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What is the probability distribution of linear formula?

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2

$begingroup$

What is the distribution name when probability values have a linear increasing of the form:

$p(i)= frac2N(N+1)i; 0 leq i leq N $

probability distributions

share|cite|improve this question

edited 4 hours ago

Lio

asked 4 hours ago

LioLio

212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While there are triangular probability distributions, I do not think the function you supplied is a probability distribution over the support $0dots N$, because the area under its curve does not equal 1.
  $endgroup$
  – Alexis
  4 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

What is the distribution name when probability values have a linear increasing of the form:

$p(i)= frac2N(N+1)i; 0 leq i leq N $

probability distributions

share|cite|improve this question

edited 4 hours ago

Lio

asked 4 hours ago

LioLio

212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While there are triangular probability distributions, I do not think the function you supplied is a probability distribution over the support $0dots N$, because the area under its curve does not equal 1.
  $endgroup$
  – Alexis
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

What is the distribution name when probability values have a linear increasing of the form:

$p(i)= frac2N(N+1)i; 0 leq i leq N $

probability distributions

share|cite|improve this question

edited 4 hours ago

Lio

asked 4 hours ago

LioLio

212

$endgroup$

share|cite|improve this question

edited 4 hours ago

Lio

asked 4 hours ago

LioLio

212

share|cite|improve this question

edited 4 hours ago

Lio

asked 4 hours ago

LioLio

212

asked 4 hours ago

LioLio

212

asked 4 hours ago

LioLio

212

2 Answers
2

active

oldest

votes

2

$begingroup$

It is a valid distribution if $i$ is integer, otherwise it isn't as @Alexis points out. When integer, if we sum over all possibilities, we get $1$:
$$sum_i=1^Nfrac2N(N+1)i=frac2N(N+1)sum_i=1^Ni=frac2N(N+1)fracN(N+1)2=1$$
Wikipedia entry for List of Probability Distributions doesn't associate this with a special name, however it is being used in some sources as Discrete Triangular Distribution, although the name has also been used to describe the probability distribution of two dice, in which the distribution is symmetric around a mean. As continuous case refers generally to asymmetric version of the triangle shape (which also includes $b=c$, i.e. right triangle), the discrete version can also do this, which makes the name valid practically.

share|cite|improve this answer

answered 2 hours ago

gunesgunes

7,6461316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) for mentioning applications. The continuous version is $mathsfBeta(2,1).$
  $endgroup$
  – BruceET
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I think you intend this to be a discrete distribution.

If $N=5$ you have $P(X = i) = frac2i30 = fraci15.$ Notice that, whatever $N$ you choose, $P(X = 0) = 0.$

You need to check that the probabilities add to $1:$ that is, $sum_i=0^5 P(X = i) = 1.$
For that, you might want to look at this Wikipedia page.

Here is a plot of the PDF (or PMF) of this distribution.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

BruceETBruceET

6,8161721

$endgroup$

add a comment | 

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2

$begingroup$

It is a valid distribution if $i$ is integer, otherwise it isn't as @Alexis points out. When integer, if we sum over all possibilities, we get $1$:
$$sum_i=1^Nfrac2N(N+1)i=frac2N(N+1)sum_i=1^Ni=frac2N(N+1)fracN(N+1)2=1$$
Wikipedia entry for List of Probability Distributions doesn't associate this with a special name, however it is being used in some sources as Discrete Triangular Distribution, although the name has also been used to describe the probability distribution of two dice, in which the distribution is symmetric around a mean. As continuous case refers generally to asymmetric version of the triangle shape (which also includes $b=c$, i.e. right triangle), the discrete version can also do this, which makes the name valid practically.

share|cite|improve this answer

answered 2 hours ago

gunesgunes

7,6461316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) for mentioning applications. The continuous version is $mathsfBeta(2,1).$
  $endgroup$
  – BruceET
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

It is a valid distribution if $i$ is integer, otherwise it isn't as @Alexis points out. When integer, if we sum over all possibilities, we get $1$:
$$sum_i=1^Nfrac2N(N+1)i=frac2N(N+1)sum_i=1^Ni=frac2N(N+1)fracN(N+1)2=1$$
Wikipedia entry for List of Probability Distributions doesn't associate this with a special name, however it is being used in some sources as Discrete Triangular Distribution, although the name has also been used to describe the probability distribution of two dice, in which the distribution is symmetric around a mean. As continuous case refers generally to asymmetric version of the triangle shape (which also includes $b=c$, i.e. right triangle), the discrete version can also do this, which makes the name valid practically.

share|cite|improve this answer

answered 2 hours ago

gunesgunes

7,6461316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) for mentioning applications. The continuous version is $mathsfBeta(2,1).$
  $endgroup$
  – BruceET
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It is a valid distribution if $i$ is integer, otherwise it isn't as @Alexis points out. When integer, if we sum over all possibilities, we get $1$:
$$sum_i=1^Nfrac2N(N+1)i=frac2N(N+1)sum_i=1^Ni=frac2N(N+1)fracN(N+1)2=1$$
Wikipedia entry for List of Probability Distributions doesn't associate this with a special name, however it is being used in some sources as Discrete Triangular Distribution, although the name has also been used to describe the probability distribution of two dice, in which the distribution is symmetric around a mean. As continuous case refers generally to asymmetric version of the triangle shape (which also includes $b=c$, i.e. right triangle), the discrete version can also do this, which makes the name valid practically.

share|cite|improve this answer

answered 2 hours ago

gunesgunes

7,6461316

$endgroup$

share|cite|improve this answer

answered 2 hours ago

gunesgunes

7,6461316

answered 2 hours ago

gunesgunes

7,6461316

answered 2 hours ago

gunesgunes

7,6461316

1

$begingroup$

I think you intend this to be a discrete distribution.

If $N=5$ you have $P(X = i) = frac2i30 = fraci15.$ Notice that, whatever $N$ you choose, $P(X = 0) = 0.$

You need to check that the probabilities add to $1:$ that is, $sum_i=0^5 P(X = i) = 1.$
For that, you might want to look at this Wikipedia page.

Here is a plot of the PDF (or PMF) of this distribution.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

BruceETBruceET

6,8161721

$endgroup$

add a comment | 

1

$begingroup$

I think you intend this to be a discrete distribution.

If $N=5$ you have $P(X = i) = frac2i30 = fraci15.$ Notice that, whatever $N$ you choose, $P(X = 0) = 0.$

You need to check that the probabilities add to $1:$ that is, $sum_i=0^5 P(X = i) = 1.$
For that, you might want to look at this Wikipedia page.

Here is a plot of the PDF (or PMF) of this distribution.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

BruceETBruceET

6,8161721

$endgroup$

add a comment | 

$begingroup$

I think you intend this to be a discrete distribution.

If $N=5$ you have $P(X = i) = frac2i30 = fraci15.$ Notice that, whatever $N$ you choose, $P(X = 0) = 0.$

You need to check that the probabilities add to $1:$ that is, $sum_i=0^5 P(X = i) = 1.$
For that, you might want to look at this Wikipedia page.

Here is a plot of the PDF (or PMF) of this distribution.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

BruceETBruceET

6,8161721

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

BruceETBruceET

6,8161721

answered 2 hours ago

BruceETBruceET

6,8161721

answered 2 hours ago

BruceETBruceET

6,8161721