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3

1

$begingroup$

Is the following distribution a still normal one?

The range of values is constrained to hard limits $ 0, 255 $. And it's generated by $ mathcalN(mu, sigma^2) operatornamemod 256 + 128^dagger $. So it's not that I'm simply truncating a normal, but folding it back into itself. Consequently the probability is never asymptotic to 0.

Normal, or is there a more appropriate term for it?

---

$dagger$ The +128 is just to centre the peak otherwise it looks silly as $mu = 0$. $sigma sim 12; < 20$. I expect that $mu$ can be ignored for the purposes of classification.

In detail, it's a plot of 128 + (sample1 - sample2) & 0xff. I wouldn't fixate on the centring too much as there's some weird +/- stuff to do with computer networking and transmission of unsigned bytes.

distributions

share|cite|improve this question

edited Apr 21 at 14:21

Paul Uszak

asked Apr 21 at 13:34

Paul UszakPaul Uszak

199116

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you give the values of $mu$ and $sigma^2$? Also, there seems to be something off: the result of a $textmod 256$ operation will be between 0 and 256, so the result after adding 128 should be between 126 and 384, unlike your picture. Did you simply not add 128 before plotting?
  $endgroup$
  – Stephan Kolassa
  Apr 21 at 13:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think it would hold for any $mu$ and $sigma$, and for some paramters it would just like subtracting a value.
  $endgroup$
  – Lerner Zhang
  Apr 21 at 14:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems related to the wrapped normal distribution also discussed here
  $endgroup$
  – jnez71
  Apr 21 at 18:54
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Is the following distribution a still normal one?

The range of values is constrained to hard limits $ 0, 255 $. And it's generated by $ mathcalN(mu, sigma^2) operatornamemod 256 + 128^dagger $. So it's not that I'm simply truncating a normal, but folding it back into itself. Consequently the probability is never asymptotic to 0.

Normal, or is there a more appropriate term for it?

---

$dagger$ The +128 is just to centre the peak otherwise it looks silly as $mu = 0$. $sigma sim 12; < 20$. I expect that $mu$ can be ignored for the purposes of classification.

In detail, it's a plot of 128 + (sample1 - sample2) & 0xff. I wouldn't fixate on the centring too much as there's some weird +/- stuff to do with computer networking and transmission of unsigned bytes.

distributions

share|cite|improve this question

edited Apr 21 at 14:21

Paul Uszak

asked Apr 21 at 13:34

Paul UszakPaul Uszak

199116

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you give the values of $mu$ and $sigma^2$? Also, there seems to be something off: the result of a $textmod 256$ operation will be between 0 and 256, so the result after adding 128 should be between 126 and 384, unlike your picture. Did you simply not add 128 before plotting?
  $endgroup$
  – Stephan Kolassa
  Apr 21 at 13:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think it would hold for any $mu$ and $sigma$, and for some paramters it would just like subtracting a value.
  $endgroup$
  – Lerner Zhang
  Apr 21 at 14:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems related to the wrapped normal distribution also discussed here
  $endgroup$
  – jnez71
  Apr 21 at 18:54
  

  
  
  
  

add a comment | 

$begingroup$

Is the following distribution a still normal one?

The range of values is constrained to hard limits $ 0, 255 $. And it's generated by $ mathcalN(mu, sigma^2) operatornamemod 256 + 128^dagger $. So it's not that I'm simply truncating a normal, but folding it back into itself. Consequently the probability is never asymptotic to 0.

Normal, or is there a more appropriate term for it?

---

$dagger$ The +128 is just to centre the peak otherwise it looks silly as $mu = 0$. $sigma sim 12; < 20$. I expect that $mu$ can be ignored for the purposes of classification.

In detail, it's a plot of 128 + (sample1 - sample2) & 0xff. I wouldn't fixate on the centring too much as there's some weird +/- stuff to do with computer networking and transmission of unsigned bytes.

distributions

share|cite|improve this question

edited Apr 21 at 14:21

Paul Uszak

asked Apr 21 at 13:34

Paul UszakPaul Uszak

199116

$endgroup$

share|cite|improve this question

edited Apr 21 at 14:21

Paul Uszak

asked Apr 21 at 13:34

Paul UszakPaul Uszak

199116

share|cite|improve this question

edited Apr 21 at 14:21

Paul Uszak

asked Apr 21 at 13:34

Paul UszakPaul Uszak

199116

asked Apr 21 at 13:34

Paul UszakPaul Uszak

199116

asked Apr 21 at 13:34

Paul UszakPaul Uszak

199116

1 Answer
1

active

oldest

votes

4

$begingroup$

1. No, by definition your distribution is not normal. The normal distribution has unbounded support, yours doesn't. QED.




2. Often enough, we don't need a normal distribution, only one that is "normal enough". Your distribution may well be normal enough for whatever you want to use it for.



3. 
   

   However, I don't think a normal distribution is really the best way to parameterize your data. Specifically, even playing around with the standard deviation, there doesn't seem to be a way to get the kurtosis your data exhibits:

   
   
   

   

   
   
   

   Note how we either get all the mass over a much smaller part of the horizontal axis for small SDs, or get much more mass in the center of the distribution than in your picture for large SDs.

