Default positive/(non-negative) probability distributionGaussian RV distributionGenerating random values from non-normal and correlated distributionsFinding the probability that a student is a random guesserProbability given X & Y are independent rand. variables and 2 p.d.f.sRenewal process large time behaviourBivariate and Marginal Probability Distributions: find the value of k that makes this a probability distributionDistribution of discrete function of continuous random variable?deriving $operatornamevar(X)=mathbbE(L)(operatornamevar(D))+mathbbE(D)^2(operatornamevar(L))$How many independent measurement of CPU-time are required such that the difference $|barX - mu|<0.1$ with probability $0.9$ at least?Conditional distribution at time t+1 given information at time t is normally distributed, showing that conditional distribution of sum is also normalUniform Distribution - Is my solution correct?
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Default positive/(non-negative) probability distribution
Gaussian RV distributionGenerating random values from non-normal and correlated distributionsFinding the probability that a student is a random guesserProbability given X & Y are independent rand. variables and 2 p.d.f.sRenewal process large time behaviourBivariate and Marginal Probability Distributions: find the value of k that makes this a probability distributionDistribution of discrete function of continuous random variable?deriving $operatornamevar(X)=mathbbE(L)(operatornamevar(D))+mathbbE(D)^2(operatornamevar(L))$How many independent measurement of CPU-time are required such that the difference $|barX - mu|<0.1$ with probability $0.9$ at least?Conditional distribution at time t+1 given information at time t is normally distributed, showing that conditional distribution of sum is also normalUniform Distribution - Is my solution correct?
$begingroup$
So if you have a random variable that corresponds to a natural phenomenon and you don't know how it is distributed, you often assume it is normally distributed. Now I have a random value that I know is strictly positive, what is the "default" assumed probability distribution for these kind of variables?
My specific case is the volume of air that a human breathes per a random unit of time, which fluctuates from time to time, thus being random if I don't know how it fluctuates. One can easily conclude this number to be strictly positive, since zero means you would be dead, and negative values would be some sort of reverse breathing (photosynthesis maybe? Hehe).
probability statistics probability-distributions
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
$endgroup$
|
show 1 more comment
$begingroup$
So if you have a random variable that corresponds to a natural phenomenon and you don't know how it is distributed, you often assume it is normally distributed. Now I have a random value that I know is strictly positive, what is the "default" assumed probability distribution for these kind of variables?
My specific case is the volume of air that a human breathes per a random unit of time, which fluctuates from time to time, thus being random if I don't know how it fluctuates. One can easily conclude this number to be strictly positive, since zero means you would be dead, and negative values would be some sort of reverse breathing (photosynthesis maybe? Hehe).
probability statistics probability-distributions
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
$endgroup$
$begingroup$
If you have some data to calculate the sample mean and sample variance, then you might take a look at this answer: stats.stackexchange.com/questions/83069/…. For a given mean and variance, the normal distribution maximizes entropy, this answer appears to provide the equivalent for a distribution with strictly nonnegative support.
$endgroup$
– Thoth
Jun 23 '15 at 5:32
$begingroup$
I do have some data, but it is very poor, and I know of several methodical errors made in the collection of the data, so I would rather avoid using it. I don't need the specific distribution btw, just the general form, so for me using the Whatever-distribution would suffice, without having the specific numbers I mean.
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:36
$begingroup$
Note that assuming a normal distribution also requires estimates for the mean and variance. I'm sure there is info online on the average volume of oxygen inhaled per minute/hour whatever.
$endgroup$
– Thoth
Jun 23 '15 at 5:41
$begingroup$
The page you linked to linked to a wiki article that stated the exponential distribution is the maximum entropy distribution with positive support. So this would be the equivalent to the normal distribution for the given support then?
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:43
$begingroup$
It appears to be a generalization of the exponential distribution for the specification of higher moments. There is no 'equivalent' distribution in the absolute sense, it appears to be equivalent in the sense of maximizing entropy. Whether or not maximizing entropy is the right call for capturing your lack of prior knowledge in the model is entirely up to your own judgement of the problem.
$endgroup$
– Thoth
Jun 23 '15 at 5:46
|
show 1 more comment
$begingroup$
So if you have a random variable that corresponds to a natural phenomenon and you don't know how it is distributed, you often assume it is normally distributed. Now I have a random value that I know is strictly positive, what is the "default" assumed probability distribution for these kind of variables?
