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Binomial to Poisson Approximation : Why does p have to be small

How Binomial and Normal distributions approximate Poisson distribution respectively?How to choose between poisson and binomial distributionsPoisson Approximation of BinomialWhy does $p$ have to be moderate in the Poisson approximation to binomial random variable?How to prove Poisson Distribution is the approximation of Binomial Distribution?Simple Question on interpretation of Poisson dist. as approximation to Binomial dist.Did computers render useless the teaching of approximating the Binomial with Poisson and Normal distribution?Probability: $Ysim P(lambda) approx B sim (n,p)$ for $n$ large, $p$ small enough.Why do probabilities have to be small in poisson distribution?Connection between the Binomial distribution, Poisson distribution and Normal distribution

3

1

$begingroup$

I understandd that as n tends towards to infinity for a Binomial distrobution, it becomes a Poisson distobution and i have completed the proof for this.

However, I am not sure why when approximating, p has to be a relativley small value. Again, i understand why n must be large, but whats the purpose/ proof that the smaller p is, the better the approximation is.

Essentially waht I'm asking is, why does p have to be small when approximamting a Poisson from a binomial.

Could someone please help explain this

Thanks

probability-distributions approximation poisson-distribution binomial-distribution

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edited 7 hours ago

Ibrahim

asked 8 hours ago

IbrahimIbrahim

162

New contributor

Ibrahim 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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3

1

$begingroup$

I understandd that as n tends towards to infinity for a Binomial distrobution, it becomes a Poisson distobution and i have completed the proof for this.

However, I am not sure why when approximating, p has to be a relativley small value. Again, i understand why n must be large, but whats the purpose/ proof that the smaller p is, the better the approximation is.

Essentially waht I'm asking is, why does p have to be small when approximamting a Poisson from a binomial.

Could someone please help explain this

Thanks

probability-distributions approximation poisson-distribution binomial-distribution

share|cite|improve this question

edited 7 hours ago

Ibrahim

asked 8 hours ago

IbrahimIbrahim

162

New contributor

Ibrahim 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 understandd that as n tends towards to infinity for a Binomial distrobution, it becomes a Poisson distobution and i have completed the proof for this.

However, I am not sure why when approximating, p has to be a relativley small value. Again, i understand why n must be large, but whats the purpose/ proof that the smaller p is, the better the approximation is.

Essentially waht I'm asking is, why does p have to be small when approximamting a Poisson from a binomial.

Could someone please help explain this

Thanks

probability-distributions approximation poisson-distribution binomial-distribution

share|cite|improve this question

edited 7 hours ago

Ibrahim

asked 8 hours ago

IbrahimIbrahim

162

New contributor

Ibrahim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 7 hours ago

Ibrahim

asked 8 hours ago

IbrahimIbrahim

162

New contributor

Ibrahim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 7 hours ago

Ibrahim

asked 8 hours ago

IbrahimIbrahim

162

New contributor

Ibrahim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

IbrahimIbrahim

162

New contributor

Ibrahim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

IbrahimIbrahim

162

2 Answers
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3

$begingroup$

In the proof, you need $npto lambda$. But if $n$ grows large and $nptolambda$, then we must have $pto 0$, or else $nptoinfty$.

share|cite|improve this answer

answered 7 hours ago

FakeAnalyst56FakeAnalyst56

7517

$endgroup$

add a comment | 

3

$begingroup$

Intuitively, for approximating $operatornameBinomial(n,p)approxoperatornamePoisson(lambda=np)$, the smaller $p$ is, the closer the variance $npq=lambda(1-p)$ to $lambda$, so you expect better approximation.

In the proof, you use
$$
fracn(n-1)dots(n-k+1)n^kleft(1-fraclambdanright)^n-kapprox e^-lambda
$$
to show
$$
binomnkp^k(1-p)^n-kapproxfrace^-np(np)^kk!
$$
If you analyse the error terms more carefully, you get some explicit bounds such as
$$
sum_k=0^infty
leftlvert
binomnkp^k(1-p)^n-k-frace^-np(np)^kk!
rightrvertleq Cp
$$
where $Cleq 4$. So this justifies the motto "smaller $p$ gives better approximation".

