Convergence almost surely - Example Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What almost sure convergence means in the context of strong law of large numbersQuestion regard almost sure convergenceProve that the series $sumlimits_n=0^inftyX_n$ converges almost surelymartingales, almost sure convergenceprove convergence almost surelyAlmost surely convergence of the sequenceProve that the sequence convergences almost surelyAlmost sure convergence to infinity of a sum of independent variablesConvergence of sum of expectationsAlmost sure convergence of posterior distribution to parameter
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Convergence almost surely - Example
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What almost sure convergence means in the context of strong law of large numbersQuestion regard almost sure convergenceProve that the series $sumlimits_n=0^inftyX_n$ converges almost surelymartingales, almost sure convergenceprove convergence almost surelyAlmost surely convergence of the sequenceProve that the sequence convergences almost surelyAlmost sure convergence to infinity of a sum of independent variablesConvergence of sum of expectationsAlmost sure convergence of posterior distribution to parameter
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Give an example of string Xn from n = 1 to infinity independent random variables with zero expected value such that arithmetic average ( sum Xn divided by n) → $ - infty$ almost surely. It is possible? I use strong law of large numbers and something is wrong :/
probability convergence
asked Mar 25 at 20:31
KingisKingis
65
$endgroup$
add a comment |
$begingroup$
Give an example of string Xn from n = 1 to infinity independent random variables with zero expected value such that arithmetic average ( sum Xn divided by n) → $ - infty$ almost surely. It is possible? I use strong law of large numbers and something is wrong :/
probability convergence
asked Mar 25 at 20:31
KingisKingis
65
$endgroup$
add a comment |
$begingroup$
Give an example of string Xn from n = 1 to infinity independent random variables with zero expected value such that arithmetic average ( sum Xn divided by n) → $ - infty$ almost surely. It is possible? I use strong law of large numbers and something is wrong :/
probability convergence
asked Mar 25 at 20:31
KingisKingis
65
$endgroup$
Give an example of string Xn from n = 1 to infinity independent random variables with zero expected value such that arithmetic average ( sum Xn divided by n) → $ - infty$ almost surely. It is possible? I use strong law of large numbers and something is wrong :/
probability convergence
probability convergence
asked Mar 25 at 20:31
KingisKingis
65
asked Mar 25 at 20:31
KingisKingis
65
asked Mar 25 at 20:31
KingisKingis
65
asked Mar 25 at 20:31
KingisKingis
65
asked Mar 25 at 20:31
KingisKingis
65
65
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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The key point is that the law of large numbers requires the random variables to be identically distributed. That assumption is lacking here.
For example, try $X_n = 2^2n$ with probability $2^-n$,
$-2^n/(1-2^-n)$ with probability $1 - 2^-n$.
Since $sum_n=1^infty 2^-n$ converges, with probability $1$ there are only finitely many positive $X_n$; when this happens, there is some finite $c$ such that $sum_n=1^N X_n le c - 2^N$, and so the average goes to $-infty$.
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
$endgroup$
$begingroup$
i don't understand the last line ? Can you explain me Why we have these c and N?
$endgroup$
– Kingis
Mar 25 at 20:55
$begingroup$
If the last positive value is $X_m$, then for $N > m$ we have $$sum_n=1^N X_n < sum_n=1^m X_n - 2^N$$
$endgroup$
– Robert Israel
Mar 26 at 3:51
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The key point is that the law of large numbers requires the random variables to be identically distributed. That assumption is lacking here.
For example, try $X_n = 2^2n$ with probability $2^-n$,
$-2^n/(1-2^-n)$ with probability $1 - 2^-n$.
Since $sum_n=1^infty 2^-n$ converges, with probability $1$ there are only finitely many positive $X_n$; when this happens, there is some finite $c$ such that $sum_n=1^N X_n le c - 2^N$, and so the average goes to $-infty$.