My specific case is the volume of air that a human breathes per a random unit of time, which fluctuates from time to time, thus being random if I don't know how it fluctuates. One can easily conclude this number to be strictly positive, since zero means you would be dead, and negative values would be some sort of reverse breathing (photosynthesis maybe? Hehe).
probability statistics probability-distributions
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
$endgroup$
So if you have a random variable that corresponds to a natural phenomenon and you don't know how it is distributed, you often assume it is normally distributed. Now I have a random value that I know is strictly positive, what is the "default" assumed probability distribution for these kind of variables?
My specific case is the volume of air that a human breathes per a random unit of time, which fluctuates from time to time, thus being random if I don't know how it fluctuates. One can easily conclude this number to be strictly positive, since zero means you would be dead, and negative values would be some sort of reverse breathing (photosynthesis maybe? Hehe).
probability statistics probability-distributions
probability statistics probability-distributions
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
edited Aug 15 '18 at 6:53
BruceET
36.2k71540
36.2k71540
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
asked Jun 23 '15 at 5:18
Andreas HagenAndreas Hagen
243211
243211
$begingroup$
If you have some data to calculate the sample mean and sample variance, then you might take a look at this answer: stats.stackexchange.com/questions/83069/…. For a given mean and variance, the normal distribution maximizes entropy, this answer appears to provide the equivalent for a distribution with strictly nonnegative support.
$endgroup$
– Thoth
Jun 23 '15 at 5:32
$begingroup$
I do have some data, but it is very poor, and I know of several methodical errors made in the collection of the data, so I would rather avoid using it. I don't need the specific distribution btw, just the general form, so for me using the Whatever-distribution would suffice, without having the specific numbers I mean.
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:36
$begingroup$
Note that assuming a normal distribution also requires estimates for the mean and variance. I'm sure there is info online on the average volume of oxygen inhaled per minute/hour whatever.
$endgroup$
– Thoth
Jun 23 '15 at 5:41
$begingroup$
The page you linked to linked to a wiki article that stated the exponential distribution is the maximum entropy distribution with positive support. So this would be the equivalent to the normal distribution for the given support then?
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:43
$begingroup$
It appears to be a generalization of the exponential distribution for the specification of higher moments. There is no 'equivalent' distribution in the absolute sense, it appears to be equivalent in the sense of maximizing entropy. Whether or not maximizing entropy is the right call for capturing your lack of prior knowledge in the model is entirely up to your own judgement of the problem.
$endgroup$
– Thoth
Jun 23 '15 at 5:46
|
show 1 more comment
$begingroup$
If you have some data to calculate the sample mean and sample variance, then you might take a look at this answer: stats.stackexchange.com/questions/83069/…. For a given mean and variance, the normal distribution maximizes entropy, this answer appears to provide the equivalent for a distribution with strictly nonnegative support.
$endgroup$
– Thoth
Jun 23 '15 at 5:32
$begingroup$
I do have some data, but it is very poor, and I know of several methodical errors made in the collection of the data, so I would rather avoid using it. I don't need the specific distribution btw, just the general form, so for me using the Whatever-distribution would suffice, without having the specific numbers I mean.
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:36
$begingroup$
Note that assuming a normal distribution also requires estimates for the mean and variance. I'm sure there is info online on the average volume of oxygen inhaled per minute/hour whatever.
$endgroup$
– Thoth
Jun 23 '15 at 5:41
$begingroup$
The page you linked to linked to a wiki article that stated the exponential distribution is the maximum entropy distribution with positive support. So this would be the equivalent to the normal distribution for the given support then?
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:43
$begingroup$
It appears to be a generalization of the exponential distribution for the specification of higher moments. There is no 'equivalent' distribution in the absolute sense, it appears to be equivalent in the sense of maximizing entropy. Whether or not maximizing entropy is the right call for capturing your lack of prior knowledge in the model is entirely up to your own judgement of the problem.
$endgroup$
– Thoth
Jun 23 '15 at 5:46
$begingroup$
If you have some data to calculate the sample mean and sample variance, then you might take a look at this answer: stats.stackexchange.com/questions/83069/…. For a given mean and variance, the normal distribution maximizes entropy, this answer appears to provide the equivalent for a distribution with strictly nonnegative support.
$endgroup$
– Thoth
Jun 23 '15 at 5:32
$begingroup$
If you have some data to calculate the sample mean and sample variance, then you might take a look at this answer: stats.stackexchange.com/questions/83069/…. For a given mean and variance, the normal distribution maximizes entropy, this answer appears to provide the equivalent for a distribution with strictly nonnegative support.
