Why is my proof of the strong law of large numbers incorrect?












1














I was wondering where I have made a mistake in the following proof:



Let $(X_n)_{n ge 1}$ be a sequence of i.i.d random variables where $mathbb E (|X_1|)<infty$. Let $S_n = sum_{k=1}^{n} X_k$. Assume that $mathbb E(X_1)=0$, since we can just replace $X_n$ with $X_n - mathbb E(X_1)$. Then $mathbb E(S_n)=sum_{k=1}^n mathbb E(X_k)=0$ so $mathbb E(S_n/n)=0$ hence $sum_{nge 1}mathbb E(S_n/n)=0$ so $mathbb E(sum_{n ge 1} (S_n/n))=0$ so $sum_{n ge 1} (S_n/n)$ converges almost surely, so $S_n/n to 0$ almost surely.



This is essentially the proof my lecturer gave but he looked at $mathbb E (S_n^4)$ instead and it was a bit more complicated, and he had to make the additional assumption that $mathbb E (X_1^4) < infty$. I am not sure why my proof is wrong though.






probability







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  • Why are you allowed to interchange expectation and series?
    – Math_QED
    Nov 25 '18 at 22:23










  • @Math_QED I guess because the first series is convergent?
    – Devilo
    Nov 25 '18 at 22:30










  • On Page 55 of statslab.cam.ac.uk/~beresty/teach/pmnotes.pdf the sum is rearranged at the end. I assume this is valid, but in my case it is not. Why is that though?
    – Devilo
    Nov 25 '18 at 22:33










  • First, please state what you are trying to prove in the beginning, not the end. Second, I believe you mean that $mathbb{E}(S_n)=sum_{k=1}^nmathbb{E}(X_k)=0$. Third, I do not understand why $sum_{k=1}^nmathbb{E}(X_k)=0$.
    – Ben W
    Nov 25 '18 at 22:56










  • @Ben Yes, sorry. That is because each of the terms is 0.
    – Devilo
    Nov 25 '18 at 23:00
















1














I was wondering where I have made a mistake in the following proof:



Let $(X_n)_{n ge 1}$ be a sequence of i.i.d random variables where $mathbb E (|X_1|)<infty$. Let $S_n = sum_{k=1}^{n} X_k$. Assume that $mathbb E(X_1)=0$, since we can just replace $X_n$ with $X_n - mathbb E(X_1)$. Then $mathbb E(S_n)=sum_{k=1}^n mathbb E(X_k)=0$ so $mathbb E(S_n/n)=0$ hence $sum_{nge 1}mathbb E(S_n/n)=0$ so $mathbb E(sum_{n ge 1} (S_n/n))=0$ so $sum_{n ge 1} (S_n/n)$ converges almost surely, so $S_n/n to 0$ almost surely.



This is essentially the proof my lecturer gave but he looked at $mathbb E (S_n^4)$ instead and it was a bit more complicated, and he had to make the additional assumption that $mathbb E (X_1^4) < infty$. I am not sure why my proof is wrong though.






probability







share|cite|improve this question
























  • Why are you allowed to interchange expectation and series?
    – Math_QED
    Nov 25 '18 at 22:23










  • @Math_QED I guess because the first series is convergent?
    – Devilo
    Nov 25 '18 at 22:30










  • On Page 55 of statslab.cam.ac.uk/~beresty/teach/pmnotes.pdf the sum is rearranged at the end. I assume this is valid, but in my case it is not. Why is that though?
    – Devilo
    Nov 25 '18 at 22:33










  • First, please state what you are trying to prove in the beginning, not the end. Second, I believe you mean that $mathbb{E}(S_n)=sum_{k=1}^nmathbb{E}(X_k)=0$. Third, I do not understand why $sum_{k=1}^nmathbb{E}(X_k)=0$.
    – Ben W
    Nov 25 '18 at 22:56










  • @Ben Yes, sorry. That is because each of the terms is 0.
    – Devilo
    Nov 25 '18 at 23:00














1












1








1







I was wondering where I have made a mistake in the following proof:



Let $(X_n)_{n ge 1}$ be a sequence of i.i.d random variables where $mathbb E (|X_1|)<infty$. Let $S_n = sum_{k=1}^{n} X_k$. Assume that $mathbb E(X_1)=0$, since we can just replace $X_n$ with $X_n - mathbb E(X_1)$. Then $mathbb E(S_n)=sum_{k=1}^n mathbb E(X_k)=0$ so $mathbb E(S_n/n)=0$ hence $sum_{nge 1}mathbb E(S_n/n)=0$ so $mathbb E(sum_{n ge 1} (S_n/n))=0$ so $sum_{n ge 1} (S_n/n)$ converges almost surely, so $S_n/n to 0$ almost surely.



