Convergence in distribution of Poisson variables.
Consider ${ X_{i} }$ are independent random variables with Poisson distribution. We want to know about convergence in distribution of $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$.
There are two cases :
Case 1 : $sum_{i}lambda_{i}$ converges. Then we can say variance and mean value of $S_{n}$ converges to some fixed value. Also we know that sum of Poisson random variables is also the Poisson random variable with parameter equals sum of previous ones.
So in my opinion $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $frac{mathrm{Pois}(c) - c}{sqrt{c}}$ , where $c = sum_{i} lambda_{i}$
Case 2 : we have that $sum_{i} lambda_{i}$ diverges. Then I guess we can try to use Central limit theorem and satisfy that $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $N(0,1)$
Am I right ? Or where have I problems to fix ?
probability-theory proof-verification weak-convergence poisson-distribution probability-limit-theorems
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
asked May 11 at 7:58
openspace
3,4452822
add a comment |
Consider ${ X_{i} }$ are independent random variables with Poisson distribution. We want to know about convergence in distribution of $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$.
There are two cases :
Case 1 : $sum_{i}lambda_{i}$ converges. Then we can say variance and mean value of $S_{n}$ converges to some fixed value. Also we know that sum of Poisson random variables is also the Poisson random variable with parameter equals sum of previous ones.
So in my opinion $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $frac{mathrm{Pois}(c) - c}{sqrt{c}}$ , where $c = sum_{i} lambda_{i}$
Case 2 : we have that $sum_{i} lambda_{i}$ diverges. Then I guess we can try to use Central limit theorem and satisfy that $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $N(0,1)$
Am I right ? Or where have I problems to fix ?
probability-theory proof-verification weak-convergence poisson-distribution probability-limit-theorems
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
asked May 11 at 7:58
openspace
3,4452822
1
How do you expand "i.i.d."? In standard terminology if $X_i$'s are i.i.d Poisson then the parameters $lambda_i$ are all the same, so CLT applies.
– Kavi Rama Murthy
May 11 at 8:04
@KaviRamaMurthy my bad!
– openspace
May 11 at 8:04
@KaviRamaMurthy changed
– openspace
May 11 at 8:15
add a comment |
Consider ${ X_{i} }$ are independent random variables with Poisson distribution. We want to know about convergence in distribution of $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$.
There are two cases :
Case 1 : $sum_{i}lambda_{i}$ converges. Then we can say variance and mean value of $S_{n}$ converges to some fixed value. Also we know that sum of Poisson random variables is also the Poisson random variable with parameter equals sum of previous ones.
So in my opinion $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $frac{mathrm{Pois}(c) - c}{sqrt{c}}$ , where $c = sum_{i} lambda_{i}$
Case 2 : we have that $sum_{i} lambda_{i}$ diverges. Then I guess we can try to use Central limit theorem and satisfy that $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $N(0,1)$
Am I right ? Or where have I problems to fix ?
probability-theory proof-verification weak-convergence poisson-distribution probability-limit-theorems
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
asked May 11 at 7:58
openspace
3,4452822
Consider ${ X_{i} }$ are independent random variables with Poisson distribution. We want to know about convergence in distribution of $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$.
There are two cases :
Case 1 : $sum_{i}lambda_{i}$ converges. Then we can say variance and mean value of $S_{n}$ converges to some fixed value. Also we know that sum of Poisson random variables is also the Poisson random variable with parameter equals sum of previous ones.
So in my opinion $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $frac{mathrm{Pois}(c) - c}{sqrt{c}}$ , where $c = sum_{i} lambda_{i}$
Case 2 : we have that $sum_{i} lambda_{i}$ diverges. Then I guess we can try to use Central limit theorem and satisfy that $frac{S_{n}-mathbb{E}(S_{n})}{sqrt{operatorname{Var}(S_{n})}}$ converges to $N(0,1)$
Am I right ? Or where have I problems to fix ?
probability-theory proof-verification weak-convergence poisson-distribution probability-limit-theorems
probability-theory proof-verification weak-convergence poisson-distribution probability-limit-theorems
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
asked May 11 at 7:58
openspace
3,4452822
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
asked May 11 at 7:58
openspace
3,4452822
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
edited Nov 24 at 22:52
Davide Giraudo
125k16150259
125k16150259
asked May 11 at 7:58
openspace
3,4452822
asked May 11 at 7:58
openspace
3,4452822
asked May 11 at 7:58
openspace
3,4452822
3,4452822
1
How do you expand "i.i.d."? In standard terminology if $X_i$'s are i.i.d Poisson then the parameters $lambda_i$ are all the same, so CLT applies.
– Kavi Rama Murthy
May 11 at 8:04
@KaviRamaMurthy my bad!
– openspace
May 11 at 8:04
@KaviRamaMurthy changed
– openspace
May 11 at 8:15
add a comment |
1
How do you expand "i.i.d."? In standard terminology if $X_i$'s are i.i.d Poisson then the parameters $lambda_i$ are all the same, so CLT applies.
– Kavi Rama Murthy
May 11 at 8:04
@KaviRamaMurthy my bad!
– openspace
May 11 at 8:04
@KaviRamaMurthy changed
– openspace
May 11 at 8:15
1
1
How do you expand "i.i.d."? In standard terminology if $X_i$'s are i.i.d Poisson then the parameters $lambda_i$ are all the same, so CLT applies.
