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Find the sum of an infinite series that involve a discrete random variable
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I need to find the sum of the following series $sumlimits_n=1^infty fracX_ne^n$ , where $X_n$ be I.I.D ,$P(X_n=1)=frac12$ or $P(X_n=-1)=frac12$
Using the ratio test, it can be shown that this series is convergent.Because,
$lim left| fraca_n+1a_n right|= lim left| frac1e right|$<1 .
But is there any method to find the actual sum ??
can i split this sum as $sumlimits_n=1^infty frac1e^n$ - $sumlimits_n=1^infty frac1e^n$ since $X_n$ can only take 1 and -1 ?
calculus probability sequences-and-series self-learning
edited Jul 24 at 0:56
Math1000
18.4k31444
asked Jul 23 at 23:35
Stat lover
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up vote
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down vote
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I need to find the sum of the following series $sumlimits_n=1^infty fracX_ne^n$ , where $X_n$ be I.I.D ,$P(X_n=1)=frac12$ or $P(X_n=-1)=frac12$
Using the ratio test, it can be shown that this series is convergent.Because,
$lim left| fraca_n+1a_n right|= lim left| frac1e right|$<1 .
But is there any method to find the actual sum ??
can i split this sum as $sumlimits_n=1^infty frac1e^n$ - $sumlimits_n=1^infty frac1e^n$ since $X_n$ can only take 1 and -1 ?
calculus probability sequences-and-series self-learning
edited Jul 24 at 0:56
Math1000
18.4k31444
asked Jul 23 at 23:35
Stat lover
4411
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to find the sum of the following series $sumlimits_n=1^infty fracX_ne^n$ , where $X_n$ be I.I.D ,$P(X_n=1)=frac12$ or $P(X_n=-1)=frac12$
Using the ratio test, it can be shown that this series is convergent.Because,
$lim left| fraca_n+1a_n right|= lim left| frac1e right|$<1 .
But is there any method to find the actual sum ??
can i split this sum as $sumlimits_n=1^infty frac1e^n$ - $sumlimits_n=1^infty frac1e^n$ since $X_n$ can only take 1 and -1 ?
calculus probability sequences-and-series self-learning
edited Jul 24 at 0:56
Math1000
18.4k31444
asked Jul 23 at 23:35
Stat lover
4411
I need to find the sum of the following series $sumlimits_n=1^infty fracX_ne^n$ , where $X_n$ be I.I.D ,$P(X_n=1)=frac12$ or $P(X_n=-1)=frac12$
Using the ratio test, it can be shown that this series is convergent.Because,
$lim left| fraca_n+1a_n right|= lim left| frac1e right|$<1 .
But is there any method to find the actual sum ??
can i split this sum as $sumlimits_n=1^infty frac1e^n$ - $sumlimits_n=1^infty frac1e^n$ since $X_n$ can only take 1 and -1 ?
calculus probability sequences-and-series self-learning
edited Jul 24 at 0:56
Math1000
18.4k31444
asked Jul 23 at 23:35
Stat lover
4411
edited Jul 24 at 0:56
Math1000
18.4k31444
edited Jul 24 at 0:56
Math1000
18.4k31444
edited Jul 24 at 0:56
Math1000
18.4k31444
18.4k31444
asked Jul 23 at 23:35
Stat lover
4411
asked Jul 23 at 23:35
Stat lover
4411
asked Jul 23 at 23:35
Stat lover
4411
4411
add a comment |Â
add a comment |Â
1 Answer
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You are only given the distribution of $X_n$'s and not a specific definition as functions on a sample space. You cannot write down the sum of the series. But one cans show that the series converges with probability $1$.
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
1
@Statlover The sum is not $0$ and you cannot evaluate the sum.
– Kavi Rama Murthy
Jul 23 at 23:45
3
@Statlover The variance of the infinite sum will be $sum_n=1^infty e^-2n = frace^-21-e^-2>0$. The sum cannot be equal to $0$ a.s...
– Clement C.
Jul 24 at 0:15
2
@Statlover The key here is that the different 'realizations' of the randomness (i.e. the different $omega$ in the sample space $Omega$) will yield different values for the $X_i$. You have seen that with probability one their sum will converge. However it will converge to different values for different realizations of the $X_i.$ The sum is a random variable, just like the $X_i.$ We can define it formally as $S(omega) = limsup_n sum_i=1^nX_i(omega)$ (we use limsup so it will always exist, even for the (possible) null set of $omega$'s for which the sum diverges.)
