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Identifying a distribution

Converting arbitrary distribution to uniform oneRandomized trace techniqueDistribution of product of sums of a setWhy does the scaling exponent of a power law fit change so radically when the data is scaled by a constant?Why is Kernel Density Estimation failing to find bandwidth?generate frequency distribution from databaseUnderstanding binomial distribution in iris recognitionMatlab: check if normal distributedBootstrap method and extensionsRe-scaling of exponentially distributed random numbers

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

$begingroup$

I am trying to understand what sort of distribution is produced by the following code.

Using the following Matlab code we can generate a distribution (normalized such that the sum of the set is equal to 1):

M=500; % Number of samples
z=5;
SUM = 1;
ns = rand(1,M).^z; % Uniformly distributed random numbers to the power of z
TOT = sum(ns);
X = (ns/TOT)*SUM; % Re-scaling
hist(X(1,:),100)

For the exponent $z=1$ the sampling distribution is essentially "flat", and it also has a small range. Enlarging $z$ gives a wider range of numbers. Here is an example:

I would like to classify this distribution and understand the underlying mathematics.

Any explanations would be greatly appreciated.

distributions matlab exponential

share|cite|improve this question

asked 9 hours ago

MerinMerin

374

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Similar method, different distribution. Not sure that makes it a duplicate.
  $endgroup$
  – BruceET
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @AdamO the question you link to is about $e^X$ where $X sim textUniform(0, 1)$ while this question is asking about $X^z$. This leads to a different distribution.
  $endgroup$
  – olooney
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I am trying to understand what sort of distribution is produced by the following code.

Using the following Matlab code we can generate a distribution (normalized such that the sum of the set is equal to 1):

M=500; % Number of samples
z=5;
SUM = 1;
ns = rand(1,M).^z; % Uniformly distributed random numbers to the power of z
TOT = sum(ns);
X = (ns/TOT)*SUM; % Re-scaling
hist(X(1,:),100)

For the exponent $z=1$ the sampling distribution is essentially "flat", and it also has a small range. Enlarging $z$ gives a wider range of numbers. Here is an example:

I would like to classify this distribution and understand the underlying mathematics.

Any explanations would be greatly appreciated.

distributions matlab exponential

share|cite|improve this question

asked 9 hours ago

MerinMerin

374

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Similar method, different distribution. Not sure that makes it a duplicate.
  $endgroup$
  – BruceET
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @AdamO the question you link to is about $e^X$ where $X sim textUniform(0, 1)$ while this question is asking about $X^z$. This leads to a different distribution.
  $endgroup$
  – olooney
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to understand what sort of distribution is produced by the following code.

Using the following Matlab code we can generate a distribution (normalized such that the sum of the set is equal to 1):

M=500; % Number of samples
z=5;
SUM = 1;
ns = rand(1,M).^z; % Uniformly distributed random numbers to the power of z
TOT = sum(ns);
X = (ns/TOT)*SUM; % Re-scaling
hist(X(1,:),100)

For the exponent $z=1$ the sampling distribution is essentially "flat", and it also has a small range. Enlarging $z$ gives a wider range of numbers. Here is an example:

I would like to classify this distribution and understand the underlying mathematics.

Any explanations would be greatly appreciated.

distributions matlab exponential

share|cite|improve this question

asked 9 hours ago

MerinMerin

374

$endgroup$

M=500; % Number of samples
z=5;
SUM = 1;
ns = rand(1,M).^z; % Uniformly distributed random numbers to the power of z
TOT = sum(ns);
X = (ns/TOT)*SUM; % Re-scaling
hist(X(1,:),100)

share|cite|improve this question

asked 9 hours ago

MerinMerin

374

share|cite|improve this question

asked 9 hours ago

MerinMerin

374

asked 9 hours ago

MerinMerin

374

asked 9 hours ago

MerinMerin

374

1 Answer
1

active

oldest

votes

5

$begingroup$

I will show this for $U^2,$ instead of $U^5.$ See @olooney's comment below my Answer.

Short answer: If $U sim mathsfUnif(0,1),$ then $X = U^2 sim mathsfBeta(frac 1 2, 1).$

Proof: For $x in (0,1),$ we have
$$F_X(x) = P(X le x) = P(U^2 le x) = P(U le x^1/2) = x^1/2.$$
Then $f_X(x) = F_X^prime(x)$ is easily seen to be the density of
$mathsfBeta(frac 12, 1).$ See Wikipedia on beta distributions.

