Strong law of large numbers for Poisson rvs with different parameter












0












$begingroup$


Let $X_n$ be independent Poisson random variables with $E[X_i] = mu_i$, and let $Y_n = X_1+...+X_n$. I want to show that if $sum_n mu_n = infty $ then $Y_n/E[Y_n] rightarrow 1$ almost surly.

What I do know is that if $X_1,...$ are independent, and $Eleft[X^4right] < infty$, then $Y_n/n rightarrow E[X]$ a.s.* So What I tried is to center the random variables: let $C_n := X_n-mu_n$. The $C$'s are independent, $E[C_n^4]< infty, E[C_n] = 0$, which means:
$frac{C_1 +...+C_n}{n} rightarrow 0$ a.s., or $frac{sum X_i - sum mu_i}{n} rightarrow 0$ a.s. which, I think, completes the proof.



However I am not sure why the requirement $sum_n mu_n = infty $ is necessary, so I suspect this proof is incorrect. Is it?



Edit: This argument is incorrect since the theorem requires $E[C^4]$ to be uniformly bounded, not just bounded. Any ideas?




  • This is the statement of the theorem from "Probability with Martingales"






probability probability-theory probability-distributions poisson-distribution law-of-large-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You cannot apply the result you recall since it requires the increments are identically distributed -- which is not true in your case (except if $mu_n$ does not depend on $mu$).
    $endgroup$
    – Did
    Dec 16 '18 at 13:04










  • $begingroup$
    To see why the $sum_nmu_n=infty$ requirement is needed consider what happens if $mu_1=1$ and $mu_n=0$ for $n>1$.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:07












  • $begingroup$
    @Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:15












  • $begingroup$
    Except you misquoted heavily.
    $endgroup$
    – Did
    Dec 16 '18 at 13:17










  • $begingroup$
    @kimchilover If $mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:17
















0












$begingroup$


Let $X_n$ be independent Poisson random variables with $E[X_i] = mu_i$, and let $Y_n = X_1+...+X_n$. I want to show that if $sum_n mu_n = infty $ then $Y_n/E[Y_n] rightarrow 1$ almost surly.

What I do know is that if $X_1,...$ are independent, and $Eleft[X^4right] < infty$, then $Y_n/n rightarrow E[X]$ a.s.* So What I tried is to center the random variables: let $C_n := X_n-mu_n$. The $C$'s are independent, $E[C_n^4]< infty, E[C_n] = 0$, which means:
$frac{C_1 +...+C_n}{n} rightarrow 0$ a.s., or $frac{sum X_i - sum mu_i}{n} rightarrow 0$ a.s. which, I think, completes the proof.



However I am not sure why the requirement $sum_n mu_n = infty $ is necessary, so I suspect this proof is incorrect. Is it?



Edit: This argument is incorrect since the theorem requires $E[C^4]$ to be uniformly bounded, not just bounded. Any ideas?




  • This is the statement of the theorem from "Probability with Martingales"






probability probability-theory probability-distributions poisson-distribution law-of-large-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You cannot apply the result you recall since it requires the increments are identically distributed -- which is not true in your case (except if $mu_n$ does not depend on $mu$).
    $endgroup$
    – Did
    Dec 16 '18 at 13:04










  • $begingroup$
    To see why the $sum_nmu_n=infty$ requirement is needed consider what happens if $mu_1=1$ and $mu_n=0$ for $n>1$.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:07












  • $begingroup$
    @Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:15












  • $begingroup$
    Except you misquoted heavily.
    $endgroup$
    – Did
    Dec 16 '18 at 13:17










  • $begingroup$
    @kimchilover If $mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:17














0












0








0





$begingroup$


Let $X_n$ be independent Poisson random variables with $E[X_i] = mu_i$, and let $Y_n = X_1+...+X_n$. I want to show that if $sum_n mu_n = infty $ then $Y_n/E[Y_n] rightarrow 1$ almost surly.

