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Example for divergent probability distribution ? 
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#1
Nov1312, 05:38 AM

PF Gold
P: 1,150

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? 


#2
Nov1312, 10:23 AM

Mentor
P: 11,617

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. 


#3
Nov1312, 11:41 AM

PF Gold
P: 1,150

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? 


#4
Nov1312, 12:19 PM

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P: 11,617

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. 


#5
Nov1312, 01:42 PM

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