Convergence in distribution of Poisson variables.












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Kavi Rama Murthy
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy my bad!
    $endgroup$
    – openspace
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy changed
    $endgroup$
    – openspace
    May 11 '18 at 8:15
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Kavi Rama Murthy
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy my bad!
    $endgroup$
    – openspace
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy changed
    $endgroup$
    – openspace
    May 11 '18 at 8:15














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 22:52









Davide Giraudo

125k16150261




125k16150261










asked May 11 '18 at 7:58









openspaceopenspace

3,4352822




3,4352822








  • 1




    $begingroup$
    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.
    $endgroup$
    – Kavi Rama Murthy
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy my bad!
    $endgroup$
    – openspace
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy changed
    $endgroup$
    – openspace
    May 11 '18 at 8:15














  • 1




    $begingroup$
    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.
    $endgroup$
    – Kavi Rama Murthy
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy my bad!
    $endgroup$
    – openspace
    May 11 '18 at 8:04










  • $begingroup$
    @KaviRamaMurthy changed
    $endgroup$
    – openspace
    May 11 '18 at 8:15








1




1




$begingroup$
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.
$endgroup$
– Kavi Rama Murthy
May 11 '18 at 8:04




$begingroup$
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.
$endgroup$
– Kavi Rama Murthy
May 11 '18 at 8:04












$begingroup$
@KaviRamaMurthy my bad!
$endgroup$
– openspace
May 11 '18 at 8:04




$begingroup$
@KaviRamaMurthy my bad!
$endgroup$
– openspace
May 11 '18 at 8:04












$begingroup$
@KaviRamaMurthy changed
$endgroup$
– openspace
May 11 '18 at 8:15




$begingroup$
@KaviRamaMurthy changed
$endgroup$
– openspace
May 11 '18 at 8:15










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.






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    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












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        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 11 '18 at 9:22









        Kavi Rama MurthyKavi Rama Murthy

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