Mixing
Sum of Poisson distributions
Clash Royale CLAN TAG#URR8PPP
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The problem I have is the following:
By applying the central limit theorem to a sequence of random variables with the Possion distribution, prove that
$$
e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)tofrac12, mbox as ntoinfty
$$
What I did so far:
If we chose $X sim Poisson(n)$, then we have that
$$
P(X = k) = fracn^kk!e^-n
$$
Hence,
$$
P(Xleq n) = e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)
$$
And, we know that $E[X] = n$ and that $Var(X) = n$, hence
$$
P(Xleq n) = Pleft (fracX-E[X]sqrtvar(X)sqrtnleq 0right )
$$
Now, I know that we cannot take the limit in the probability as $X$ is not, yet a sum of independent random variables, but if we could write $X$ as a sum of independent random variables, then we would be able to have $P(Xleq n)to phi(0) = frac12$.
My understanding of the question is that it is required to find $X_1, X_2, dots$ random variables, each a particular type of a Poisson distribution, so that added up, namely $X_1 + dots + X_n = X$. Also, I know that a Poisson distribution can be written as a sum of independent Poisson distributions, however given that we have $X sim Poisson(n)$ can we find exactly $n$ such distributions so that they add up to $X$ ? If so:
How do I find these Poisson distributions?
probability probability-distributions central-limit-theorem
asked Jul 19 at 11:00
Andrei Crisan
3219
add a comment |Â
up vote
1
down vote
favorite
The problem I have is the following:
By applying the central limit theorem to a sequence of random variables with the Possion distribution, prove that
$$
e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)tofrac12, mbox as ntoinfty
$$
What I did so far:
If we chose $X sim Poisson(n)$, then we have that
$$
P(X = k) = fracn^kk!e^-n
$$
Hence,
$$
P(Xleq n) = e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)
$$
And, we know that $E[X] = n$ and that $Var(X) = n$, hence
$$
P(Xleq n) = Pleft (fracX-E[X]sqrtvar(X)sqrtnleq 0right )
$$
Now, I know that we cannot take the limit in the probability as $X$ is not, yet a sum of independent random variables, but if we could write $X$ as a sum of independent random variables, then we would be able to have $P(Xleq n)to phi(0) = frac12$.
My understanding of the question is that it is required to find $X_1, X_2, dots$ random variables, each a particular type of a Poisson distribution, so that added up, namely $X_1 + dots + X_n = X$. Also, I know that a Poisson distribution can be written as a sum of independent Poisson distributions, however given that we have $X sim Poisson(n)$ can we find exactly $n$ such distributions so that they add up to $X$ ? If so:
How do I find these Poisson distributions?
probability probability-distributions central-limit-theorem
asked Jul 19 at 11:00
Andrei Crisan
3219
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The problem I have is the following:
By applying the central limit theorem to a sequence of random variables with the Possion distribution, prove that
$$
e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)tofrac12, mbox as ntoinfty
$$
What I did so far:
If we chose $X sim Poisson(n)$, then we have that
$$
P(X = k) = fracn^kk!e^-n
$$
Hence,
$$
P(Xleq n) = e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)
$$
And, we know that $E[X] = n$ and that $Var(X) = n$, hence
$$
P(Xleq n) = Pleft (fracX-E[X]sqrtvar(X)sqrtnleq 0right )
$$
Now, I know that we cannot take the limit in the probability as $X$ is not, yet a sum of independent random variables, but if we could write $X$ as a sum of independent random variables, then we would be able to have $P(Xleq n)to phi(0) = frac12$.
My understanding of the question is that it is required to find $X_1, X_2, dots$ random variables, each a particular type of a Poisson distribution, so that added up, namely $X_1 + dots + X_n = X$. Also, I know that a Poisson distribution can be written as a sum of independent Poisson distributions, however given that we have $X sim Poisson(n)$ can we find exactly $n$ such distributions so that they add up to $X$ ? If so:
How do I find these Poisson distributions?
probability probability-distributions central-limit-theorem
asked Jul 19 at 11:00
Andrei Crisan
3219
The problem I have is the following:
By applying the central limit theorem to a sequence of random variables with the Possion distribution, prove that
$$
e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)tofrac12, mbox as ntoinfty
$$
What I did so far:
If we chose $X sim Poisson(n)$, then we have that
$$
P(X = k) = fracn^kk!e^-n
$$
Hence,
$$
P(Xleq n) = e^-nleft(1+n+fracn^22! + dots + fracn^nn!right)
$$
And, we know that $E[X] = n$ and that $Var(X) = n$, hence
$$
P(Xleq n) = Pleft (fracX-E[X]sqrtvar(X)sqrtnleq 0right )
$$
Now, I know that we cannot take the limit in the probability as $X$ is not, yet a sum of independent random variables, but if we could write $X$ as a sum of independent random variables, then we would be able to have $P(Xleq n)to phi(0) = frac12$.
My understanding of the question is that it is required to find $X_1, X_2, dots$ random variables, each a particular type of a Poisson distribution, so that added up, namely $X_1 + dots + X_n = X$. Also, I know that a Poisson distribution can be written as a sum of independent Poisson distributions, however given that we have $X sim Poisson(n)$ can we find exactly $n$ such distributions so that they add up to $X$ ? If so:
How do I find these Poisson distributions?
probability probability-distributions central-limit-theorem
asked Jul 19 at 11:00
Andrei Crisan
3219
asked Jul 19 at 11:00
Andrei Crisan
3219
asked Jul 19 at 11:00
Andrei Crisan
3219
asked Jul 19 at 11:00
Andrei Crisan
3219
3219
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The idea is to take a sequence of independent random variables
$X_1,X_2,ldots$ each Poisson with mean $1$. Then $S_n=X_1+cdots+X_n$
is Poisson with mean $n$ and one can then apply CLT.
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
Thank you! It was so obvious, no idea how I missed it.
– Andrei Crisan
Jul 19 at 12:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The idea is to take a sequence of independent random variables
$X_1,X_2,ldots$ each Poisson with mean $1$. Then $S_n=X_1+cdots+X_n$
is Poisson with mean $n$ and one can then apply CLT.
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
Thank you! It was so obvious, no idea how I missed it.
– Andrei Crisan
Jul 19 at 12:07
add a comment |Â
up vote
2
down vote
accepted
The idea is to take a sequence of independent random variables
$X_1,X_2,ldots$ each Poisson with mean $1$. Then $S_n=X_1+cdots+X_n$
is Poisson with mean $n$ and one can then apply CLT.
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
Thank you! It was so obvious, no idea how I missed it.
– Andrei Crisan
Jul 19 at 12:07
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The idea is to take a sequence of independent random variables
$X_1,X_2,ldots$ each Poisson with mean $1$. Then $S_n=X_1+cdots+X_n$
is Poisson with mean $n$ and one can then apply CLT.
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
The idea is to take a sequence of independent random variables
$X_1,X_2,ldots$ each Poisson with mean $1$. Then $S_n=X_1+cdots+X_n$
is Poisson with mean $n$ and one can then apply CLT.
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:12
Lord Shark the Unknown
85.5k951112
85.5k951112
Thank you! It was so obvious, no idea how I missed it.
– Andrei Crisan
Jul 19 at 12:07
add a comment |Â
Thank you! It was so obvious, no idea how I missed it.
– Andrei Crisan
Jul 19 at 12:07
Thank you! It was so obvious, no idea how I missed it.
– Andrei Crisan
Jul 19 at 12:07
Thank you! It was so obvious, no idea how I missed it.
– Andrei Crisan
Jul 19 at 12:07
add a comment |Â
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