   
   
   

   mu <- 128
   ss <- c(5,10,20,40)
   nn <- 1e7
   
   par(mfrow=c(2,2),mai=c(.5,.5,.5,.1))
   for ( ii in 1:4 ) 
    set.seed(1)
    hist(rnorm(nn,mu,ss[ii])%%256,col="grey",
    freq=FALSE,xlim=c(0,256),xlab="",ylab="",main=paste("SD =",ss[ii]))
   
   

   
   
   

   I also tried a $t$ distribution by varying the degrees of freedom, but that didn't look very good, either.

   
   

So, if your distribution is not normal enough for your purposes, you may want to look at something like a Pearson type VII distribution. If you truncate this using the modulo operator, it will again not be a Pearson VII strictly speaking, but it may again be "sufficiently Pearson VII".

share|cite|improve this answer

answered Apr 21 at 14:19

Stephan KolassaStephan Kolassa

48.5k8102182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I called it Normal as I assumed that '(sample - sample)' = Normal, even if 'sample' itself is not Normal. 'sample' actually looks more like log-Normal. Would that explain the kurtosis problem?
  $endgroup$
  – Paul Uszak
  Apr 21 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps it could be normal in the wrapped distribution sense of directional statistics?
  $endgroup$
  – jnez71
  Apr 21 at 19:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is possible. I'm afraid you have lost me with the sample, and its log-normality. Could you give a few more details?
  $endgroup$
  – Stephan Kolassa
  Apr 21 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My understanding of it isn't fantastic (otherwise I'd post an answer) but I think the idea is that when you have a probability space with a sample set that is not R but rather some cyclic group like SO2 or perhaps R mod 256, the mathematically accepted way to generalize any distribution on R to these spaces is to, well, do what the OP did and "identify" (via modulo) various samples as each other, summing the probabilities ad infinitum, like the wiki article describes: p_wrapped(x) = sum_over_all_k_in_N(p(x + k*r)) where r is the mod value (so 256 for the OP)
  $endgroup$
  – jnez71
  Apr 21 at 21:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the normal distribution you get this which I suspect retains most of the properties we'd hope for (though I'm not 100% sure). Like perhaps a central limit theorem for this mod space?
  $endgroup$
  – jnez71
  Apr 21 at 21:29
  

  
  
  
  

add a comment | 

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4

$begingroup$

1. No, by definition your distribution is not normal. The normal distribution has unbounded support, yours doesn't. QED.




2. Often enough, we don't need a normal distribution, only one that is "normal enough". Your distribution may well be normal enough for whatever you want to use it for.



3. 
   

   However, I don't think a normal distribution is really the best way to parameterize your data. Specifically, even playing around with the standard deviation, there doesn't seem to be a way to get the kurtosis your data exhibits:

   
   
   

   

   
   
   

   Note how we either get all the mass over a much smaller part of the horizontal axis for small SDs, or get much more mass in the center of the distribution than in your picture for large SDs.

   
   
   

   mu <- 128
   ss <- c(5,10,20,40)
   nn <- 1e7
   
   par(mfrow=c(2,2),mai=c(.5,.5,.5,.1))
   for ( ii in 1:4 ) 
    set.seed(1)
    hist(rnorm(nn,mu,ss[ii])%%256,col="grey",
    freq=FALSE,xlim=c(0,256),xlab="",ylab="",main=paste("SD =",ss[ii]))
   
   

   
   
   

   I also tried a $t$ distribution by varying the degrees of freedom, but that didn't look very good, either.

   
   

So, if your distribution is not normal enough for your purposes, you may want to look at something like a Pearson type VII distribution. If you truncate this using the modulo operator, it will again not be a Pearson VII strictly speaking, but it may again be "sufficiently Pearson VII".

share|cite|improve this answer

answered Apr 21 at 14:19

Stephan KolassaStephan Kolassa

48.5k8102182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I called it Normal as I assumed that '(sample - sample)' = Normal, even if 'sample' itself is not Normal. 'sample' actually looks more like log-Normal. Would that explain the kurtosis problem?
  $endgroup$
  – Paul Uszak
  Apr 21 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps it could be normal in the wrapped distribution sense of directional statistics?
  $endgroup$
  – jnez71
  Apr 21 at 19:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is possible. I'm afraid you have lost me with the sample, and its log-normality. Could you give a few more details?
  $endgroup$
  – Stephan Kolassa
  Apr 21 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My understanding of it isn't fantastic (otherwise I'd post an answer) but I think the idea is that when you have a probability space with a sample set that is not R but rather some cyclic group like SO2 or perhaps R mod 256, the mathematically accepted way to generalize any distribution on R to these spaces is to, well, do what the OP did and "identify" (via modulo) various samples as each other, summing the probabilities ad infinitum, like the wiki article describes: p_wrapped(x) = sum_over_all_k_in_N(p(x + k*r)) where r is the mod value (so 256 for the OP)
  $endgroup$
  – jnez71
  Apr 21 at 21:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the normal distribution you get this which I suspect retains most of the properties we'd hope for (though I'm not 100% sure). Like perhaps a central limit theorem for this mod space?
  $endgroup$
  – jnez71
  Apr 21 at 21:29
  

  
  
  
  

add a comment | 

4

$begingroup$

1. No, by definition your distribution is not normal. The normal distribution has unbounded support, yours doesn't. QED.




2. Often enough, we don't need a normal distribution, only one that is "normal enough". Your distribution may well be normal enough for whatever you want to use it for.



