What is the name of this formula derived from the Poisson distribution?
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I am learning about the Poisson distribution in this document and its link reference.
This is the key formula to compute the Poisson distribution:
$$
f(k; lambda)=frac{lambda^k e^{-lambda}}{k!}
$$
I saw another related formula somewhere.
$$
sumlimits_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
Is there a name for this formula?
probability
edited 19 hours ago
Peter Mortensen
563310
asked yesterday
shiqangpanshiqangpan
152
New contributor
shiqangpan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I am learning about the Poisson distribution in this document and its link reference.
This is the key formula to compute the Poisson distribution:
$$
f(k; lambda)=frac{lambda^k e^{-lambda}}{k!}
$$
I saw another related formula somewhere.
$$
sumlimits_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
Is there a name for this formula?
probability
edited 19 hours ago
Peter Mortensen
563310
asked yesterday
shiqangpanshiqangpan
152
New contributor
shiqangpan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am learning about the Poisson distribution in this document and its link reference.
This is the key formula to compute the Poisson distribution:
$$
f(k; lambda)=frac{lambda^k e^{-lambda}}{k!}
$$
I saw another related formula somewhere.
$$
sumlimits_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
Is there a name for this formula?
probability
edited 19 hours ago
Peter Mortensen
563310
asked yesterday
shiqangpanshiqangpan
152
New contributor
shiqangpan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am learning about the Poisson distribution in this document and its link reference.
This is the key formula to compute the Poisson distribution:
$$
f(k; lambda)=frac{lambda^k e^{-lambda}}{k!}
$$
I saw another related formula somewhere.
$$
sumlimits_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
Is there a name for this formula?
probability
probability
edited 19 hours ago
Peter Mortensen
563310
asked yesterday
shiqangpanshiqangpan
152
New contributor
shiqangpan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 hours ago
Peter Mortensen
563310
asked yesterday
shiqangpanshiqangpan
152
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edited 19 hours ago
Peter Mortensen
563310
edited 19 hours ago
Peter Mortensen
563310
edited 19 hours ago
Peter Mortensen
563310
563310
asked yesterday
shiqangpanshiqangpan
152
New contributor
shiqangpan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
shiqangpanshiqangpan
152
asked yesterday
shiqangpanshiqangpan
152
152
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shiqangpan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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4 Answers
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The first formula is the probability mass function (PMF) of the Poisson distribution and the second one is the survival function of this distribution. The first one gives you the probability that the Poisson random variable (which can take integer values) will equal $k$ and the second one the probability that it will be greater than or equal to $k$.
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
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The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_{k=0}^infty frac{lambda^k}{k!}.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_{k=0}^infty P(X = k) = sum_{k=0}^infty e^{-lambda} frac{lambda^k}{k!} = 1.$$
answered yesterday
angryavianangryavian
42.3k23481
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To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_{k = 0}^{x-1}
frac{lambda^k e^{-lambda}}{k!} = 1 - sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
.
$$
answered yesterday
ErtxiemErtxiem
38410
$endgroup$
add a comment |
$begingroup$
To add on the answers by angryavian, Rohit Pandey and Ertxiem, the second formula is the complement of the CDF of Poisson PMF for "x-1"
$$
1 - F(x-1) =
sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
which means the probability of at least $x$ observations
answered yesterday
YongYong
111
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first formula is the probability mass function (PMF) of the Poisson distribution and the second one is the survival function of this distribution. The first one gives you the probability that the Poisson random variable (which can take integer values) will equal $k$ and the second one the probability that it will be greater than or equal to $k$.
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
$endgroup$
add a comment |
$begingroup$
The first formula is the probability mass function (PMF) of the Poisson distribution and the second one is the survival function of this distribution. The first one gives you the probability that the Poisson random variable (which can take integer values) will equal $k$ and the second one the probability that it will be greater than or equal to $k$.
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
$endgroup$
add a comment |
$begingroup$
The first formula is the probability mass function (PMF) of the Poisson distribution and the second one is the survival function of this distribution. The first one gives you the probability that the Poisson random variable (which can take integer values) will equal $k$ and the second one the probability that it will be greater than or equal to $k$.
