Show as surely convergence (Borel Cantelli Lemma)
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Let $X_n$ be independent r.v. taking values $n^2-1$ and $-1$ with $P(X_n=n^2-1)=1/n^2$ and $P(X_n=-1)=1-frac1n^2$. Show that if $S_k=X_1+...+X_k$ then $S_k/k$ converges to -1 a.s but $E(S_k)=0$.
Attempt: I can see that $P(X_n = n^2 -1i.o.)=0$ as the sum $1/n^2$ is finite. After this how to proceed? What do I do with $P(X_n=-1)=1-frac1n^2$?
probability-theory
asked Aug 10 at 21:46
Dom
1587
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up vote
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down vote
favorite
Let $X_n$ be independent r.v. taking values $n^2-1$ and $-1$ with $P(X_n=n^2-1)=1/n^2$ and $P(X_n=-1)=1-frac1n^2$. Show that if $S_k=X_1+...+X_k$ then $S_k/k$ converges to -1 a.s but $E(S_k)=0$.
Attempt: I can see that $P(X_n = n^2 -1i.o.)=0$ as the sum $1/n^2$ is finite. After this how to proceed? What do I do with $P(X_n=-1)=1-frac1n^2$?
probability-theory
asked Aug 10 at 21:46
Dom
1587
Compute $E(S_k) = E(X_1)+dots+E(X_k)$.
– GEdgar
Aug 10 at 21:52
Is E(X_k) = 0 for all k?
– Dom
Aug 10 at 21:53
From the definition, it should be easy to compute $E(X_k)$.
– GEdgar
Aug 10 at 21:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X_n$ be independent r.v. taking values $n^2-1$ and $-1$ with $P(X_n=n^2-1)=1/n^2$ and $P(X_n=-1)=1-frac1n^2$. Show that if $S_k=X_1+...+X_k$ then $S_k/k$ converges to -1 a.s but $E(S_k)=0$.
Attempt: I can see that $P(X_n = n^2 -1i.o.)=0$ as the sum $1/n^2$ is finite. After this how to proceed? What do I do with $P(X_n=-1)=1-frac1n^2$?
probability-theory
asked Aug 10 at 21:46
Dom
1587
Let $X_n$ be independent r.v. taking values $n^2-1$ and $-1$ with $P(X_n=n^2-1)=1/n^2$ and $P(X_n=-1)=1-frac1n^2$. Show that if $S_k=X_1+...+X_k$ then $S_k/k$ converges to -1 a.s but $E(S_k)=0$.
Attempt: I can see that $P(X_n = n^2 -1i.o.)=0$ as the sum $1/n^2$ is finite. After this how to proceed? What do I do with $P(X_n=-1)=1-frac1n^2$?
probability-theory
asked Aug 10 at 21:46
Dom
1587
asked Aug 10 at 21:46
Dom
1587
asked Aug 10 at 21:46
Dom
1587
asked Aug 10 at 21:46
Dom
1587
1587
Compute $E(S_k) = E(X_1)+dots+E(X_k)$.
– GEdgar
Aug 10 at 21:52
Is E(X_k) = 0 for all k?
– Dom
Aug 10 at 21:53
From the definition, it should be easy to compute $E(X_k)$.
– GEdgar
Aug 10 at 21:54
add a comment |Â
Compute $E(S_k) = E(X_1)+dots+E(X_k)$.
– GEdgar
Aug 10 at 21:52
Is E(X_k) = 0 for all k?
– Dom
Aug 10 at 21:53
From the definition, it should be easy to compute $E(X_k)$.
– GEdgar
Aug 10 at 21:54
Compute $E(S_k) = E(X_1)+dots+E(X_k)$.
– GEdgar
Aug 10 at 21:52
Compute $E(S_k) = E(X_1)+dots+E(X_k)$.
– GEdgar
Aug 10 at 21:52
Is E(X_k) = 0 for all k?
– Dom
Aug 10 at 21:53
Is E(X_k) = 0 for all k?
– Dom
Aug 10 at 21:53
From the definition, it should be easy to compute $E(X_k)$.
– GEdgar
Aug 10 at 21:54
From the definition, it should be easy to compute $E(X_k)$.
– GEdgar
Aug 10 at 21:54
add a comment |Â
1 Answer
1
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Note that $A_n, i.o.$ is identical to $limsuplimits_nto infty A_n$.
Let $A_n:= omega : X_n(omega) = n^2-1$. Then, $P(liminflimits_nto infty A_n^c) = 1- P(A_n ,i.o.) = 1$.
For (a.a.) $omega in liminflimits_nto infty A_n^c$, by definition there exists $N=N(omega)$ s.d. $forall ngeq N$, $X_n(omega) = -1$. Hence, $X_n(omega) to -1$. Now it's just real analysis and a classic result of Cesaro sum that if a sequence $a_n to a$, then $sumlimits_i=1^n a_i /n to a$.
It suffices to show $E(X_n) = 0$. In that endeavor, $E(X_n) = (n^2-1) cdot frac1n^2 - (1-frac1n^2) = 0$. Then $E(S_k) = frac1k sumlimits_i=1^k E(X_i) = 0$.
answered Aug 10 at 22:04
James Yang
4449
And the E(S_k) part? It seems simple but I do not see it.