What I do know is that if $X_1,...$ are independent, and $Eleft[X^4right] < infty$, then $Y_n/n rightarrow E[X]$ a.s.* So What I tried is to center the random variables: let $C_n := X_n-mu_n$. The $C$'s are independent, $E[C_n^4]< infty, E[C_n] = 0$, which means:
$frac{C_1 +...+C_n}{n} rightarrow 0$ a.s., or $frac{sum X_i - sum mu_i}{n} rightarrow 0$ a.s. which, I think, completes the proof.



However I am not sure why the requirement $sum_n mu_n = infty $ is necessary, so I suspect this proof is incorrect. Is it?



Edit: This argument is incorrect since the theorem requires $E[C^4]$ to be uniformly bounded, not just bounded. Any ideas?




  • This is the statement of the theorem from "Probability with Martingales"






probability probability-theory probability-distributions poisson-distribution law-of-large-numbers







share|cite|improve this question











$endgroup$




Let $X_n$ be independent Poisson random variables with $E[X_i] = mu_i$, and let $Y_n = X_1+...+X_n$. I want to show that if $sum_n mu_n = infty $ then $Y_n/E[Y_n] rightarrow 1$ almost surly.

What I do know is that if $X_1,...$ are independent, and $Eleft[X^4right] < infty$, then $Y_n/n rightarrow E[X]$ a.s.* So What I tried is to center the random variables: let $C_n := X_n-mu_n$. The $C$'s are independent, $E[C_n^4]< infty, E[C_n] = 0$, which means:
$frac{C_1 +...+C_n}{n} rightarrow 0$ a.s., or $frac{sum X_i - sum mu_i}{n} rightarrow 0$ a.s. which, I think, completes the proof.



However I am not sure why the requirement $sum_n mu_n = infty $ is necessary, so I suspect this proof is incorrect. Is it?



Edit: This argument is incorrect since the theorem requires $E[C^4]$ to be uniformly bounded, not just bounded. Any ideas?




  • This is the statement of the theorem from "Probability with Martingales"






probability probability-theory probability-distributions poisson-distribution law-of-large-numbers




probability probability-theory probability-distributions poisson-distribution law-of-large-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 13:51







SomeoneHAHA

















asked Dec 16 '18 at 12:49









SomeoneHAHASomeoneHAHA

83




83








  • 1




    $begingroup$
    You cannot apply the result you recall since it requires the increments are identically distributed -- which is not true in your case (except if $mu_n$ does not depend on $mu$).
    $endgroup$
    – Did
    Dec 16 '18 at 13:04










  • $begingroup$
    To see why the $sum_nmu_n=infty$ requirement is needed consider what happens if $mu_1=1$ and $mu_n=0$ for $n>1$.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:07












  • $begingroup$
    @Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:15












  • $begingroup$
    Except you misquoted heavily.
    $endgroup$
    – Did
    Dec 16 '18 at 13:17










  • $begingroup$
    @kimchilover If $mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:17














  • 1




    $begingroup$
    You cannot apply the result you recall since it requires the increments are identically distributed -- which is not true in your case (except if $mu_n$ does not depend on $mu$).
    $endgroup$
    – Did
    Dec 16 '18 at 13:04










  • $begingroup$
    To see why the $sum_nmu_n=infty$ requirement is needed consider what happens if $mu_1=1$ and $mu_n=0$ for $n>1$.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:07












  • $begingroup$
    @Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:15












  • $begingroup$
    Except you misquoted heavily.
    $endgroup$
    – Did
    Dec 16 '18 at 13:17










  • $begingroup$
    @kimchilover If $mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:17








1




1




$begingroup$
You cannot apply the result you recall since it requires the increments are identically distributed -- which is not true in your case (except if $mu_n$ does not depend on $mu$).
$endgroup$
– Did
Dec 16 '18 at 13:04




$begingroup$
You cannot apply the result you recall since it requires the increments are identically distributed -- which is not true in your case (except if $mu_n$ does not depend on $mu$).
$endgroup$
– Did
Dec 16 '18 at 13:04