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
$endgroup$
The first formula is the probability mass function (PMF) of the Poisson distribution and the second one is the survival function of this distribution. The first one gives you the probability that the Poisson random variable (which can take integer values) will equal $k$ and the second one the probability that it will be greater than or equal to $k$.
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
answered yesterday
Rohit PandeyRohit Pandey
1,6281023
1,6281023
add a comment |
add a comment |
$begingroup$
The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_{k=0}^infty frac{lambda^k}{k!}.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_{k=0}^infty P(X = k) = sum_{k=0}^infty e^{-lambda} frac{lambda^k}{k!} = 1.$$
answered yesterday
angryavianangryavian
42.3k23481
$endgroup$
add a comment |
$begingroup$
The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_{k=0}^infty frac{lambda^k}{k!}.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_{k=0}^infty P(X = k) = sum_{k=0}^infty e^{-lambda} frac{lambda^k}{k!} = 1.$$
answered yesterday
angryavianangryavian
42.3k23481
$endgroup$
add a comment |
$begingroup$
The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_{k=0}^infty frac{lambda^k}{k!}.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_{k=0}^infty P(X = k) = sum_{k=0}^infty e^{-lambda} frac{lambda^k}{k!} = 1.$$
answered yesterday
angryavianangryavian
42.3k23481
$endgroup$
The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_{k=0}^infty frac{lambda^k}{k!}.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_{k=0}^infty P(X = k) = sum_{k=0}^infty e^{-lambda} frac{lambda^k}{k!} = 1.$$
answered yesterday
angryavianangryavian
42.3k23481
answered yesterday
angryavianangryavian
42.3k23481
answered yesterday
angryavianangryavian
42.3k23481
answered yesterday
angryavianangryavian
42.3k23481
42.3k23481
add a comment |
add a comment |
$begingroup$
To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_{k = 0}^{x-1}
frac{lambda^k e^{-lambda}}{k!} = 1 - sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
.
$$
answered yesterday
ErtxiemErtxiem
38410
$endgroup$
add a comment |
$begingroup$
To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_{k = 0}^{x-1}
frac{lambda^k e^{-lambda}}{k!} = 1 - sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
.
$$
answered yesterday
ErtxiemErtxiem
38410
$endgroup$
add a comment |
$begingroup$
To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_{k = 0}^{x-1}
frac{lambda^k e^{-lambda}}{k!} = 1 - sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
.
$$
answered yesterday
ErtxiemErtxiem
38410
$endgroup$
To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_{k = 0}^{x-1}
frac{lambda^k e^{-lambda}}{k!} = 1 - sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
.
$$
answered yesterday
ErtxiemErtxiem
38410
answered yesterday
ErtxiemErtxiem
38410
answered yesterday
ErtxiemErtxiem
38410
answered yesterday
ErtxiemErtxiem
38410
38410
add a comment |
add a comment |
$begingroup$
To add on the answers by angryavian, Rohit Pandey and Ertxiem, the second formula is the complement of the CDF of Poisson PMF for "x-1"
$$
1 - F(x-1) =
sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
which means the probability of at least $x$ observations
answered yesterday
YongYong
111
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
To add on the answers by angryavian, Rohit Pandey and Ertxiem, the second formula is the complement of the CDF of Poisson PMF for "x-1"
$$
1 - F(x-1) =
sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
which means the probability of at least $x$ observations
answered yesterday
YongYong
111
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
To add on the answers by angryavian, Rohit Pandey and Ertxiem, the second formula is the complement of the CDF of Poisson PMF for "x-1"
$$
1 - F(x-1) =
sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
which means the probability of at least $x$ observations
answered yesterday
YongYong
111
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
To add on the answers by angryavian, Rohit Pandey and Ertxiem, the second formula is the complement of the CDF of Poisson PMF for "x-1"
$$
1 - F(x-1) =
sum_{k = x}^{+ infty}
frac{lambda^k e^{-lambda}}{k!}
$$
which means the probability of at least $x$ observations
answered yesterday
YongYong
111
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
YongYong
111
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
YongYong
111
answered yesterday
YongYong
111
111
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
shiqangpan is a new contributor. Be nice, and check out our Code of Conduct.
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