What is the probability distribution of linear formula?
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What is the distribution name when probability values have a linear increasing of the form:
$p(i)= frac{2}{N(N+1)}i; 0 leq i leq N $
probability distributions
asked yesterday
LioLio
234
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add a comment |
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What is the distribution name when probability values have a linear increasing of the form:
$p(i)= frac{2}{N(N+1)}i; 0 leq i leq N $
probability distributions
asked yesterday
LioLio
234
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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.
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– Alexis
yesterday
add a comment |
$begingroup$
What is the distribution name when probability values have a linear increasing of the form:
$p(i)= frac{2}{N(N+1)}i; 0 leq i leq N $
probability distributions
asked yesterday
LioLio
234
$endgroup$
What is the distribution name when probability values have a linear increasing of the form:
$p(i)= frac{2}{N(N+1)}i; 0 leq i leq N $
probability distributions
probability distributions
asked yesterday
LioLio
234
asked yesterday
LioLio
234
edited yesterday
Lio
asked yesterday
LioLio
234
asked yesterday
LioLio
234
asked yesterday
LioLio
234
234
$begingroup$
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
yesterday
add a comment |
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
add a comment |
2 Answers
2
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votes
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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 $1$:
$$sum_{i=1}^N{frac{2}{N(N+1)}}i=frac{2}{N(N+1)}sum_{i=1}^N{}i=frac{2}{N(N+1)}frac{N(N+1)}{2}=1$$
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.
answered yesterday
gunesgunes
7,6961316
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$begingroup$
(+1) for mentioning applications. The continuous version is $mathsf{Beta}(2,1).$
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– BruceET
yesterday
add a comment |
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I think you intend this to be a discrete distribution.
If $N=5$ you have $P(X = i) = frac{2i}{30} = frac{i}{15}.$ 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.
answered yesterday
BruceETBruceET
6,8561721
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $1$:
$$sum_{i=1}^N{frac{2}{N(N+1)}}i=frac{2}{N(N+1)}sum_{i=1}^N{}i=frac{2}{N(N+1)}frac{N(N+1)}{2}=1$$
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.
answered yesterday
gunesgunes
7,6961316
$endgroup$
$begingroup$
(+1) for mentioning applications. The continuous version is $mathsf{Beta}(2,1).$
$endgroup$
– BruceET
yesterday
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 $1$:
$$sum_{i=1}^N{frac{2}{N(N+1)}}i=frac{2}{N(N+1)}sum_{i=1}^N{}i=frac{2}{N(N+1)}frac{N(N+1)}{2}=1$$
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.
answered yesterday
gunesgunes
7,6961316
$endgroup$
$begingroup$
(+1) for mentioning applications. The continuous version is $mathsf{Beta}(2,1).$
$endgroup$
– BruceET
yesterday
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 $1$:
$$sum_{i=1}^N{frac{2}{N(N+1)}}i=frac{2}{N(N+1)}sum_{i=1}^N{}i=frac{2}{N(N+1)}frac{N(N+1)}{2}=1$$
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.
answered yesterday
gunesgunes
7,6961316
$endgroup$
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 $1$:
$$sum_{i=1}^N{frac{2}{N(N+1)}}i=frac{2}{N(N+1)}sum_{i=1}^N{}i=frac{2}{N(N+1)}frac{N(N+1)}{2}=1$$
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.
answered yesterday
gunesgunes
7,6961316
answered yesterday
gunesgunes
7,6961316
answered yesterday
gunesgunes
7,6961316
answered yesterday
gunesgunes
7,6961316
7,6961316
$begingroup$
(+1) for mentioning applications. The continuous version is $mathsf{Beta}(2,1).$
$endgroup$
– BruceET
yesterday
add a comment |
$begingroup$
(+1) for mentioning applications. The continuous version is $mathsf{Beta}(2,1).$
$endgroup$
– BruceET
yesterday
$begingroup$
(+1) for mentioning applications. The continuous version is $mathsf{Beta}(2,1).$
$endgroup$
– BruceET
yesterday
$begingroup$
(+1) for mentioning applications. The continuous version is $mathsf{Beta}(2,1).$
$endgroup$
– BruceET
yesterday
add a comment |
$begingroup$
I think you intend this to be a discrete distribution.
If $N=5$ you have $P(X = i) = frac{2i}{30} = frac{i}{15}.$ 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.
answered yesterday
BruceETBruceET
6,8561721
$endgroup$
add a comment |
$begingroup$
I think you intend this to be a discrete distribution.
If $N=5$ you have $P(X = i) = frac{2i}{30} = frac{i}{15}.$ 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.
answered yesterday
BruceETBruceET
6,8561721
$endgroup$
add a comment |
$begingroup$
I think you intend this to be a discrete distribution.
If $N=5$ you have $P(X = i) = frac{2i}{30} = frac{i}{15}.$ 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.
answered yesterday
BruceETBruceET
6,8561721
$endgroup$
I think you intend this to be a discrete distribution.
If $N=5$ you have $P(X = i) = frac{2i}{30} = frac{i}{15}.$ 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.
answered yesterday
BruceETBruceET
6,8561721
edited yesterday
answered yesterday
BruceETBruceET
6,8561721
answered yesterday
BruceETBruceET
6,8561721
answered yesterday
BruceETBruceET
6,8561721
6,8561721
add a comment |
add a comment |
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$begingroup$
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
yesterday