Demonstration by simulation:

set.seed(618) 
u = runif(10^6); x = u^2
mean(x)
[1] 0.3330528 # aprx E(X) = 1/3 from simulation
.5/(.5+1)
[1] 0.3333333 # exact E(X) from formula

cutp.u = seq(0, 1, by = 0.1); cutp.x = cutp.u^2
frb = rainbow(12)
par(mfrow = c(1,3))
 hist(u, prob=T, br=cutp.u, col=frb, ylim=c(0,10), main="UNIF(0,1)")
 curve(dunif(x), 0, 1, add=T, n=10001, lwd=2)
 hist(x, prob=T, br=cutp.x, col=frb, main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, lwd=2)
 hist(x, prob=T, br=40, col="skyblue2", main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, n=10001, col="red")
par(mfrow = c(1,1))

In the first and second histograms, each bar contains about 100,000 simulated values. For the beta distribution in the middle histogram,
each bar is the image of a bar of the same color in the first histogram.
Ordinarily, it is not useful to make histograms with bins of unequal width, so the third histogram shows the simulated beta distribution
(and its density function in red) in a more familiar way.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

BruceETBruceET

9,2231824

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice! However, because the question concerns the $z=5$ power rather than the square, it would be better to generalize your answer a little.
  $endgroup$
  – whuber♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Generalizes to $X^z sim textBeta(1/z, 1)$, where $X sim textUniform(0, 1)$.
  $endgroup$
  – olooney
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Both absolutely correct. I'm still fussing with this. Will edit accordingly: (a) Not sure if it's homework, so want to show method without just being an answerbook. (b) Difficult to show clear graphs for $z = 5.$
  $endgroup$
  – BruceET
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for the explanations. No, this is not a homework problem. I was working on a physical simulation looking for a distribution that can model the distribution of light in multimode optical fibers. So I came across this distribution and I wasn't quite sure what it was.
  $endgroup$
  – Merin
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems nobody (including the OP) cares about the summation performed after the transformation. The resulting RVs are of the form $$fracX_isum_i X_i$$
  $endgroup$
  – gunes
  37 mins ago
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

I will show this for $U^2,$ instead of $U^5.$ See @olooney's comment below my Answer.

Short answer: If $U sim mathsfUnif(0,1),$ then $X = U^2 sim mathsfBeta(frac 1 2, 1).$

Proof: For $x in (0,1),$ we have
$$F_X(x) = P(X le x) = P(U^2 le x) = P(U le x^1/2) = x^1/2.$$
Then $f_X(x) = F_X^prime(x)$ is easily seen to be the density of
$mathsfBeta(frac 12, 1).$ See Wikipedia on beta distributions.

Demonstration by simulation:

set.seed(618) 
u = runif(10^6); x = u^2
mean(x)
[1] 0.3330528 # aprx E(X) = 1/3 from simulation
.5/(.5+1)
[1] 0.3333333 # exact E(X) from formula

cutp.u = seq(0, 1, by = 0.1); cutp.x = cutp.u^2
frb = rainbow(12)
par(mfrow = c(1,3))
 hist(u, prob=T, br=cutp.u, col=frb, ylim=c(0,10), main="UNIF(0,1)")
 curve(dunif(x), 0, 1, add=T, n=10001, lwd=2)
 hist(x, prob=T, br=cutp.x, col=frb, main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, lwd=2)
 hist(x, prob=T, br=40, col="skyblue2", main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, n=10001, col="red")
par(mfrow = c(1,1))

In the first and second histograms, each bar contains about 100,000 simulated values. For the beta distribution in the middle histogram,
each bar is the image of a bar of the same color in the first histogram.
Ordinarily, it is not useful to make histograms with bins of unequal width, so the third histogram shows the simulated beta distribution
(and its density function in red) in a more familiar way.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

BruceETBruceET

9,2231824

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice! However, because the question concerns the $z=5$ power rather than the square, it would be better to generalize your answer a little.
  $endgroup$
  – whuber♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Generalizes to $X^z sim textBeta(1/z, 1)$, where $X sim textUniform(0, 1)$.
  $endgroup$
  – olooney
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Both absolutely correct. I'm still fussing with this. Will edit accordingly: (a) Not sure if it's homework, so want to show method without just being an answerbook. (b) Difficult to show clear graphs for $z = 5.$
  $endgroup$
  – BruceET
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for the explanations. No, this is not a homework problem. I was working on a physical simulation looking for a distribution that can model the distribution of light in multimode optical fibers. So I came across this distribution and I wasn't quite sure what it was.
  $endgroup$
  – Merin
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems nobody (including the OP) cares about the summation performed after the transformation. The resulting RVs are of the form $$fracX_isum_i X_i$$
  $endgroup$
  – gunes
  37 mins ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

I will show this for $U^2,$ instead of $U^5.$ See @olooney's comment below my Answer.

Short answer: If $U sim mathsfUnif(0,1),$ then $X = U^2 sim mathsfBeta(frac 1 2, 1).$

Proof: For $x in (0,1),$ we have
$$F_X(x) = P(X le x) = P(U^2 le x) = P(U le x^1/2) = x^1/2.$$
Then $f_X(x) = F_X^prime(x)$ is easily seen to be the density of
$mathsfBeta(frac 12, 1).$ See Wikipedia on beta distributions.