– Dom
Aug 10 at 22:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that $A_n, i.o.$ is identical to $limsuplimits_nto infty A_n$.
Let $A_n:= omega : X_n(omega) = n^2-1$. Then, $P(liminflimits_nto infty A_n^c) = 1- P(A_n ,i.o.) = 1$.
For (a.a.) $omega in liminflimits_nto infty A_n^c$, by definition there exists $N=N(omega)$ s.d. $forall ngeq N$, $X_n(omega) = -1$. Hence, $X_n(omega) to -1$. Now it's just real analysis and a classic result of Cesaro sum that if a sequence $a_n to a$, then $sumlimits_i=1^n a_i /n to a$.
It suffices to show $E(X_n) = 0$. In that endeavor, $E(X_n) = (n^2-1) cdot frac1n^2 - (1-frac1n^2) = 0$. Then $E(S_k) = frac1k sumlimits_i=1^k E(X_i) = 0$.
answered Aug 10 at 22:04
James Yang
4449
And the E(S_k) part? It seems simple but I do not see it.
– Dom
Aug 10 at 22:10
add a comment |Â
up vote
1
down vote
accepted
Note that $A_n, i.o.$ is identical to $limsuplimits_nto infty A_n$.
Let $A_n:= omega : X_n(omega) = n^2-1$. Then, $P(liminflimits_nto infty A_n^c) = 1- P(A_n ,i.o.) = 1$.
For (a.a.) $omega in liminflimits_nto infty A_n^c$, by definition there exists $N=N(omega)$ s.d. $forall ngeq N$, $X_n(omega) = -1$. Hence, $X_n(omega) to -1$. Now it's just real analysis and a classic result of Cesaro sum that if a sequence $a_n to a$, then $sumlimits_i=1^n a_i /n to a$.
It suffices to show $E(X_n) = 0$. In that endeavor, $E(X_n) = (n^2-1) cdot frac1n^2 - (1-frac1n^2) = 0$. Then $E(S_k) = frac1k sumlimits_i=1^k E(X_i) = 0$.
answered Aug 10 at 22:04
James Yang
4449
And the E(S_k) part? It seems simple but I do not see it.
– Dom
Aug 10 at 22:10
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that $A_n, i.o.$ is identical to $limsuplimits_nto infty A_n$.
Let $A_n:= omega : X_n(omega) = n^2-1$. Then, $P(liminflimits_nto infty A_n^c) = 1- P(A_n ,i.o.) = 1$.
For (a.a.) $omega in liminflimits_nto infty A_n^c$, by definition there exists $N=N(omega)$ s.d. $forall ngeq N$, $X_n(omega) = -1$. Hence, $X_n(omega) to -1$. Now it's just real analysis and a classic result of Cesaro sum that if a sequence $a_n to a$, then $sumlimits_i=1^n a_i /n to a$.
It suffices to show $E(X_n) = 0$. In that endeavor, $E(X_n) = (n^2-1) cdot frac1n^2 - (1-frac1n^2) = 0$. Then $E(S_k) = frac1k sumlimits_i=1^k E(X_i) = 0$.
answered Aug 10 at 22:04
James Yang
4449
Note that $A_n, i.o.$ is identical to $limsuplimits_nto infty A_n$.
Let $A_n:= omega : X_n(omega) = n^2-1$. Then, $P(liminflimits_nto infty A_n^c) = 1- P(A_n ,i.o.) = 1$.
For (a.a.) $omega in liminflimits_nto infty A_n^c$, by definition there exists $N=N(omega)$ s.d. $forall ngeq N$, $X_n(omega) = -1$. Hence, $X_n(omega) to -1$. Now it's just real analysis and a classic result of Cesaro sum that if a sequence $a_n to a$, then $sumlimits_i=1^n a_i /n to a$.
It suffices to show $E(X_n) = 0$. In that endeavor, $E(X_n) = (n^2-1) cdot frac1n^2 - (1-frac1n^2) = 0$. Then $E(S_k) = frac1k sumlimits_i=1^k E(X_i) = 0$.
answered Aug 10 at 22:04
James Yang
4449
edited Aug 10 at 22:19
answered Aug 10 at 22:04
James Yang
4449
answered Aug 10 at 22:04
James Yang
4449
answered Aug 10 at 22:04
James Yang
4449
4449
And the E(S_k) part? It seems simple but I do not see it.
– Dom
Aug 10 at 22:10
add a comment |Â
And the E(S_k) part? It seems simple but I do not see it.
– Dom
Aug 10 at 22:10
And the E(S_k) part? It seems simple but I do not see it.
– Dom
Aug 10 at 22:10
And the E(S_k) part? It seems simple but I do not see it.
– Dom
Aug 10 at 22:10
add a comment |Â
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Compute $E(S_k) = E(X_1)+dots+E(X_k)$.
– GEdgar
Aug 10 at 21:52
Is E(X_k) = 0 for all k?
– Dom
Aug 10 at 21:53
From the definition, it should be easy to compute $E(X_k)$.
– GEdgar
Aug 10 at 21:54