If $text Pleft[|X_n|>n^-alpharight]to0$ as $ntoinfty$ for some $alpha>0$, does $(X_n)_ninmathbb N$ converge in probability?Show that $sum_n=1^inftymathbbPleft(frac1nlvert X_nrvert>varepsilonright)<infty$.$sum_n=1^inftyPleft( |X_n-X|>epsilonright)<infty$ for each $epsilon>0$ $implies$ $X_nto X$ a.s.If $X_nto0$ in probability then $X_n/r_nto0$ in probability, for some $r_nto0$If $τ_n$ are stopping times with $τ_nuparrow T$ and $X_n$ are random variable, then $sum_n1_leftτ_n-1<trightX_n$ is convergentShow that $suplimits_tleft|X_t^nright|to0$ in probability iff $limlimits_ttoinfty[X^n]_tto0$ in probability when $ntoinfty$If $M$ is a local martingale and $τ:=infleftt≥0:$, then $text P[[M]_∞≥δ]≤text P[τ<∞]+text P[[M]_τ≥δ]$Does $X_nxrightarrow[nto+infty]law 0$ imply$mathbbEleft(log |1-X_n| right)xrightarrow[nto +infty] 0$?If $X_nto X$ in probability, are we able to conclude $sup_nge m|X_n-X|to 0$ in probability?Dominated convergence theorem for the conditional expectation under the assumption of convergence in probabilityHow is the weak law of large numbers applied here?
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If $text Pleft[|X_n|>n^-alpharight]to0$ as $ntoinfty$ for some $alpha>0$, does $(X_n)_ninmathbb N$ converge in probability?
Show that $sum_n=1^inftymathbbPleft(frac1nlvert X_nrvert>varepsilonright)<infty$.$sum_n=1^inftyPleft( |X_n-X|>epsilonright)<infty$ for each $epsilon>0$ $implies$ $X_nto X$ a.s.If $X_nto0$ in probability then $X_n/r_nto0$ in probability, for some $r_nto0$If $τ_n$ are stopping times with $τ_nuparrow T$ and $X_n$ are random variable, then $sum_n1_leftτ_n-1<trightX_n$ is convergentShow that $suplimits_tleft|X_t^nright|to0$ in probability iff $limlimits_ttoinfty[X^n]_tto0$ in probability when $ntoinfty$If $M$ is a local martingale and $τ:=infleftt≥0:$, then $text P[[M]_∞≥δ]≤text P[τ<∞]+text P[[M]_τ≥δ]$Does $X_nxrightarrow[nto+infty]law 0$ imply$mathbbEleft(log |1-X_n| right)xrightarrow[nto +infty] 0$?If $X_nto X$ in probability, are we able to conclude $sup_nge m|X_n-X|to 0$ in probability?Dominated convergence theorem for the conditional expectation under the assumption of convergence in probabilityHow is the weak law of large numbers applied here?
$begingroup$
Let $(X_n)_ninmathbb N$ be a sequence of real-valued random variables on a probability space $(Omega,mathcal A,operatorname P)$ and $alpha>0$. Is there some relation between convergence in probability, i.e. $$operatorname Pleft[|X_n|>varepsilonright]xrightarrowntoinfty0;;;textfor all varepsilon>0tag1$$ and $$operatorname Pleft[|X_n|>frac1n^alpharight]xrightarrowntoinfty0?tag2$$ It seems like neither implies the other. If so, can we deduce any other mode of convergence from $(2)$?
probability-theory measure-theory weak-convergence
edited Mar 29 at 21:54
Henning Makholm
243k17309554
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
$endgroup$
add a comment |
$begingroup$
Let $(X_n)_ninmathbb N$ be a sequence of real-valued random variables on a probability space $(Omega,mathcal A,operatorname P)$ and $alpha>0$. Is there some relation between convergence in probability, i.e. $$operatorname Pleft[|X_n|>varepsilonright]xrightarrowntoinfty0;;;textfor all varepsilon>0tag1$$ and $$operatorname Pleft[|X_n|>frac1n^alpharight]xrightarrowntoinfty0?tag2$$ It seems like neither implies the other. If so, can we deduce any other mode of convergence from $(2)$?
probability-theory measure-theory weak-convergence
edited Mar 29 at 21:54
Henning Makholm
243k17309554
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
$endgroup$
add a comment |
$begingroup$
Let $(X_n)_ninmathbb N$ be a sequence of real-valued random variables on a probability space $(Omega,mathcal A,operatorname P)$ and $alpha>0$. Is there some relation between convergence in probability, i.e. $$operatorname Pleft[|X_n|>varepsilonright]xrightarrowntoinfty0;;;textfor all varepsilon>0tag1$$ and $$operatorname Pleft[|X_n|>frac1n^alpharight]xrightarrowntoinfty0?tag2$$ It seems like neither implies the other. If so, can we deduce any other mode of convergence from $(2)$?
probability-theory measure-theory weak-convergence
edited Mar 29 at 21:54
Henning Makholm
243k17309554
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
$endgroup$
Let $(X_n)_ninmathbb N$ be a sequence of real-valued random variables on a probability space $(Omega,mathcal A,operatorname P)$ and $alpha>0$. Is there some relation between convergence in probability, i.e. $$operatorname Pleft[|X_n|>varepsilonright]xrightarrowntoinfty0;;;textfor all varepsilon>0tag1$$ and $$operatorname Pleft[|X_n|>frac1n^alpharight]xrightarrowntoinfty0?tag2$$ It seems like neither implies the other. If so, can we deduce any other mode of convergence from $(2)$?