Demonstration by simulation:

set.seed(618) 
u = runif(10^6); x = u^2
mean(x)
[1] 0.3330528 # aprx E(X) = 1/3 from simulation
.5/(.5+1)
[1] 0.3333333 # exact E(X) from formula

cutp.u = seq(0, 1, by = 0.1); cutp.x = cutp.u^2
frb = rainbow(12)
par(mfrow = c(1,3))
 hist(u, prob=T, br=cutp.u, col=frb, ylim=c(0,10), main="UNIF(0,1)")
 curve(dunif(x), 0, 1, add=T, n=10001, lwd=2)
 hist(x, prob=T, br=cutp.x, col=frb, main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, lwd=2)
 hist(x, prob=T, br=40, col="skyblue2", main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, n=10001, col="red")
par(mfrow = c(1,1))

In the first and second histograms, each bar contains about 100,000 simulated values. For the beta distribution in the middle histogram,
each bar is the image of a bar of the same color in the first histogram.
Ordinarily, it is not useful to make histograms with bins of unequal width, so the third histogram shows the simulated beta distribution
(and its density function in red) in a more familiar way.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

BruceETBruceET

9,2231824

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice! However, because the question concerns the $z=5$ power rather than the square, it would be better to generalize your answer a little.
  $endgroup$
  – whuber♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Generalizes to $X^z sim textBeta(1/z, 1)$, where $X sim textUniform(0, 1)$.
  $endgroup$
  – olooney
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Both absolutely correct. I'm still fussing with this. Will edit accordingly: (a) Not sure if it's homework, so want to show method without just being an answerbook. (b) Difficult to show clear graphs for $z = 5.$
  $endgroup$
  – BruceET
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for the explanations. No, this is not a homework problem. I was working on a physical simulation looking for a distribution that can model the distribution of light in multimode optical fibers. So I came across this distribution and I wasn't quite sure what it was.
  $endgroup$
  – Merin
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems nobody (including the OP) cares about the summation performed after the transformation. The resulting RVs are of the form $$fracX_isum_i X_i$$
  $endgroup$
  – gunes
  37 mins ago
  
  

  
  
  
  

add a comment | 

$begingroup$

I will show this for $U^2,$ instead of $U^5.$ See @olooney's comment below my Answer.

Short answer: If $U sim mathsfUnif(0,1),$ then $X = U^2 sim mathsfBeta(frac 1 2, 1).$

Proof: For $x in (0,1),$ we have
$$F_X(x) = P(X le x) = P(U^2 le x) = P(U le x^1/2) = x^1/2.$$
Then $f_X(x) = F_X^prime(x)$ is easily seen to be the density of
$mathsfBeta(frac 12, 1).$ See Wikipedia on beta distributions.

Demonstration by simulation:

set.seed(618) 
u = runif(10^6); x = u^2
mean(x)
[1] 0.3330528 # aprx E(X) = 1/3 from simulation
.5/(.5+1)
[1] 0.3333333 # exact E(X) from formula

cutp.u = seq(0, 1, by = 0.1); cutp.x = cutp.u^2
frb = rainbow(12)
par(mfrow = c(1,3))
 hist(u, prob=T, br=cutp.u, col=frb, ylim=c(0,10), main="UNIF(0,1)")
 curve(dunif(x), 0, 1, add=T, n=10001, lwd=2)
 hist(x, prob=T, br=cutp.x, col=frb, main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, lwd=2)
 hist(x, prob=T, br=40, col="skyblue2", main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, n=10001, col="red")
par(mfrow = c(1,1))

In the first and second histograms, each bar contains about 100,000 simulated values. For the beta distribution in the middle histogram,
each bar is the image of a bar of the same color in the first histogram.
Ordinarily, it is not useful to make histograms with bins of unequal width, so the third histogram shows the simulated beta distribution
(and its density function in red) in a more familiar way.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

BruceETBruceET

9,2231824

$endgroup$

set.seed(618) 
u = runif(10^6); x = u^2
mean(x)
[1] 0.3330528 # aprx E(X) = 1/3 from simulation
.5/(.5+1)
[1] 0.3333333 # exact E(X) from formula

cutp.u = seq(0, 1, by = 0.1); cutp.x = cutp.u^2
frb = rainbow(12)
par(mfrow = c(1,3))
 hist(u, prob=T, br=cutp.u, col=frb, ylim=c(0,10), main="UNIF(0,1)")
 curve(dunif(x), 0, 1, add=T, n=10001, lwd=2)
 hist(x, prob=T, br=cutp.x, col=frb, main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, lwd=2)
 hist(x, prob=T, br=40, col="skyblue2", main="BETA(.5,1)")
 curve(dbeta(x, .5, 1), add=T, n=10001, col="red")
par(mfrow = c(1,1))

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

BruceETBruceET

9,2231824

answered 8 hours ago

BruceETBruceET

9,2231824

answered 8 hours ago

BruceETBruceET

9,2231824