3. 
   

   However, I don't think a normal distribution is really the best way to parameterize your data. Specifically, even playing around with the standard deviation, there doesn't seem to be a way to get the kurtosis your data exhibits:

   
   
   

   

   
   
   

   Note how we either get all the mass over a much smaller part of the horizontal axis for small SDs, or get much more mass in the center of the distribution than in your picture for large SDs.

   
   
   

   mu <- 128
   ss <- c(5,10,20,40)
   nn <- 1e7
   
   par(mfrow=c(2,2),mai=c(.5,.5,.5,.1))
   for ( ii in 1:4 ) 
    set.seed(1)
    hist(rnorm(nn,mu,ss[ii])%%256,col="grey",
    freq=FALSE,xlim=c(0,256),xlab="",ylab="",main=paste("SD =",ss[ii]))
   
   

   
   
   

   I also tried a $t$ distribution by varying the degrees of freedom, but that didn't look very good, either.

   
   

So, if your distribution is not normal enough for your purposes, you may want to look at something like a Pearson type VII distribution. If you truncate this using the modulo operator, it will again not be a Pearson VII strictly speaking, but it may again be "sufficiently Pearson VII".

share|cite|improve this answer

answered Apr 21 at 14:19

Stephan KolassaStephan Kolassa

48.5k8102182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I called it Normal as I assumed that '(sample - sample)' = Normal, even if 'sample' itself is not Normal. 'sample' actually looks more like log-Normal. Would that explain the kurtosis problem?
  $endgroup$
  – Paul Uszak
  Apr 21 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps it could be normal in the wrapped distribution sense of directional statistics?
  $endgroup$
  – jnez71
  Apr 21 at 19:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is possible. I'm afraid you have lost me with the sample, and its log-normality. Could you give a few more details?
  $endgroup$
  – Stephan Kolassa
  Apr 21 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My understanding of it isn't fantastic (otherwise I'd post an answer) but I think the idea is that when you have a probability space with a sample set that is not R but rather some cyclic group like SO2 or perhaps R mod 256, the mathematically accepted way to generalize any distribution on R to these spaces is to, well, do what the OP did and "identify" (via modulo) various samples as each other, summing the probabilities ad infinitum, like the wiki article describes: p_wrapped(x) = sum_over_all_k_in_N(p(x + k*r)) where r is the mod value (so 256 for the OP)
  $endgroup$
  – jnez71
  Apr 21 at 21:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the normal distribution you get this which I suspect retains most of the properties we'd hope for (though I'm not 100% sure). Like perhaps a central limit theorem for this mod space?
  $endgroup$
  – jnez71
  Apr 21 at 21:29
  

  
  
  
  

add a comment | 

$begingroup$

1. No, by definition your distribution is not normal. The normal distribution has unbounded support, yours doesn't. QED.




2. Often enough, we don't need a normal distribution, only one that is "normal enough". Your distribution may well be normal enough for whatever you want to use it for.



3. 
   

   However, I don't think a normal distribution is really the best way to parameterize your data. Specifically, even playing around with the standard deviation, there doesn't seem to be a way to get the kurtosis your data exhibits:

   
   
   

   

   
   
   

   Note how we either get all the mass over a much smaller part of the horizontal axis for small SDs, or get much more mass in the center of the distribution than in your picture for large SDs.

   
   
   

   mu <- 128
   ss <- c(5,10,20,40)
   nn <- 1e7
   
   par(mfrow=c(2,2),mai=c(.5,.5,.5,.1))
   for ( ii in 1:4 ) 
    set.seed(1)
    hist(rnorm(nn,mu,ss[ii])%%256,col="grey",
    freq=FALSE,xlim=c(0,256),xlab="",ylab="",main=paste("SD =",ss[ii]))
   
   

   
   
   

   I also tried a $t$ distribution by varying the degrees of freedom, but that didn't look very good, either.

   
   

So, if your distribution is not normal enough for your purposes, you may want to look at something like a Pearson type VII distribution. If you truncate this using the modulo operator, it will again not be a Pearson VII strictly speaking, but it may again be "sufficiently Pearson VII".

share|cite|improve this answer

answered Apr 21 at 14:19

Stephan KolassaStephan Kolassa

48.5k8102182

$endgroup$

share|cite|improve this answer

answered Apr 21 at 14:19

Stephan KolassaStephan Kolassa

48.5k8102182

answered Apr 21 at 14:19

Stephan KolassaStephan Kolassa

48.5k8102182

answered Apr 21 at 14:19

Stephan KolassaStephan Kolassa

48.5k8102182