probability-theory measure-theory weak-convergence
probability-theory measure-theory weak-convergence
edited Mar 29 at 21:54
Henning Makholm
243k17309554
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
edited Mar 29 at 21:54
Henning Makholm
243k17309554
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
edited Mar 29 at 21:54
Henning Makholm
243k17309554
edited Mar 29 at 21:54
Henning Makholm
243k17309554
edited Mar 29 at 21:54
Henning Makholm
243k17309554
243k17309554
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
asked Mar 29 at 21:17
0xbadf00d0xbadf00d
1,74141534
1,74141534
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$(2)Rightarrow(1)$: Given $epsilon>0$, then for any $n$ large enough so $n^-alpha<epsilon$, we have
$$
P(|X_n|>epsilon)<P(|X_n|>n^-alpha)to 0.
$$$(1);requirecancelnRightarrow; (2)$. Fix $beta$ so $0<beta<alpha$, and let $X_n=n^-beta$ deterministically.
$(2)$ does not imply almost sure convergence. Let $X_n$ be independent Bernoulli random variables with $P(X_n=1)=1/n$, then apply Borel-Cantelli.
$(2)$ does not implie $L_1$ convergence. On the probability space $[0,1]$ with Lebesgue measure, let $X_n=n^alphabf 1_(0,n^-alpha)$. Then $P(|X_n|>n^-alpha)=n^-alphato 0$, yet $E[|X_n|]=1notto 0$.
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
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oldest
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active
oldest
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$begingroup$
$(2)Rightarrow(1)$: Given $epsilon>0$, then for any $n$ large enough so $n^-alpha<epsilon$, we have
$$
P(|X_n|>epsilon)<P(|X_n|>n^-alpha)to 0.
$$$(1);requirecancelnRightarrow; (2)$. Fix $beta$ so $0<beta<alpha$, and let $X_n=n^-beta$ deterministically.
$(2)$ does not imply almost sure convergence. Let $X_n$ be independent Bernoulli random variables with $P(X_n=1)=1/n$, then apply Borel-Cantelli.
$(2)$ does not implie $L_1$ convergence. On the probability space $[0,1]$ with Lebesgue measure, let $X_n=n^alphabf 1_(0,n^-alpha)$. Then $P(|X_n|>n^-alpha)=n^-alphato 0$, yet $E[|X_n|]=1notto 0$.
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
$endgroup$
add a comment |
$begingroup$
$(2)Rightarrow(1)$: Given $epsilon>0$, then for any $n$ large enough so $n^-alpha<epsilon$, we have
$$
P(|X_n|>epsilon)<P(|X_n|>n^-alpha)to 0.
$$$(1);requirecancelnRightarrow; (2)$. Fix $beta$ so $0<beta<alpha$, and let $X_n=n^-beta$ deterministically.
$(2)$ does not imply almost sure convergence. Let $X_n$ be independent Bernoulli random variables with $P(X_n=1)=1/n$, then apply Borel-Cantelli.
$(2)$ does not implie $L_1$ convergence. On the probability space $[0,1]$ with Lebesgue measure, let $X_n=n^alphabf 1_(0,n^-alpha)$. Then $P(|X_n|>n^-alpha)=n^-alphato 0$, yet $E[|X_n|]=1notto 0$.
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
$endgroup$
add a comment |
$begingroup$
$(2)Rightarrow(1)$: Given $epsilon>0$, then for any $n$ large enough so $n^-alpha<epsilon$, we have
$$
P(|X_n|>epsilon)<P(|X_n|>n^-alpha)to 0.
$$$(1);requirecancelnRightarrow; (2)$. Fix $beta$ so $0<beta<alpha$, and let $X_n=n^-beta$ deterministically.
$(2)$ does not imply almost sure convergence. Let $X_n$ be independent Bernoulli random variables with $P(X_n=1)=1/n$, then apply Borel-Cantelli.
$(2)$ does not implie $L_1$ convergence. On the probability space $[0,1]$ with Lebesgue measure, let $X_n=n^alphabf 1_(0,n^-alpha)$. Then $P(|X_n|>n^-alpha)=n^-alphato 0$, yet $E[|X_n|]=1notto 0$.
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
$endgroup$
$(2)Rightarrow(1)$: Given $epsilon>0$, then for any $n$ large enough so $n^-alpha<epsilon$, we have
$$
P(|X_n|>epsilon)<P(|X_n|>n^-alpha)to 0.
$$$(1);requirecancelnRightarrow; (2)$. Fix $beta$ so $0<beta<alpha$, and let $X_n=n^-beta$ deterministically.
$(2)$ does not imply almost sure convergence. Let $X_n$ be independent Bernoulli random variables with $P(X_n=1)=1/n$, then apply Borel-Cantelli.
$(2)$ does not implie $L_1$ convergence. On the probability space $[0,1]$ with Lebesgue measure, let $X_n=n^alphabf 1_(0,n^-alpha)$. Then $P(|X_n|>n^-alpha)=n^-alphato 0$, yet $E[|X_n|]=1notto 0$.
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
answered Mar 29 at 21:58
Mike EarnestMike Earnest
27.1k22152
27.1k22152
add a comment |
add a comment |
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