What is the name of this formula derived from the Poisson distribution?
$begingroup$
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 Mar 22 at 0:42
Peter Mortensen
565310
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
$endgroup$
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 Mar 22 at 0:42
Peter Mortensen
565310
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
$endgroup$
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 Mar 22 at 0:42
Peter Mortensen
565310
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
$endgroup$
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 Mar 22 at 0:42
Peter Mortensen
565310
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
edited Mar 22 at 0:42
Peter Mortensen
565310
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
edited Mar 22 at 0:42
Peter Mortensen
565310
edited Mar 22 at 0:42
Peter Mortensen
565310
edited Mar 22 at 0:42
Peter Mortensen
565310
565310
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
asked Mar 21 at 6:50
shiqangpanshiqangpan
152
152
add a comment |
add a comment |
4 Answers
4
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 Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
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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 Mar 21 at 6:55
angryavianangryavian
42.5k23481
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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 Mar 21 at 7:07
ErtxiemErtxiem
60112
$endgroup$
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 Mar 21 at 7:45
YongYong
111
$endgroup$
add a comment |
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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 Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
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 Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
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 Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
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 Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
answered Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
answered Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
answered Mar 21 at 6:56
Rohit PandeyRohit Pandey
1,6581024
1,6581024
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 Mar 21 at 6:55
angryavianangryavian
42.5k23481
$endgroup$
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 Mar 21 at 6:55
angryavianangryavian
42.5k23481
$endgroup$
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 Mar 21 at 6:55
angryavianangryavian
42.5k23481
$endgroup$
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 Mar 21 at 6:55
angryavianangryavian
42.5k23481
answered Mar 21 at 6:55
angryavianangryavian
42.5k23481
answered Mar 21 at 6:55
angryavianangryavian
42.5k23481
answered Mar 21 at 6:55
angryavianangryavian
42.5k23481
42.5k23481
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 Mar 21 at 7:07
ErtxiemErtxiem
60112
$endgroup$
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 Mar 21 at 7:07
ErtxiemErtxiem
60112
$endgroup$
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 Mar 21 at 7:07
ErtxiemErtxiem
60112
$endgroup$
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 Mar 21 at 7:07
ErtxiemErtxiem
60112
answered Mar 21 at 7:07
ErtxiemErtxiem
60112
answered Mar 21 at 7:07
ErtxiemErtxiem
60112
answered Mar 21 at 7:07
ErtxiemErtxiem
60112
60112
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 Mar 21 at 7:45
YongYong
111
$endgroup$
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 Mar 21 at 7:45
YongYong
111
$endgroup$
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 Mar 21 at 7:45
YongYong
111
$endgroup$
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 Mar 21 at 7:45
YongYong
111
answered Mar 21 at 7:45
YongYong
111
answered Mar 21 at 7:45
YongYong
111
answered Mar 21 at 7:45
YongYong
111
111
add a comment |
add a comment |
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