Poisson and convergence in distribution











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Let $X_1,X_2,...$ be iid with mean $mu$ and variance $sigma^2$. Let $N_{lambda}$ be poisson($lambda$) independent of the $X_i$'s.



(a) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-N_{lambda}mu}{sqrt{lambda}}.$$



(b) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-lambdamu}{sqrt{lambda}}.$$



(c) For which random variables $X$ will the two limits be the same?



My solution:
Let $Y_{lambda}$ be the quotient given in the problem. Then $mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}(phi_{X'_1}(frac{t}{lambda} ))^nfrac{e^{-lambda} lambda^n}{n!}=sum_{n=0}^{infty}(1-sigma^2frac{t^2}{2lambda^2}+o(frac{t^2}{lambda^2}))^n frac{e^{-lambda}lambda^n}{n!}$



where $X'=X-mu$. Can the above sum be simplified as $lambda$ goes to $infty$?










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    Let $X_1,X_2,...$ be iid with mean $mu$ and variance $sigma^2$. Let $N_{lambda}$ be poisson($lambda$) independent of the $X_i$'s.



    (a) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-N_{lambda}mu}{sqrt{lambda}}.$$



    (b) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-lambdamu}{sqrt{lambda}}.$$



    (c) For which random variables $X$ will the two limits be the same?



    My solution:
    Let $Y_{lambda}$ be the quotient given in the problem. Then $mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}(phi_{X'_1}(frac{t}{lambda} ))^nfrac{e^{-lambda} lambda^n}{n!}=sum_{n=0}^{infty}(1-sigma^2frac{t^2}{2lambda^2}+o(frac{t^2}{lambda^2}))^n frac{e^{-lambda}lambda^n}{n!}$



    where $X'=X-mu$. Can the above sum be simplified as $lambda$ goes to $infty$?










    share|cite|improve this question
























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      Let $X_1,X_2,...$ be iid with mean $mu$ and variance $sigma^2$. Let $N_{lambda}$ be poisson($lambda$) independent of the $X_i$'s.



      (a) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-N_{lambda}mu}{sqrt{lambda}}.$$



      (b) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-lambdamu}{sqrt{lambda}}.$$



      (c) For which random variables $X$ will the two limits be the same?



      My solution:
      Let $Y_{lambda}$ be the quotient given in the problem. Then $mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}(phi_{X'_1}(frac{t}{lambda} ))^nfrac{e^{-lambda} lambda^n}{n!}=sum_{n=0}^{infty}(1-sigma^2frac{t^2}{2lambda^2}+o(frac{t^2}{lambda^2}))^n frac{e^{-lambda}lambda^n}{n!}$



      where $X'=X-mu$. Can the above sum be simplified as $lambda$ goes to $infty$?










      share|cite|improve this question













      Let $X_1,X_2,...$ be iid with mean $mu$ and variance $sigma^2$. Let $N_{lambda}$ be poisson($lambda$) independent of the $X_i$'s.



      (a) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-N_{lambda}mu}{sqrt{lambda}}.$$



      (b) Find the limit in distribution as $lambda rightarrow infty$ for $$frac{sum_{i=1}^{N_{lambda}}X_i-lambdamu}{sqrt{lambda}}.$$



      (c) For which random variables $X$ will the two limits be the same?



      My solution:
      Let $Y_{lambda}$ be the quotient given in the problem. Then $mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}(phi_{X'_1}(frac{t}{lambda} ))^nfrac{e^{-lambda} lambda^n}{n!}=sum_{n=0}^{infty}(1-sigma^2frac{t^2}{2lambda^2}+o(frac{t^2}{lambda^2}))^n frac{e^{-lambda}lambda^n}{n!}$



      where $X'=X-mu$. Can the above sum be simplified as $lambda$ goes to $infty$?







      real-analysis probability-theory characteristic-functions poisson-process






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      asked 9 hours ago









      S_Alex

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