– spaceisdarkgreen
Jul 24 at 0:28
1
Let $S_n=sum_k=1^n e^-kX_k$ and $S=lim_ntoinfty S_n$.. Note that $$sum_k=2^infty e^-k = frac1e(e-1) approx 0.214, $$ so $$mathbb P(S >1/2) geqslant mathbb P(X_1=1) = frac12$$ and similarly $mathbb P(S<-1/2)=1/2$. So $S_n$ does not converge almost surely to zero.
– Math1000
Jul 24 at 1:26
1
@Statlover Well a Vitali set is not measurable, but yes this concentrates on a Cantor-like set of measure zero. The seeds of this are actually in Math1000's hint above where we see that the $S$ is excluded from the interval $(-.154, .154).$ (They actually make a small error here when they say $P(S>1/2)$... it should be $P(S>1/e-.214)$)
– spaceisdarkgreen
Jul 24 at 2:07
 |Â
show 8 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You are only given the distribution of $X_n$'s and not a specific definition as functions on a sample space. You cannot write down the sum of the series. But one cans show that the series converges with probability $1$.
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
1
@Statlover The sum is not $0$ and you cannot evaluate the sum.
– Kavi Rama Murthy
Jul 23 at 23:45
3
@Statlover The variance of the infinite sum will be $sum_n=1^infty e^-2n = frace^-21-e^-2>0$. The sum cannot be equal to $0$ a.s...
– Clement C.
Jul 24 at 0:15
2
@Statlover The key here is that the different 'realizations' of the randomness (i.e. the different $omega$ in the sample space $Omega$) will yield different values for the $X_i$. You have seen that with probability one their sum will converge. However it will converge to different values for different realizations of the $X_i.$ The sum is a random variable, just like the $X_i.$ We can define it formally as $S(omega) = limsup_n sum_i=1^nX_i(omega)$ (we use limsup so it will always exist, even for the (possible) null set of $omega$'s for which the sum diverges.)
– spaceisdarkgreen
Jul 24 at 0:28
1
Let $S_n=sum_k=1^n e^-kX_k$ and $S=lim_ntoinfty S_n$.. Note that $$sum_k=2^infty e^-k = frac1e(e-1) approx 0.214, $$ so $$mathbb P(S >1/2) geqslant mathbb P(X_1=1) = frac12$$ and similarly $mathbb P(S<-1/2)=1/2$. So $S_n$ does not converge almost surely to zero.
– Math1000
Jul 24 at 1:26
1
@Statlover Well a Vitali set is not measurable, but yes this concentrates on a Cantor-like set of measure zero. The seeds of this are actually in Math1000's hint above where we see that the $S$ is excluded from the interval $(-.154, .154).$ (They actually make a small error here when they say $P(S>1/2)$... it should be $P(S>1/e-.214)$)
– spaceisdarkgreen
Jul 24 at 2:07
 |Â
show 8 more comments
up vote
0
down vote
You are only given the distribution of $X_n$'s and not a specific definition as functions on a sample space. You cannot write down the sum of the series. But one cans show that the series converges with probability $1$.
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
1
@Statlover The sum is not $0$ and you cannot evaluate the sum.
– Kavi Rama Murthy
Jul 23 at 23:45
3
@Statlover The variance of the infinite sum will be $sum_n=1^infty e^-2n = frace^-21-e^-2>0$. The sum cannot be equal to $0$ a.s...
– Clement C.
Jul 24 at 0:15
2
@Statlover The key here is that the different 'realizations' of the randomness (i.e. the different $omega$ in the sample space $Omega$) will yield different values for the $X_i$. You have seen that with probability one their sum will converge. However it will converge to different values for different realizations of the $X_i.$ The sum is a random variable, just like the $X_i.$ We can define it formally as $S(omega) = limsup_n sum_i=1^nX_i(omega)$ (we use limsup so it will always exist, even for the (possible) null set of $omega$'s for which the sum diverges.)