This is essentially the proof my lecturer gave but he looked at $mathbb E (S_n^4)$ instead and it was a bit more complicated, and he had to make the additional assumption that $mathbb E (X_1^4) < infty$. I am not sure why my proof is wrong though.






probability







share|cite|improve this question















I was wondering where I have made a mistake in the following proof:



Let $(X_n)_{n ge 1}$ be a sequence of i.i.d random variables where $mathbb E (|X_1|)<infty$. Let $S_n = sum_{k=1}^{n} X_k$. Assume that $mathbb E(X_1)=0$, since we can just replace $X_n$ with $X_n - mathbb E(X_1)$. Then $mathbb E(S_n)=sum_{k=1}^n mathbb E(X_k)=0$ so $mathbb E(S_n/n)=0$ hence $sum_{nge 1}mathbb E(S_n/n)=0$ so $mathbb E(sum_{n ge 1} (S_n/n))=0$ so $sum_{n ge 1} (S_n/n)$ converges almost surely, so $S_n/n to 0$ almost surely.



This is essentially the proof my lecturer gave but he looked at $mathbb E (S_n^4)$ instead and it was a bit more complicated, and he had to make the additional assumption that $mathbb E (X_1^4) < infty$. I am not sure why my proof is wrong though.






probability




probability






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share|cite|improve this question













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edited Nov 25 '18 at 22:59

























asked Nov 25 '18 at 22:20









Devilo

11718




11718












  • Why are you allowed to interchange expectation and series?
    – Math_QED
    Nov 25 '18 at 22:23










  • @Math_QED I guess because the first series is convergent?
    – Devilo
    Nov 25 '18 at 22:30










  • On Page 55 of statslab.cam.ac.uk/~beresty/teach/pmnotes.pdf the sum is rearranged at the end. I assume this is valid, but in my case it is not. Why is that though?
    – Devilo
    Nov 25 '18 at 22:33










  • First, please state what you are trying to prove in the beginning, not the end. Second, I believe you mean that $mathbb{E}(S_n)=sum_{k=1}^nmathbb{E}(X_k)=0$. Third, I do not understand why $sum_{k=1}^nmathbb{E}(X_k)=0$.
    – Ben W
    Nov 25 '18 at 22:56










  • @Ben Yes, sorry. That is because each of the terms is 0.
    – Devilo
    Nov 25 '18 at 23:00


















  • Why are you allowed to interchange expectation and series?
    – Math_QED
    Nov 25 '18 at 22:23










  • @Math_QED I guess because the first series is convergent?
    – Devilo
    Nov 25 '18 at 22:30










  • On Page 55 of statslab.cam.ac.uk/~beresty/teach/pmnotes.pdf the sum is rearranged at the end. I assume this is valid, but in my case it is not. Why is that though?
    – Devilo
    Nov 25 '18 at 22:33










  • First, please state what you are trying to prove in the beginning, not the end. Second, I believe you mean that $mathbb{E}(S_n)=sum_{k=1}^nmathbb{E}(X_k)=0$. Third, I do not understand why $sum_{k=1}^nmathbb{E}(X_k)=0$.
    – Ben W
    Nov 25 '18 at 22:56










  • @Ben Yes, sorry. That is because each of the terms is 0.
    – Devilo
    Nov 25 '18 at 23:00
















Why are you allowed to interchange expectation and series?
– Math_QED
Nov 25 '18 at 22:23




Why are you allowed to interchange expectation and series?
– Math_QED
Nov 25 '18 at 22:23












@Math_QED I guess because the first series is convergent?
– Devilo
Nov 25 '18 at 22:30




@Math_QED I guess because the first series is convergent?
– Devilo
Nov 25 '18 at 22:30












On Page 55 of statslab.cam.ac.uk/~beresty/teach/pmnotes.pdf the sum is rearranged at the end. I assume this is valid, but in my case it is not. Why is that though?
– Devilo
Nov 25 '18 at 22:33