$endgroup$
– Thoth
Jun 23 '15 at 5:32
$begingroup$
I do have some data, but it is very poor, and I know of several methodical errors made in the collection of the data, so I would rather avoid using it. I don't need the specific distribution btw, just the general form, so for me using the Whatever-distribution would suffice, without having the specific numbers I mean.
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:36
$begingroup$
I do have some data, but it is very poor, and I know of several methodical errors made in the collection of the data, so I would rather avoid using it. I don't need the specific distribution btw, just the general form, so for me using the Whatever-distribution would suffice, without having the specific numbers I mean.
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:36
$begingroup$
Note that assuming a normal distribution also requires estimates for the mean and variance. I'm sure there is info online on the average volume of oxygen inhaled per minute/hour whatever.
$endgroup$
– Thoth
Jun 23 '15 at 5:41
$begingroup$
Note that assuming a normal distribution also requires estimates for the mean and variance. I'm sure there is info online on the average volume of oxygen inhaled per minute/hour whatever.
$endgroup$
– Thoth
Jun 23 '15 at 5:41
$begingroup$
The page you linked to linked to a wiki article that stated the exponential distribution is the maximum entropy distribution with positive support. So this would be the equivalent to the normal distribution for the given support then?
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:43
$begingroup$
The page you linked to linked to a wiki article that stated the exponential distribution is the maximum entropy distribution with positive support. So this would be the equivalent to the normal distribution for the given support then?
$endgroup$
– Andreas Hagen
Jun 23 '15 at 5:43
$begingroup$
It appears to be a generalization of the exponential distribution for the specification of higher moments. There is no 'equivalent' distribution in the absolute sense, it appears to be equivalent in the sense of maximizing entropy. Whether or not maximizing entropy is the right call for capturing your lack of prior knowledge in the model is entirely up to your own judgement of the problem.
$endgroup$
– Thoth
Jun 23 '15 at 5:46
$begingroup$
It appears to be a generalization of the exponential distribution for the specification of higher moments. There is no 'equivalent' distribution in the absolute sense, it appears to be equivalent in the sense of maximizing entropy. Whether or not maximizing entropy is the right call for capturing your lack of prior knowledge in the model is entirely up to your own judgement of the problem.
$endgroup$
– Thoth
Jun 23 '15 at 5:46
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The normal distribution is often used to model phenomena that give strictly positive results. Example: Heights of women in a particular population with mean $mu = 67$ inches and standard deviation (SD) $sigma = 3.5.$
Technically, using a normal is 'wrong' because a normal distribution has a left
tail that extends to negative infinity, and obviously no women have
negative heights. However, almost all probability is contained within 3 or 4 SD of the mean, so zero (almost 20 SD below) plays no practical role.
Similarly, SAT (and other) exam scores are sometimes described as normal, totally ignoring the minuscule probability that would correspond to nonexistent negative scores.
The gamma family of distributions places all its probability on
the positive half-line. Gamma distributions are right-skewed with
mode < median < mean. The shape is governed by a shape parameter.
For larger shape parameters
the distribution has a left sided tail and a somewhat more pronounced
right sided tail. See graphs of various gamma densities in the Wikipedia article. (A scale or rate parameter helps to determine the
variance of a gamma distribution.)
Gamma distributions are used to model waiting times, and many other phenomena in social, biological, and physical sciences. The gamma family of distributions is not as widely used as the normal family, but if
any family of continuous distributions can be described as the 'default' non-negative family, the gamma family would be the prime candidate.
In queueing theory (concerned with the
behavior of waiting lines), gamma distributions with integer
shape parameters are called Erlang distributions. The chi-squared family is another sub-family of gamma.
Exponential distributions are members of the gamma family with shape parameter 1; strongly skewed with no left-sided tail.
A few other particularly useful continuous non-negative families, among many, are the lognormal (taking logs of data
produces normality), Weibull, Rayleigh and Pareto. (Also discussed in Wikipedia articles.)
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
$endgroup$
1
$begingroup$
Thanks, I'll have a look at the gamma distributions then. =)
$endgroup$
– Andreas Hagen
Jun 23 '15 at 8:30
add a comment |
Your Answer
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1 Answer
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$begingroup$
The normal distribution is often used to model phenomena that give strictly positive results. Example: Heights of women in a particular population with mean $mu = 67$ inches and standard deviation (SD) $sigma = 3.5.$
Technically, using a normal is 'wrong' because a normal distribution has a left
tail that extends to negative infinity, and obviously no women have
negative heights. However, almost all probability is contained within 3 or 4 SD of the mean, so zero (almost 20 SD below) plays no practical role.
