1/n[/itex]?

In quantum theory, it was suggested that divergent probability distribution still gives relative probabilities correctly. But I am unsure about this concept of "relative probability". Does it make sense for divergent distribution?

mfb
#4
Nov13-12, 12:19 PM
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Example for divergent probability distribution ?

I don't think this works for countable sets. There is no probability distribution with those ratios.
It can work for uncountable sets as a probability density function, if the integral is bounded.
micromass
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Nov13-12, 01:42 PM
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Quote Quote by Jano L. View Post
Thank you. I was thinking mainly of the case when the distribution cannot be normalized. Can I work with relative probabilities given by [itex]p_n = 1/n[/itex]?

In quantum theory, it was suggested that divergent probability distribution still gives relative probabilities correctly. But I am unsure about this concept of "relative probability". Does it make sense for divergent distribution?
I'm sure it makes sense if you define things properly. Defining a probability [itex]\mu[/itex] on [itex]\mathbb{N}[/itex] with [itex]\mu\{n\}=1/n[/itex] will not give you a probability distribution and cannot be renormalized to give you one. However, if you want to interpret it as a probability distribution and if you want to say that 1 is twice as probably as 2, then you can do that. You can interpret most definitions and theorems for this "infinite" distribution (however not all theorems are going to hold). So there is nothing wrong with interpreting [itex]\mu[/itex] as an probability distribution (however,strictly speaking it is not one). How useful this interpretation is depends on the context. It might make things a lot clearer, or it might not.


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