$begingroup$
To see why the $sum_nmu_n=infty$ requirement is needed consider what happens if $mu_1=1$ and $mu_n=0$ for $n>1$.
$endgroup$
– kimchi lover
Dec 16 '18 at 13:07






$begingroup$
To see why the $sum_nmu_n=infty$ requirement is needed consider what happens if $mu_1=1$ and $mu_n=0$ for $n>1$.
$endgroup$
– kimchi lover
Dec 16 '18 at 13:07














$begingroup$
@Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:15






$begingroup$
@Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:15














$begingroup$
Except you misquoted heavily.
$endgroup$
– Did
Dec 16 '18 at 13:17




$begingroup$
Except you misquoted heavily.
$endgroup$
– Did
Dec 16 '18 at 13:17












$begingroup$
@kimchilover If $mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:17




$begingroup$
@kimchilover If $mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:17










1 Answer
1






active

oldest

votes


















0












$begingroup$

If, in your argument, $M=sum_n mu_n<infty$ then your $Y_n$ converge in distribution to a Poisson rv. with expectation $M$, and your ratio $Y_n/EY_n$ has a non trivial limit distribution, violating a.s. convergence to a constant.



And, the theorem you quote needs a uniform 4th moment bound $EX_k^4le K$, which need not hold given your hypotheses.



A better way to prove your result might be to write your $Y_n$ as $Y_n=Y(M_n)$, where $M_n=sum_{tle n} mu_t$ and where $Y(t)$ is a Poisson process, as described in the wikipedia article. You might be able to show $Y(t)/tto 1$ almost surely.



Alternatively, you could use your argument with the added hypothesis that $mu_nle1$ for all $n$ and then use that to prove the general result by representing each $mu_n$ as a sum of quantities that are individually bounded by $1$, and your corresponding $Y_n$ as sums of independent Poissons whose expectations are bounded and add up to the right thing.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n rightarrow 0$ implies the claim only if $M = infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks!
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:38










  • $begingroup$
    Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process?
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:49










  • $begingroup$
    I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:58










  • $begingroup$
    Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 14:55











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

If, in your argument, $M=sum_n mu_n<infty$ then your $Y_n$ converge in distribution to a Poisson rv. with expectation $M$, and your ratio $Y_n/EY_n$ has a non trivial limit distribution, violating a.s. convergence to a constant.



And, the theorem you quote needs a uniform 4th moment bound $EX_k^4le K$, which need not hold given your hypotheses.



A better way to prove your result might be to write your $Y_n$ as $Y_n=Y(M_n)$, where $M_n=sum_{tle n} mu_t$ and where $Y(t)$ is a Poisson process, as described in the wikipedia article. You might be able to show $Y(t)/tto 1$ almost surely.



Alternatively, you could use your argument with the added hypothesis that $mu_nle1$ for all $n$ and then use that to prove the general result by representing each $mu_n$ as a sum of quantities that are individually bounded by $1$, and your corresponding $Y_n$ as sums of independent Poissons whose expectations are bounded and add up to the right thing.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n rightarrow 0$ implies the claim only if $M = infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks!
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:38










  • $begingroup$
    Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process?
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:49










  • $begingroup$
    I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:58










  • $begingroup$
    Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 14:55
















0












$begingroup$

If, in your argument, $M=sum_n mu_n<infty$ then your $Y_n$ converge in distribution to a Poisson rv. with expectation $M$, and your ratio $Y_n/EY_n$ has a non trivial limit distribution, violating a.s. convergence to a constant.



And, the theorem you quote needs a uniform 4th moment bound $EX_k^4le K$, which need not hold given your hypotheses.



A better way to prove your result might be to write your $Y_n$ as $Y_n=Y(M_n)$, where $M_n=sum_{tle n} mu_t$ and where $Y(t)$ is a Poisson process, as described in the wikipedia article. You might be able to show $Y(t)/tto 1$ almost surely.