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

13k21125

$endgroup$

add a comment | 

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3

$begingroup$

In the proof, you need $npto lambda$. But if $n$ grows large and $nptolambda$, then we must have $pto 0$, or else $nptoinfty$.

share|cite|improve this answer

answered 7 hours ago

FakeAnalyst56FakeAnalyst56

7517

$endgroup$

add a comment | 

3

$begingroup$

In the proof, you need $npto lambda$. But if $n$ grows large and $nptolambda$, then we must have $pto 0$, or else $nptoinfty$.

share|cite|improve this answer

answered 7 hours ago

FakeAnalyst56FakeAnalyst56

7517

$endgroup$

add a comment | 

$begingroup$

In the proof, you need $npto lambda$. But if $n$ grows large and $nptolambda$, then we must have $pto 0$, or else $nptoinfty$.

share|cite|improve this answer

answered 7 hours ago

FakeAnalyst56FakeAnalyst56

7517

$endgroup$

share|cite|improve this answer

answered 7 hours ago

FakeAnalyst56FakeAnalyst56

7517

answered 7 hours ago

FakeAnalyst56FakeAnalyst56

7517

answered 7 hours ago

FakeAnalyst56FakeAnalyst56

7517

3

$begingroup$

Intuitively, for approximating $operatornameBinomial(n,p)approxoperatornamePoisson(lambda=np)$, the smaller $p$ is, the closer the variance $npq=lambda(1-p)$ to $lambda$, so you expect better approximation.

In the proof, you use
$$
fracn(n-1)dots(n-k+1)n^kleft(1-fraclambdanright)^n-kapprox e^-lambda
$$
to show
$$
binomnkp^k(1-p)^n-kapproxfrace^-np(np)^kk!
$$
If you analyse the error terms more carefully, you get some explicit bounds such as
$$
sum_k=0^infty
leftlvert
binomnkp^k(1-p)^n-k-frace^-np(np)^kk!
rightrvertleq Cp
$$
where $Cleq 4$. So this justifies the motto "smaller $p$ gives better approximation".

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

13k21125

$endgroup$

add a comment | 

3

$begingroup$

Intuitively, for approximating $operatornameBinomial(n,p)approxoperatornamePoisson(lambda=np)$, the smaller $p$ is, the closer the variance $npq=lambda(1-p)$ to $lambda$, so you expect better approximation.

In the proof, you use
$$
fracn(n-1)dots(n-k+1)n^kleft(1-fraclambdanright)^n-kapprox e^-lambda
$$
to show
$$
binomnkp^k(1-p)^n-kapproxfrace^-np(np)^kk!
$$
If you analyse the error terms more carefully, you get some explicit bounds such as
$$
sum_k=0^infty
leftlvert
binomnkp^k(1-p)^n-k-frace^-np(np)^kk!
rightrvertleq Cp
$$
where $Cleq 4$. So this justifies the motto "smaller $p$ gives better approximation".

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

13k21125

$endgroup$

add a comment | 

$begingroup$

Intuitively, for approximating $operatornameBinomial(n,p)approxoperatornamePoisson(lambda=np)$, the smaller $p$ is, the closer the variance $npq=lambda(1-p)$ to $lambda$, so you expect better approximation.

In the proof, you use
$$
fracn(n-1)dots(n-k+1)n^kleft(1-fraclambdanright)^n-kapprox e^-lambda
$$
to show
$$
binomnkp^k(1-p)^n-kapproxfrace^-np(np)^kk!
$$
If you analyse the error terms more carefully, you get some explicit bounds such as
$$
sum_k=0^infty
leftlvert
binomnkp^k(1-p)^n-k-frace^-np(np)^kk!
rightrvertleq Cp
$$
where $Cleq 4$. So this justifies the motto "smaller $p$ gives better approximation".

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

13k21125

$endgroup$

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

13k21125

answered 7 hours ago

user10354138user10354138

13k21125

answered 7 hours ago

user10354138user10354138

13k21125