On Page 55 of statslab.cam.ac.uk/~beresty/teach/pmnotes.pdf the sum is rearranged at the end. I assume this is valid, but in my case it is not. Why is that though?
– Devilo
Nov 25 '18 at 22:33












First, please state what you are trying to prove in the beginning, not the end. Second, I believe you mean that $mathbb{E}(S_n)=sum_{k=1}^nmathbb{E}(X_k)=0$. Third, I do not understand why $sum_{k=1}^nmathbb{E}(X_k)=0$.
– Ben W
Nov 25 '18 at 22:56




First, please state what you are trying to prove in the beginning, not the end. Second, I believe you mean that $mathbb{E}(S_n)=sum_{k=1}^nmathbb{E}(X_k)=0$. Third, I do not understand why $sum_{k=1}^nmathbb{E}(X_k)=0$.
– Ben W
Nov 25 '18 at 22:56












@Ben Yes, sorry. That is because each of the terms is 0.
– Devilo
Nov 25 '18 at 23:00




@Ben Yes, sorry. That is because each of the terms is 0.
– Devilo
Nov 25 '18 at 23:00










1 Answer
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You cannot just switch expectation and summation (try it, for instance, with $f_k(x) = 1_{[k,k+1]}-1$). In the comments, you write "the sum is rearranged at the end" of page 55, but it is not the same sum as the one you have. There, the sum is of a series with non-negative terms whose series of expectations is convergent. A well known theorem is implicitly used to justify convergence.



In general this is why even moments are used to prove large number type theorems with independence assumptions: even moments give rise to series with non-negative terms where elementary sum/integral exchange criteria can be applied.






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  • Thank you. I understand
    – Devilo
    Nov 25 '18 at 23:26











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1 Answer
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1 Answer
1






active

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active

oldest

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active

oldest

votes









2














You cannot just switch expectation and summation (try it, for instance, with $f_k(x) = 1_{[k,k+1]}-1$). In the comments, you write "the sum is rearranged at the end" of page 55, but it is not the same sum as the one you have. There, the sum is of a series with non-negative terms whose series of expectations is convergent. A well known theorem is implicitly used to justify convergence.



In general this is why even moments are used to prove large number type theorems with independence assumptions: even moments give rise to series with non-negative terms where elementary sum/integral exchange criteria can be applied.






share|cite|improve this answer





















  • Thank you. I understand
    – Devilo
    Nov 25 '18 at 23:26
















2














You cannot just switch expectation and summation (try it, for instance, with $f_k(x) = 1_{[k,k+1]}-1$). In the comments, you write "the sum is rearranged at the end" of page 55, but it is not the same sum as the one you have. There, the sum is of a series with non-negative terms whose series of expectations is convergent. A well known theorem is implicitly used to justify convergence.



In general this is why even moments are used to prove large number type theorems with independence assumptions: even moments give rise to series with non-negative terms where elementary sum/integral exchange criteria can be applied.






share|cite|improve this answer





















  • Thank you. I understand
    – Devilo
    Nov 25 '18 at 23:26














2












2








2






You cannot just switch expectation and summation (try it, for instance, with $f_k(x) = 1_{[k,k+1]}-1$). In the comments, you write "the sum is rearranged at the end" of page 55, but it is not the same sum as the one you have. There, the sum is of a series with non-negative terms whose series of expectations is convergent. A well known theorem is implicitly used to justify convergence.



In general this is why even moments are used to prove large number type theorems with independence assumptions: even moments give rise to series with non-negative terms where elementary sum/integral exchange criteria can be applied.






share|cite|improve this answer












You cannot just switch expectation and summation (try it, for instance, with $f_k(x) = 1_{[k,k+1]}-1$). In the comments, you write "the sum is rearranged at the end" of page 55, but it is not the same sum as the one you have. There, the sum is of a series with non-negative terms whose series of expectations is convergent. A well known theorem is implicitly used to justify convergence.



In general this is why even moments are used to prove large number type theorems with independence assumptions: even moments give rise to series with non-negative terms where elementary sum/integral exchange criteria can be applied.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 23:16









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  • Thank you. I understand
    – Devilo
    Nov 25 '18 at 23:26


















  • Thank you. I understand
    – Devilo
    Nov 25 '18 at 23:26
















Thank you. I understand
– Devilo
Nov 25 '18 at 23:26




Thank you. I understand
– Devilo
Nov 25 '18 at 23:26


















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