– spaceisdarkgreen
Jul 24 at 0:28
1
Let $S_n=sum_k=1^n e^-kX_k$ and $S=lim_ntoinfty S_n$.. Note that $$sum_k=2^infty e^-k = frac1e(e-1) approx 0.214, $$ so $$mathbb P(S >1/2) geqslant mathbb P(X_1=1) = frac12$$ and similarly $mathbb P(S<-1/2)=1/2$. So $S_n$ does not converge almost surely to zero.
– Math1000
Jul 24 at 1:26
1
@Statlover Well a Vitali set is not measurable, but yes this concentrates on a Cantor-like set of measure zero. The seeds of this are actually in Math1000's hint above where we see that the $S$ is excluded from the interval $(-.154, .154).$ (They actually make a small error here when they say $P(S>1/2)$... it should be $P(S>1/e-.214)$)
– spaceisdarkgreen
Jul 24 at 2:07
 |Â
show 8 more comments
up vote
0
down vote
up vote
0
down vote
You are only given the distribution of $X_n$'s and not a specific definition as functions on a sample space. You cannot write down the sum of the series. But one cans show that the series converges with probability $1$.
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
You are only given the distribution of $X_n$'s and not a specific definition as functions on a sample space. You cannot write down the sum of the series. But one cans show that the series converges with probability $1$.
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
answered Jul 23 at 23:38
Kavi Rama Murthy
20.2k2829
20.2k2829
1
@Statlover The sum is not $0$ and you cannot evaluate the sum.
– Kavi Rama Murthy
Jul 23 at 23:45
3
@Statlover The variance of the infinite sum will be $sum_n=1^infty e^-2n = frace^-21-e^-2>0$. The sum cannot be equal to $0$ a.s...
– Clement C.
Jul 24 at 0:15
2
@Statlover The key here is that the different 'realizations' of the randomness (i.e. the different $omega$ in the sample space $Omega$) will yield different values for the $X_i$. You have seen that with probability one their sum will converge. However it will converge to different values for different realizations of the $X_i.$ The sum is a random variable, just like the $X_i.$ We can define it formally as $S(omega) = limsup_n sum_i=1^nX_i(omega)$ (we use limsup so it will always exist, even for the (possible) null set of $omega$'s for which the sum diverges.)
– spaceisdarkgreen
Jul 24 at 0:28
1
Let $S_n=sum_k=1^n e^-kX_k$ and $S=lim_ntoinfty S_n$.. Note that $$sum_k=2^infty e^-k = frac1e(e-1) approx 0.214, $$ so $$mathbb P(S >1/2) geqslant mathbb P(X_1=1) = frac12$$ and similarly $mathbb P(S<-1/2)=1/2$. So $S_n$ does not converge almost surely to zero.
– Math1000
Jul 24 at 1:26
1
@Statlover Well a Vitali set is not measurable, but yes this concentrates on a Cantor-like set of measure zero. The seeds of this are actually in Math1000's hint above where we see that the $S$ is excluded from the interval $(-.154, .154).$ (They actually make a small error here when they say $P(S>1/2)$... it should be $P(S>1/e-.214)$)
– spaceisdarkgreen
Jul 24 at 2:07
 |Â
show 8 more comments
1
@Statlover The sum is not $0$ and you cannot evaluate the sum.
– Kavi Rama Murthy
Jul 23 at 23:45
3
@Statlover The variance of the infinite sum will be $sum_n=1^infty e^-2n = frace^-21-e^-2>0$. The sum cannot be equal to $0$ a.s...
– Clement C.
Jul 24 at 0:15
2
@Statlover The key here is that the different 'realizations' of the randomness (i.e. the different $omega$ in the sample space $Omega$) will yield different values for the $X_i$. You have seen that with probability one their sum will converge. However it will converge to different values for different realizations of the $X_i.$ The sum is a random variable, just like the $X_i.$ We can define it formally as $S(omega) = limsup_n sum_i=1^nX_i(omega)$ (we use limsup so it will always exist, even for the (possible) null set of $omega$'s for which the sum diverges.)