Similarly, SAT (and other) exam scores are sometimes described as normal, totally ignoring the minuscule probability that would correspond to nonexistent negative scores.
The gamma family of distributions places all its probability on
the positive half-line. Gamma distributions are right-skewed with
mode < median < mean. The shape is governed by a shape parameter.
For larger shape parameters
the distribution has a left sided tail and a somewhat more pronounced
right sided tail. See graphs of various gamma densities in the Wikipedia article. (A scale or rate parameter helps to determine the
variance of a gamma distribution.)
Gamma distributions are used to model waiting times, and many other phenomena in social, biological, and physical sciences. The gamma family of distributions is not as widely used as the normal family, but if
any family of continuous distributions can be described as the 'default' non-negative family, the gamma family would be the prime candidate.
In queueing theory (concerned with the
behavior of waiting lines), gamma distributions with integer
shape parameters are called Erlang distributions. The chi-squared family is another sub-family of gamma.
Exponential distributions are members of the gamma family with shape parameter 1; strongly skewed with no left-sided tail.
A few other particularly useful continuous non-negative families, among many, are the lognormal (taking logs of data
produces normality), Weibull, Rayleigh and Pareto. (Also discussed in Wikipedia articles.)
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
$endgroup$
1
$begingroup$
Thanks, I'll have a look at the gamma distributions then. =)
$endgroup$
– Andreas Hagen
Jun 23 '15 at 8:30
add a comment |
$begingroup$
The normal distribution is often used to model phenomena that give strictly positive results. Example: Heights of women in a particular population with mean $mu = 67$ inches and standard deviation (SD) $sigma = 3.5.$
Technically, using a normal is 'wrong' because a normal distribution has a left
tail that extends to negative infinity, and obviously no women have
negative heights. However, almost all probability is contained within 3 or 4 SD of the mean, so zero (almost 20 SD below) plays no practical role.
Similarly, SAT (and other) exam scores are sometimes described as normal, totally ignoring the minuscule probability that would correspond to nonexistent negative scores.
The gamma family of distributions places all its probability on
the positive half-line. Gamma distributions are right-skewed with
mode < median < mean. The shape is governed by a shape parameter.
For larger shape parameters
the distribution has a left sided tail and a somewhat more pronounced
right sided tail. See graphs of various gamma densities in the Wikipedia article. (A scale or rate parameter helps to determine the
variance of a gamma distribution.)
Gamma distributions are used to model waiting times, and many other phenomena in social, biological, and physical sciences. The gamma family of distributions is not as widely used as the normal family, but if
any family of continuous distributions can be described as the 'default' non-negative family, the gamma family would be the prime candidate.
In queueing theory (concerned with the
behavior of waiting lines), gamma distributions with integer
shape parameters are called Erlang distributions. The chi-squared family is another sub-family of gamma.
Exponential distributions are members of the gamma family with shape parameter 1; strongly skewed with no left-sided tail.
A few other particularly useful continuous non-negative families, among many, are the lognormal (taking logs of data
produces normality), Weibull, Rayleigh and Pareto. (Also discussed in Wikipedia articles.)
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
$endgroup$
1
$begingroup$
Thanks, I'll have a look at the gamma distributions then. =)
$endgroup$
– Andreas Hagen
Jun 23 '15 at 8:30
add a comment |
$begingroup$
The normal distribution is often used to model phenomena that give strictly positive results. Example: Heights of women in a particular population with mean $mu = 67$ inches and standard deviation (SD) $sigma = 3.5.$
Technically, using a normal is 'wrong' because a normal distribution has a left
tail that extends to negative infinity, and obviously no women have
negative heights. However, almost all probability is contained within 3 or 4 SD of the mean, so zero (almost 20 SD below) plays no practical role.
Similarly, SAT (and other) exam scores are sometimes described as normal, totally ignoring the minuscule probability that would correspond to nonexistent negative scores.
The gamma family of distributions places all its probability on
the positive half-line. Gamma distributions are right-skewed with
mode < median < mean. The shape is governed by a shape parameter.
For larger shape parameters
the distribution has a left sided tail and a somewhat more pronounced
right sided tail. See graphs of various gamma densities in the Wikipedia article. (A scale or rate parameter helps to determine the
variance of a gamma distribution.)
