Poisson and convergence in distribution
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
asked Nov 19 '18 at 6:11
S_Alex
1109
add a comment |
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
asked Nov 19 '18 at 6:11
S_Alex
1109
add a comment |
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
asked Nov 19 '18 at 6:11
S_Alex
1109
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
real-analysis probability-theory characteristic-functions poisson-process
asked Nov 19 '18 at 6:11
S_Alex
1109
asked Nov 19 '18 at 6:11
S_Alex
1109
asked Nov 19 '18 at 6:11
S_Alex
1109
asked Nov 19 '18 at 6:11
S_Alex
1109
asked Nov 19 '18 at 6:11
S_Alex
1109
1109
add a comment |
add a comment |
1 Answer
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Note that $sum_{n=0}^infty z^nmathbb P(N_lambda =n)=e^{lambda(z-1)}$ is a probability generating function of $N_lambda$. You need also to replace $lambda$ in denominator by $sqrt{lambda}$. Then
$$mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}left(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)right)^nfrac{e^{-lambda} lambda^n}{n!}
$$
$$
=e^{lambdaleft(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)-1right)}=
e^{lambdaleft(1-sigma^2frac{t^2}{2lambda}+oleft(frac{t^2}{lambda}right)-1right)} to e^{-frac{t^2sigma^2}{2}}.
$$
answered Nov 20 '18 at 2:40
NCh
6,2682723
add a comment |
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1 Answer
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1 Answer
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Note that $sum_{n=0}^infty z^nmathbb P(N_lambda =n)=e^{lambda(z-1)}$ is a probability generating function of $N_lambda$. You need also to replace $lambda$ in denominator by $sqrt{lambda}$. Then
$$mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}left(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)right)^nfrac{e^{-lambda} lambda^n}{n!}
$$
$$
=e^{lambdaleft(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)-1right)}=
e^{lambdaleft(1-sigma^2frac{t^2}{2lambda}+oleft(frac{t^2}{lambda}right)-1right)} to e^{-frac{t^2sigma^2}{2}}.
$$
answered Nov 20 '18 at 2:40
NCh
6,2682723
add a comment |
Note that $sum_{n=0}^infty z^nmathbb P(N_lambda =n)=e^{lambda(z-1)}$ is a probability generating function of $N_lambda$. You need also to replace $lambda$ in denominator by $sqrt{lambda}$. Then
$$mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}left(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)right)^nfrac{e^{-lambda} lambda^n}{n!}
$$
$$
=e^{lambdaleft(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)-1right)}=
e^{lambdaleft(1-sigma^2frac{t^2}{2lambda}+oleft(frac{t^2}{lambda}right)-1right)} to e^{-frac{t^2sigma^2}{2}}.
$$
answered Nov 20 '18 at 2:40
NCh
6,2682723
add a comment |
Note that $sum_{n=0}^infty z^nmathbb P(N_lambda =n)=e^{lambda(z-1)}$ is a probability generating function of $N_lambda$. You need also to replace $lambda$ in denominator by $sqrt{lambda}$. Then
$$mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}left(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)right)^nfrac{e^{-lambda} lambda^n}{n!}
$$
$$
=e^{lambdaleft(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)-1right)}=
e^{lambdaleft(1-sigma^2frac{t^2}{2lambda}+oleft(frac{t^2}{lambda}right)-1right)} to e^{-frac{t^2sigma^2}{2}}.
$$
answered Nov 20 '18 at 2:40
NCh
6,2682723
Note that $sum_{n=0}^infty z^nmathbb P(N_lambda =n)=e^{lambda(z-1)}$ is a probability generating function of $N_lambda$. You need also to replace $lambda$ in denominator by $sqrt{lambda}$. Then
$$mathbb{E}e^{it Y_{lambda}}=sum_{n=0}^{infty}mathbb{E}(e^{itY_{lambda}}1_{N_{lambda}=n})=sum_{n=0}^{infty}left(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)right)^nfrac{e^{-lambda} lambda^n}{n!}
$$
$$
=e^{lambdaleft(phi_{X'_1}left(frac{t}{sqrt{lambda}}right)-1right)}=
e^{lambdaleft(1-sigma^2frac{t^2}{2lambda}+oleft(frac{t^2}{lambda}right)-1right)} to e^{-frac{t^2sigma^2}{2}}.
$$
answered Nov 20 '18 at 2:40
NCh
6,2682723
answered Nov 20 '18 at 2:40
NCh
6,2682723
answered Nov 20 '18 at 2:40
NCh
6,2682723
answered Nov 20 '18 at 2:40
NCh
6,2682723
6,2682723
add a comment |
add a comment |
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