Alternatively, you could use your argument with the added hypothesis that $mu_nle1$ for all $n$ and then use that to prove the general result by representing each $mu_n$ as a sum of quantities that are individually bounded by $1$, and your corresponding $Y_n$ as sums of independent Poissons whose expectations are bounded and add up to the right thing.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n rightarrow 0$ implies the claim only if $M = infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks!
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:38










  • $begingroup$
    Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process?
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:49










  • $begingroup$
    I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:58










  • $begingroup$
    Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 14:55














0












0








0





$begingroup$

If, in your argument, $M=sum_n mu_n<infty$ then your $Y_n$ converge in distribution to a Poisson rv. with expectation $M$, and your ratio $Y_n/EY_n$ has a non trivial limit distribution, violating a.s. convergence to a constant.



And, the theorem you quote needs a uniform 4th moment bound $EX_k^4le K$, which need not hold given your hypotheses.



A better way to prove your result might be to write your $Y_n$ as $Y_n=Y(M_n)$, where $M_n=sum_{tle n} mu_t$ and where $Y(t)$ is a Poisson process, as described in the wikipedia article. You might be able to show $Y(t)/tto 1$ almost surely.



Alternatively, you could use your argument with the added hypothesis that $mu_nle1$ for all $n$ and then use that to prove the general result by representing each $mu_n$ as a sum of quantities that are individually bounded by $1$, and your corresponding $Y_n$ as sums of independent Poissons whose expectations are bounded and add up to the right thing.






share|cite|improve this answer











$endgroup$



If, in your argument, $M=sum_n mu_n<infty$ then your $Y_n$ converge in distribution to a Poisson rv. with expectation $M$, and your ratio $Y_n/EY_n$ has a non trivial limit distribution, violating a.s. convergence to a constant.



And, the theorem you quote needs a uniform 4th moment bound $EX_k^4le K$, which need not hold given your hypotheses.



A better way to prove your result might be to write your $Y_n$ as $Y_n=Y(M_n)$, where $M_n=sum_{tle n} mu_t$ and where $Y(t)$ is a Poisson process, as described in the wikipedia article. You might be able to show $Y(t)/tto 1$ almost surely.



Alternatively, you could use your argument with the added hypothesis that $mu_nle1$ for all $n$ and then use that to prove the general result by representing each $mu_n$ as a sum of quantities that are individually bounded by $1$, and your corresponding $Y_n$ as sums of independent Poissons whose expectations are bounded and add up to the right thing.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 15:03

























answered Dec 16 '18 at 13:23









kimchi loverkimchi lover

9,70631128




9,70631128












  • $begingroup$
    I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n rightarrow 0$ implies the claim only if $M = infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks!
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:38










  • $begingroup$
    Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process?
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:49










  • $begingroup$
    I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:58










  • $begingroup$
    Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 14:55


















  • $begingroup$
    I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n rightarrow 0$ implies the claim only if $M = infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks!
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:38










  • $begingroup$
    Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process?
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 13:49










  • $begingroup$
    I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent.
    $endgroup$
    – kimchi lover
    Dec 16 '18 at 13:58










  • $begingroup$
    Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here
    $endgroup$
    – SomeoneHAHA
    Dec 16 '18 at 14:55
















$begingroup$
I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n rightarrow 0$ implies the claim only if $M = infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks!
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:38




$begingroup$
I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n rightarrow 0$ implies the claim only if $M = infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks!
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:38












$begingroup$
Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process?
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:49




$begingroup$
Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process?
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 13:49












$begingroup$
I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent.
$endgroup$
– kimchi lover
Dec 16 '18 at 13:58




$begingroup$
I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent.
$endgroup$
– kimchi lover
Dec 16 '18 at 13:58












$begingroup$
Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 14:55




$begingroup$
Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here
$endgroup$
– SomeoneHAHA
Dec 16 '18 at 14:55


















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