Hfrxdjt

Why aren't air breathing engines used as small first stages

Is the Standard Deduction better than Itemized when both are the same amount?

How to compare two different files line by line in unix?

Extracting terms with certain heads in a function

Is it common practice to audition new musicians 1-2-1 before rehearsing with the entire band?

Why do we bend a book to keep it straight?

What is this building called? (It was built in 2002)

Can an alien society believe that their star system is the universe?

Can anything be seen from the center of the Boötes void? How dark would it be?

What is the escape velocity of a neutron particle (not neutron star)

Does classifying an integer as a discrete log require it be part of a multiplicative group?

Ports Showing Closed/Filtered in Nmap Scans

How to react to hostile behavior from a senior developer?

Denied boarding although I have proper visa and documentation. To whom should I make a complaint?

Is it fair for a professor to grade us on the possession of past papers?

Why wasn't DOSKEY integrated with command.com?

Is there any way for the UK Prime Minister to make a motion directly dependent on Government confidence?

Crossing US/Canada Border for less than 24 hours

For a new assistant professor in CS, how to build/manage a publication pipeline

2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?

Using et al. for a last / senior author rather than for a first author

Can a party unilaterally change candidates in preparation for a General election?

How come Sam didn't become Lord of Horn Hill?

Should I use a zero-interest credit card for a large one-time purchase?

What is the probability distribution of linear formula?

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

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 5 hours ago

Lio

asked 5 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
  5 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 5 hours ago

Lio

asked 5 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
  5 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 5 hours ago

Lio

asked 5 hours ago

LioLio

212

$endgroup$

share|cite|improve this question

edited 5 hours ago

Lio

asked 5 hours ago

LioLio

212

share|cite|improve this question

edited 5 hours ago

Lio

asked 5 hours ago

LioLio

212

asked 5 hours ago

LioLio

212

asked 5 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 3 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 3 hours ago

answered 3 hours ago

BruceETBruceET

6,8161721

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "65"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f403611%2fwhat-is-the-probability-distribution-of-linear-formula%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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 3 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 3 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 3 hours ago

gunesgunes

7,6461316

$endgroup$

share|cite|improve this answer

answered 3 hours ago

gunesgunes

7,6461316

answered 3 hours ago

gunesgunes

7,6461316

answered 3 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 3 hours ago

answered 3 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 3 hours ago

answered 3 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 3 hours ago

answered 3 hours ago

BruceETBruceET

6,8161721

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

BruceETBruceET

6,8161721

answered 3 hours ago

BruceETBruceET

6,8161721

answered 3 hours ago

BruceETBruceET

6,8161721