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
$endgroup$
$begingroup$
i don't understand the last line ? Can you explain me Why we have these c and N?
$endgroup$
– Kingis
Mar 25 at 20:55
$begingroup$
If the last positive value is $X_m$, then for $N > m$ we have $$sum_n=1^N X_n < sum_n=1^m X_n - 2^N$$
$endgroup$
– Robert Israel
Mar 26 at 3:51
add a comment |
$begingroup$
The key point is that the law of large numbers requires the random variables to be identically distributed. That assumption is lacking here.
For example, try $X_n = 2^2n$ with probability $2^-n$,
$-2^n/(1-2^-n)$ with probability $1 - 2^-n$.
Since $sum_n=1^infty 2^-n$ converges, with probability $1$ there are only finitely many positive $X_n$; when this happens, there is some finite $c$ such that $sum_n=1^N X_n le c - 2^N$, and so the average goes to $-infty$.
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
$endgroup$
$begingroup$
i don't understand the last line ? Can you explain me Why we have these c and N?
$endgroup$
– Kingis
Mar 25 at 20:55
$begingroup$
If the last positive value is $X_m$, then for $N > m$ we have $$sum_n=1^N X_n < sum_n=1^m X_n - 2^N$$
$endgroup$
– Robert Israel
Mar 26 at 3:51
add a comment |
$begingroup$
The key point is that the law of large numbers requires the random variables to be identically distributed. That assumption is lacking here.
For example, try $X_n = 2^2n$ with probability $2^-n$,
$-2^n/(1-2^-n)$ with probability $1 - 2^-n$.
Since $sum_n=1^infty 2^-n$ converges, with probability $1$ there are only finitely many positive $X_n$; when this happens, there is some finite $c$ such that $sum_n=1^N X_n le c - 2^N$, and so the average goes to $-infty$.
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
$endgroup$
The key point is that the law of large numbers requires the random variables to be identically distributed. That assumption is lacking here.
For example, try $X_n = 2^2n$ with probability $2^-n$,
$-2^n/(1-2^-n)$ with probability $1 - 2^-n$.
Since $sum_n=1^infty 2^-n$ converges, with probability $1$ there are only finitely many positive $X_n$; when this happens, there is some finite $c$ such that $sum_n=1^N X_n le c - 2^N$, and so the average goes to $-infty$.
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
answered Mar 25 at 20:43
Robert IsraelRobert Israel
331k23221478
331k23221478
$begingroup$
i don't understand the last line ? Can you explain me Why we have these c and N?
$endgroup$
– Kingis
Mar 25 at 20:55
$begingroup$
If the last positive value is $X_m$, then for $N > m$ we have $$sum_n=1^N X_n < sum_n=1^m X_n - 2^N$$
$endgroup$
– Robert Israel
Mar 26 at 3:51
add a comment |
$begingroup$
i don't understand the last line ? Can you explain me Why we have these c and N?
$endgroup$
– Kingis
Mar 25 at 20:55
$begingroup$
If the last positive value is $X_m$, then for $N > m$ we have $$sum_n=1^N X_n < sum_n=1^m X_n - 2^N$$
$endgroup$
– Robert Israel
Mar 26 at 3:51
$begingroup$
i don't understand the last line ? Can you explain me Why we have these c and N?
$endgroup$
– Kingis
Mar 25 at 20:55
$begingroup$
i don't understand the last line ? Can you explain me Why we have these c and N?
$endgroup$
– Kingis
Mar 25 at 20:55
$begingroup$
If the last positive value is $X_m$, then for $N > m$ we have $$sum_n=1^N X_n < sum_n=1^m X_n - 2^N$$
$endgroup$
– Robert Israel
Mar 26 at 3:51
$begingroup$
If the last positive value is $X_m$, then for $N > m$ we have $$sum_n=1^N X_n < sum_n=1^m X_n - 2^N$$
$endgroup$
– Robert Israel
Mar 26 at 3:51
add a comment |
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