– spaceisdarkgreen
Jul 24 at 0:28
1
Let $S_n=sum_k=1^n e^-kX_k$ and $S=lim_ntoinfty S_n$.. Note that $$sum_k=2^infty e^-k = frac1e(e-1) approx 0.214, $$ so $$mathbb P(S >1/2) geqslant mathbb P(X_1=1) = frac12$$ and similarly $mathbb P(S<-1/2)=1/2$. So $S_n$ does not converge almost surely to zero.
– Math1000
Jul 24 at 1:26
1
@Statlover Well a Vitali set is not measurable, but yes this concentrates on a Cantor-like set of measure zero. The seeds of this are actually in Math1000's hint above where we see that the $S$ is excluded from the interval $(-.154, .154).$ (They actually make a small error here when they say $P(S>1/2)$... it should be $P(S>1/e-.214)$)
– spaceisdarkgreen
Jul 24 at 2:07
1
1
@Statlover The sum is not $0$ and you cannot evaluate the sum.
– Kavi Rama Murthy
Jul 23 at 23:45
@Statlover The sum is not $0$ and you cannot evaluate the sum.
– Kavi Rama Murthy
Jul 23 at 23:45
3
3
@Statlover The variance of the infinite sum will be $sum_n=1^infty e^-2n = frace^-21-e^-2>0$. The sum cannot be equal to $0$ a.s...
– Clement C.
Jul 24 at 0:15
@Statlover The variance of the infinite sum will be $sum_n=1^infty e^-2n = frace^-21-e^-2>0$. The sum cannot be equal to $0$ a.s...
– Clement C.
Jul 24 at 0:15
2
2
@Statlover The key here is that the different 'realizations' of the randomness (i.e. the different $omega$ in the sample space $Omega$) will yield different values for the $X_i$. You have seen that with probability one their sum will converge. However it will converge to different values for different realizations of the $X_i.$ The sum is a random variable, just like the $X_i.$ We can define it formally as $S(omega) = limsup_n sum_i=1^nX_i(omega)$ (we use limsup so it will always exist, even for the (possible) null set of $omega$'s for which the sum diverges.)
– spaceisdarkgreen
Jul 24 at 0:28
@Statlover The key here is that the different 'realizations' of the randomness (i.e. the different $omega$ in the sample space $Omega$) will yield different values for the $X_i$. You have seen that with probability one their sum will converge. However it will converge to different values for different realizations of the $X_i.$ The sum is a random variable, just like the $X_i.$ We can define it formally as $S(omega) = limsup_n sum_i=1^nX_i(omega)$ (we use limsup so it will always exist, even for the (possible) null set of $omega$'s for which the sum diverges.)
– spaceisdarkgreen
Jul 24 at 0:28
1
1
Let $S_n=sum_k=1^n e^-kX_k$ and $S=lim_ntoinfty S_n$.. Note that $$sum_k=2^infty e^-k = frac1e(e-1) approx 0.214, $$ so $$mathbb P(S >1/2) geqslant mathbb P(X_1=1) = frac12$$ and similarly $mathbb P(S<-1/2)=1/2$. So $S_n$ does not converge almost surely to zero.
– Math1000
Jul 24 at 1:26
Let $S_n=sum_k=1^n e^-kX_k$ and $S=lim_ntoinfty S_n$.. Note that $$sum_k=2^infty e^-k = frac1e(e-1) approx 0.214, $$ so $$mathbb P(S >1/2) geqslant mathbb P(X_1=1) = frac12$$ and similarly $mathbb P(S<-1/2)=1/2$. So $S_n$ does not converge almost surely to zero.
– Math1000
Jul 24 at 1:26
1
1
@Statlover Well a Vitali set is not measurable, but yes this concentrates on a Cantor-like set of measure zero. The seeds of this are actually in Math1000's hint above where we see that the $S$ is excluded from the interval $(-.154, .154).$ (They actually make a small error here when they say $P(S>1/2)$... it should be $P(S>1/e-.214)$)
– spaceisdarkgreen
Jul 24 at 2:07
@Statlover Well a Vitali set is not measurable, but yes this concentrates on a Cantor-like set of measure zero. The seeds of this are actually in Math1000's hint above where we see that the $S$ is excluded from the interval $(-.154, .154).$ (They actually make a small error here when they say $P(S>1/2)$... it should be $P(S>1/e-.214)$)
– spaceisdarkgreen
Jul 24 at 2:07
 |Â
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