Gamma distributions are used to model waiting times, and many other phenomena in social, biological, and physical sciences. The gamma family of distributions is not as widely used as the normal family, but if
any family of continuous distributions can be described as the 'default' non-negative family, the gamma family would be the prime candidate.
In queueing theory (concerned with the
behavior of waiting lines), gamma distributions with integer
shape parameters are called Erlang distributions. The chi-squared family is another sub-family of gamma.
Exponential distributions are members of the gamma family with shape parameter 1; strongly skewed with no left-sided tail.
A few other particularly useful continuous non-negative families, among many, are the lognormal (taking logs of data
produces normality), Weibull, Rayleigh and Pareto. (Also discussed in Wikipedia articles.)
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
$endgroup$
The normal distribution is often used to model phenomena that give strictly positive results. Example: Heights of women in a particular population with mean $mu = 67$ inches and standard deviation (SD) $sigma = 3.5.$
Technically, using a normal is 'wrong' because a normal distribution has a left
tail that extends to negative infinity, and obviously no women have
negative heights. However, almost all probability is contained within 3 or 4 SD of the mean, so zero (almost 20 SD below) plays no practical role.
Similarly, SAT (and other) exam scores are sometimes described as normal, totally ignoring the minuscule probability that would correspond to nonexistent negative scores.
The gamma family of distributions places all its probability on
the positive half-line. Gamma distributions are right-skewed with
mode < median < mean. The shape is governed by a shape parameter.
For larger shape parameters
the distribution has a left sided tail and a somewhat more pronounced
right sided tail. See graphs of various gamma densities in the Wikipedia article. (A scale or rate parameter helps to determine the
variance of a gamma distribution.)
Gamma distributions are used to model waiting times, and many other phenomena in social, biological, and physical sciences. The gamma family of distributions is not as widely used as the normal family, but if
any family of continuous distributions can be described as the 'default' non-negative family, the gamma family would be the prime candidate.
In queueing theory (concerned with the
behavior of waiting lines), gamma distributions with integer
shape parameters are called Erlang distributions. The chi-squared family is another sub-family of gamma.
Exponential distributions are members of the gamma family with shape parameter 1; strongly skewed with no left-sided tail.
A few other particularly useful continuous non-negative families, among many, are the lognormal (taking logs of data
produces normality), Weibull, Rayleigh and Pareto. (Also discussed in Wikipedia articles.)
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
edited Mar 21 at 17:11
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
answered Jun 23 '15 at 6:36
BruceETBruceET
36.2k71540
36.2k71540
1
$begingroup$
Thanks, I'll have a look at the gamma distributions then. =)
$endgroup$
– Andreas Hagen
Jun 23 '15 at 8:30
add a comment |
1
$begingroup$
Thanks, I'll have a look at the gamma distributions then. =)
$endgroup$
– Andreas Hagen
Jun 23 '15 at 8:30
1
1
$begingroup$
Thanks, I'll have a look at the gamma distributions then. =)
$endgroup$
– Andreas Hagen
Jun 23 '15 at 8:30
$begingroup$
Thanks, I'll have a look at the gamma distributions then. =)
$endgroup$
– Andreas Hagen
Jun 23 '15 at 8:30
add a comment |
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If you have some data to calculate the sample mean and sample variance, then you might take a look at this answer: stats.stackexchange.com/questions/83069/…. For a given mean and variance, the normal distribution maximizes entropy, this answer appears to provide the equivalent for a distribution with strictly nonnegative support.
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– Thoth
Jun 23 '15 at 5:32
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I do have some data, but it is very poor, and I know of several methodical errors made in the collection of the data, so I would rather avoid using it. I don't need the specific distribution btw, just the general form, so for me using the Whatever-distribution would suffice, without having the specific numbers I mean.
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– Andreas Hagen
Jun 23 '15 at 5:36
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Note that assuming a normal distribution also requires estimates for the mean and variance. I'm sure there is info online on the average volume of oxygen inhaled per minute/hour whatever.
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– Thoth
Jun 23 '15 at 5:41
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The page you linked to linked to a wiki article that stated the exponential distribution is the maximum entropy distribution with positive support. So this would be the equivalent to the normal distribution for the given support then?
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– Andreas Hagen
Jun 23 '15 at 5:43
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It appears to be a generalization of the exponential distribution for the specification of higher moments. There is no 'equivalent' distribution in the absolute sense, it appears to be equivalent in the sense of maximizing entropy. Whether or not maximizing entropy is the right call for capturing your lack of prior knowledge in the model is entirely up to your own judgement of the problem.
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– Thoth
Jun 23 '15 at 5:46