– Kavi Rama Murthy
May 11 at 8:04
How do you expand "i.i.d."? In standard terminology if $X_i$'s are i.i.d Poisson then the parameters $lambda_i$ are all the same, so CLT applies.
– Kavi Rama Murthy
May 11 at 8:04
@KaviRamaMurthy my bad!
– openspace
May 11 at 8:04
@KaviRamaMurthy my bad!
– openspace
May 11 at 8:04
@KaviRamaMurthy changed
– openspace
May 11 at 8:15
@KaviRamaMurthy changed
– openspace
May 11 at 8:15
add a comment |
1 Answer
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The answer is easy if you compute the characteristic function explicitly. Let $u_n=lambda_1+lambda_2++...+lambda_n$. Then $Ee^{it(S_n-ES_n)/sqrt (Var(S_n)}=e^{-itsqrt (u_n)}prod_{j=1}^{n}(Ee^{itX_j /sqrt(u_n)}$. The Poisson characteristic function (paramater $lambda $) is $e^{-lambda (1- e^{it})}$. We get $e^{-itsqrt (u_n)}e^{-u_n (1-e^{it/sqrt (u_n)})}$. Using the Taylor expansion of $e^{it/sqrt (u_n)}$ up to the term in $t^{2}$ you will see that that characteristic function indeed converges to $e^{-t^{2}/2}$ if $sum lambda_j =infty$. The characteristic function also converges when $sum lambda_j <infty$ and you can write down the limiting characteristic function explictly.
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
add a comment |
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1 Answer
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1 Answer
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active
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The answer is easy if you compute the characteristic function explicitly. Let $u_n=lambda_1+lambda_2++...+lambda_n$. Then $Ee^{it(S_n-ES_n)/sqrt (Var(S_n)}=e^{-itsqrt (u_n)}prod_{j=1}^{n}(Ee^{itX_j /sqrt(u_n)}$. The Poisson characteristic function (paramater $lambda $) is $e^{-lambda (1- e^{it})}$. We get $e^{-itsqrt (u_n)}e^{-u_n (1-e^{it/sqrt (u_n)})}$. Using the Taylor expansion of $e^{it/sqrt (u_n)}$ up to the term in $t^{2}$ you will see that that characteristic function indeed converges to $e^{-t^{2}/2}$ if $sum lambda_j =infty$. The characteristic function also converges when $sum lambda_j <infty$ and you can write down the limiting characteristic function explictly.
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
add a comment |
The answer is easy if you compute the characteristic function explicitly. Let $u_n=lambda_1+lambda_2++...+lambda_n$. Then $Ee^{it(S_n-ES_n)/sqrt (Var(S_n)}=e^{-itsqrt (u_n)}prod_{j=1}^{n}(Ee^{itX_j /sqrt(u_n)}$. The Poisson characteristic function (paramater $lambda $) is $e^{-lambda (1- e^{it})}$. We get $e^{-itsqrt (u_n)}e^{-u_n (1-e^{it/sqrt (u_n)})}$. Using the Taylor expansion of $e^{it/sqrt (u_n)}$ up to the term in $t^{2}$ you will see that that characteristic function indeed converges to $e^{-t^{2}/2}$ if $sum lambda_j =infty$. The characteristic function also converges when $sum lambda_j <infty$ and you can write down the limiting characteristic function explictly.
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
add a comment |
The answer is easy if you compute the characteristic function explicitly. Let $u_n=lambda_1+lambda_2++...+lambda_n$. Then $Ee^{it(S_n-ES_n)/sqrt (Var(S_n)}=e^{-itsqrt (u_n)}prod_{j=1}^{n}(Ee^{itX_j /sqrt(u_n)}$. The Poisson characteristic function (paramater $lambda $) is $e^{-lambda (1- e^{it})}$. We get $e^{-itsqrt (u_n)}e^{-u_n (1-e^{it/sqrt (u_n)})}$. Using the Taylor expansion of $e^{it/sqrt (u_n)}$ up to the term in $t^{2}$ you will see that that characteristic function indeed converges to $e^{-t^{2}/2}$ if $sum lambda_j =infty$. The characteristic function also converges when $sum lambda_j <infty$ and you can write down the limiting characteristic function explictly.
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
The answer is easy if you compute the characteristic function explicitly. Let $u_n=lambda_1+lambda_2++...+lambda_n$. Then $Ee^{it(S_n-ES_n)/sqrt (Var(S_n)}=e^{-itsqrt (u_n)}prod_{j=1}^{n}(Ee^{itX_j /sqrt(u_n)}$. The Poisson characteristic function (paramater $lambda $) is $e^{-lambda (1- e^{it})}$. We get $e^{-itsqrt (u_n)}e^{-u_n (1-e^{it/sqrt (u_n)})}$. Using the Taylor expansion of $e^{it/sqrt (u_n)}$ up to the term in $t^{2}$ you will see that that characteristic function indeed converges to $e^{-t^{2}/2}$ if $sum lambda_j =infty$. The characteristic function also converges when $sum lambda_j <infty$ and you can write down the limiting characteristic function explictly.
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
answered May 11 at 9:22
Kavi Rama Murthy
49.9k31854
49.9k31854
add a comment |
add a comment |
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1
How do you expand "i.i.d."? In standard terminology if $X_i$'s are i.i.d Poisson then the parameters $lambda_i$ are all the same, so CLT applies.
– Kavi Rama Murthy
May 11 at 8:04
@KaviRamaMurthy my bad!
– openspace
May 11 at 8:04
@KaviRamaMurthy changed
– openspace
May 11 at 8:15