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What is the name of this formula derived from the Poisson distribution?

Going from binomial distribution to Poisson distributionPoisson Distribution?Is there a way to standardize the Poisson distribution?Compute the mean of $(1 + X)^-1$ where $X$ is Poisson$(lambda)$The normal approximation of Poisson distributionCan we prove that the Poisson distribution is independent (starting from the definition given here)?Poisson Distribution*Proof of Poisson distribution for the “continuous time arrival model”What is this exponential distribution called?Name of a Particular Distribution Family

4

$begingroup$

I am learning about the Poisson distribution in this document and its link reference.

This is the key formula to compute the Poisson distribution:

$$
f(k; lambda)=fraclambda^k e^-lambdak!
$$

I saw another related formula somewhere.

$$
sumlimits_k = x^+ infty
fraclambda^k e^-lambdak!
$$

Is there a name for this formula?

probability

share|cite|improve this question

edited Mar 22 at 0:42

Peter Mortensen

561310

asked Mar 21 at 6:50

shiqangpanshiqangpan

152

$endgroup$

add a comment | 

4

$begingroup$

I am learning about the Poisson distribution in this document and its link reference.

This is the key formula to compute the Poisson distribution:

$$
f(k; lambda)=fraclambda^k e^-lambdak!
$$

I saw another related formula somewhere.

$$
sumlimits_k = x^+ infty
fraclambda^k e^-lambdak!
$$

Is there a name for this formula?

probability

share|cite|improve this question

edited Mar 22 at 0:42

Peter Mortensen

561310

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:

$$
f(k; lambda)=fraclambda^k e^-lambdak!
$$

I saw another related formula somewhere.

$$
sumlimits_k = x^+ infty
fraclambda^k e^-lambdak!
$$

Is there a name for this formula?

probability

share|cite|improve this question

edited Mar 22 at 0:42

Peter Mortensen

561310

asked Mar 21 at 6:50

shiqangpanshiqangpan

152

$endgroup$

share|cite|improve this question

edited Mar 22 at 0:42

Peter Mortensen

561310

asked Mar 21 at 6:50

shiqangpanshiqangpan

152

share|cite|improve this question

edited Mar 22 at 0:42

Peter Mortensen

561310

asked Mar 21 at 6:50

shiqangpanshiqangpan

152

edited Mar 22 at 0:42

Peter Mortensen

561310

edited Mar 22 at 0:42

Peter Mortensen

561310

asked Mar 21 at 6:50

shiqangpanshiqangpan

152

asked Mar 21 at 6:50

shiqangpanshiqangpan

152

4 Answers
4

active

oldest

votes

6

$begingroup$

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 $k$.

share|cite|improve this answer

answered Mar 21 at 6:56

Rohit PandeyRohit Pandey

1,6581024

$endgroup$

add a comment | 

3

$begingroup$

The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_k=0^infty fraclambda^kk!.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_k=0^infty P(X = k) = sum_k=0^infty e^-lambda fraclambda^kk! = 1.$$

share|cite|improve this answer

answered Mar 21 at 6:55

angryavianangryavian

42.5k23481

$endgroup$

add a comment | 

3

$begingroup$

To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_k = 0^x-1
fraclambda^k e^-lambdak! = 1 - sum_k = x^+ infty
fraclambda^k e^-lambdak!
.
$$

share|cite|improve this answer

answered Mar 21 at 7:07

ErtxiemErtxiem

661112

$endgroup$

add a comment | 

1

$begingroup$

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"

$$
1 - F(x-1) =
sum_k = x^+ infty
fraclambda^k e^-lambdak!
$$

which means the probability of at least $x$ observations

share|cite|improve this answer

answered Mar 21 at 7:45

YongYong

111

$endgroup$

add a comment | 

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6

$begingroup$

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 $k$.

share|cite|improve this answer

answered Mar 21 at 6:56

Rohit PandeyRohit Pandey

1,6581024

$endgroup$

add a comment | 

6

$begingroup$

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 $k$.

share|cite|improve this answer

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 $k$.

share|cite|improve this answer

answered Mar 21 at 6:56

Rohit PandeyRohit Pandey

1,6581024

$endgroup$

share|cite|improve this answer

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

3

$begingroup$

The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_k=0^infty fraclambda^kk!.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_k=0^infty P(X = k) = sum_k=0^infty e^-lambda fraclambda^kk! = 1.$$

share|cite|improve this answer

answered Mar 21 at 6:55

angryavianangryavian

42.5k23481

$endgroup$

add a comment | 

3

$begingroup$

The Taylor series for the function $g(lambda) = e^lambda$ is
$$e^lambda = sum_k=0^infty fraclambda^kk!.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_k=0^infty P(X = k) = sum_k=0^infty e^-lambda fraclambda^kk! = 1.$$

share|cite|improve this answer

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
$$e^lambda = sum_k=0^infty fraclambda^kk!.$$
By rearranging, you can verify that the sum of the probabilities of all outcomes for a Poisson($lambda$) random variable $X$ is $1$.
$$sum_k=0^infty P(X = k) = sum_k=0^infty e^-lambda fraclambda^kk! = 1.$$

share|cite|improve this answer

answered Mar 21 at 6:55

angryavianangryavian

42.5k23481

$endgroup$

share|cite|improve this answer

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

3

$begingroup$

To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_k = 0^x-1
fraclambda^k e^-lambdak! = 1 - sum_k = x^+ infty
fraclambda^k e^-lambdak!
.
$$

share|cite|improve this answer

answered Mar 21 at 7:07

ErtxiemErtxiem

661112

$endgroup$

add a comment | 

3

$begingroup$

To add on the answers by angryavian and Rohit Pandey, the complementary of the second formula is also the cumulative distribution function (CDF):
$$ F(x-1) = P(X leq x-1) = sum_k = 0^x-1
fraclambda^k e^-lambdak! = 1 - sum_k = x^+ infty
fraclambda^k e^-lambdak!
.
$$

share|cite|improve this answer

answered Mar 21 at 7:07

ErtxiemErtxiem

661112

$endgroup$

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):
$$ F(x-1) = P(X leq x-1) = sum_k = 0^x-1
fraclambda^k e^-lambdak! = 1 - sum_k = x^+ infty
fraclambda^k e^-lambdak!
.
$$

share|cite|improve this answer

answered Mar 21 at 7:07

ErtxiemErtxiem

661112

$endgroup$

share|cite|improve this answer

answered Mar 21 at 7:07

ErtxiemErtxiem

661112

answered Mar 21 at 7:07

ErtxiemErtxiem

661112

answered Mar 21 at 7:07

ErtxiemErtxiem

661112

1

$begingroup$

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"

$$
1 - F(x-1) =
sum_k = x^+ infty
fraclambda^k e^-lambdak!
$$

which means the probability of at least $x$ observations

share|cite|improve this answer

answered Mar 21 at 7:45

YongYong

111

$endgroup$

add a comment | 

1

$begingroup$

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"

$$
1 - F(x-1) =
sum_k = x^+ infty
fraclambda^k e^-lambdak!
$$

which means the probability of at least $x$ observations

share|cite|improve this answer

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"

$$
1 - F(x-1) =
sum_k = x^+ infty
fraclambda^k e^-lambdak!
$$

which means the probability of at least $x$ observations

share|cite|improve this answer

answered Mar 21 at 7:45

YongYong

111

$endgroup$

share|cite|improve this answer

answered Mar 21 at 7:45

YongYong

111

answered Mar 21 at 7:45

YongYong

111

answered Mar 21 at 7:45

YongYong

111