Property of uniformly tight random variables?

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I'm stumped on the following question, which is problem 1.3.9 in the book Weak Convergence and Empirical Proceses by van der Vaart and Wellner. It is based on the following notion of asymptotic tightness:
Definition: A sequence of random variables $X_1, X_2, ldots$ taking values in a metric space with metric $d$ is asymptotically tight if for every $epsilon > 0$ there exists a compact $K_0$ such that for every $delta > 0$ we have $varliminf_n to infty P(X_n in K_0^delta) ge 1 - epsilon$ where $K_0^delta = x: d(x, K_0) < delta$ is the $delta$-enlargement of $K_0$.
For convenience, the definition of uniform tightness is
Definition: A sequence of random variables $X_1, X_2, ldots$ is uniformly tight if for every $epsilon > 0$ there exists a compact $K$ such that $P(X_n in K) ge 1 - epsilon$ for every $K$.
The problem is:
Claim: A sequence $X_n$ is uniformly tight if and only if it is asymptotically tight and each $X_n$ is tight.
I cannot seem to prove the non-trivial direction of this claim. The following hint is given:
Hint: Fix $epsilon > 0$. Take a compact $K_0$ with $varliminf P(X_n in K_0^delta) ge 1 - epsilon$ for every $delta$. Choose $n_1 < n_2 < cdots $ such that $P(X_n in K_0^1/m) ge 1 - 2 epsilon$ for $n ge n_m$. For $n_m < n le n_m+1$ choose a compact $K_n$ with $P(X_n in K_0^1/m - K_n) < epsilon$. Now $K = bigcup_n = 0 ^ infty K_n$ is compact and $P(X_n in K) ge 1 - 3 epsilon$.
Now, I understand everything in the hint except for the claim that $K = bigcup_n = 0 ^ infty K_n$ is compact. I assume I'm missing something in how $K_n$ is constructed that makes the countable union compact, since obviously a countable union of compact sets isn't necessarily compact. I also can't seem to find a proof of this claim anywhere other than this book, except a proof in the book Introduction to Empirical Processes and Semiparametric Inference which apparently just copies this hint and leaves the claim that $K$ is compact as "an exercise left to the reader."
EDIT: I think I have a sketch now for compactness, so if anyone could verify it I would be appreciative. The $K_n$ can be chosen so that $K_0 subseteq K_n subseteq K_0^1/m$ for $n_m < n le n_m+1$. Given an open cover $U_alpha$ of $K$, find a finite subcover of $K_0$, $U_1, ldots, U_K$. Next, find $m$ such that $K_0^1/m subseteq bigcup_k = 1 ^K U_k$ (exists because $K_0$ is compact and $bigcup_k = 1 ^ K U_k$ is open). Hence $U_1, ldots, U_K$ is an open cover for $K_n$ for all $n > n_m$. Now, just use compactness of $K_1, ldots, K_n_m$ extend the open cover $K_n_m + 1, K_n_m + 2, ldots$ to one which covers all $K_n$.
probability measure-theory probability-theory
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I'm stumped on the following question, which is problem 1.3.9 in the book Weak Convergence and Empirical Proceses by van der Vaart and Wellner. It is based on the following notion of asymptotic tightness:
Definition: A sequence of random variables $X_1, X_2, ldots$ taking values in a metric space with metric $d$ is asymptotically tight if for every $epsilon > 0$ there exists a compact $K_0$ such that for every $delta > 0$ we have $varliminf_n to infty P(X_n in K_0^delta) ge 1 - epsilon$ where $K_0^delta = x: d(x, K_0) < delta$ is the $delta$-enlargement of $K_0$.
For convenience, the definition of uniform tightness is
Definition: A sequence of random variables $X_1, X_2, ldots$ is uniformly tight if for every $epsilon > 0$ there exists a compact $K$ such that $P(X_n in K) ge 1 - epsilon$ for every $K$.
The problem is:
Claim: A sequence $X_n$ is uniformly tight if and only if it is asymptotically tight and each $X_n$ is tight.
I cannot seem to prove the non-trivial direction of this claim. The following hint is given:
Hint: Fix $epsilon > 0$. Take a compact $K_0$ with $varliminf P(X_n in K_0^delta) ge 1 - epsilon$ for every $delta$. Choose $n_1 < n_2 < cdots $ such that $P(X_n in K_0^1/m) ge 1 - 2 epsilon$ for $n ge n_m$. For $n_m < n le n_m+1$ choose a compact $K_n$ with $P(X_n in K_0^1/m - K_n) < epsilon$. Now $K = bigcup_n = 0 ^ infty K_n$ is compact and $P(X_n in K) ge 1 - 3 epsilon$.
Now, I understand everything in the hint except for the claim that $K = bigcup_n = 0 ^ infty K_n$ is compact. I assume I'm missing something in how $K_n$ is constructed that makes the countable union compact, since obviously a countable union of compact sets isn't necessarily compact. I also can't seem to find a proof of this claim anywhere other than this book, except a proof in the book Introduction to Empirical Processes and Semiparametric Inference which apparently just copies this hint and leaves the claim that $K$ is compact as "an exercise left to the reader."
EDIT: I think I have a sketch now for compactness, so if anyone could verify it I would be appreciative. The $K_n$ can be chosen so that $K_0 subseteq K_n subseteq K_0^1/m$ for $n_m < n le n_m+1$. Given an open cover $U_alpha$ of $K$, find a finite subcover of $K_0$, $U_1, ldots, U_K$. Next, find $m$ such that $K_0^1/m subseteq bigcup_k = 1 ^K U_k$ (exists because $K_0$ is compact and $bigcup_k = 1 ^ K U_k$ is open). Hence $U_1, ldots, U_K$ is an open cover for $K_n$ for all $n > n_m$. Now, just use compactness of $K_1, ldots, K_n_m$ extend the open cover $K_n_m + 1, K_n_m + 2, ldots$ to one which covers all $K_n$.
probability measure-theory probability-theory
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm stumped on the following question, which is problem 1.3.9 in the book Weak Convergence and Empirical Proceses by van der Vaart and Wellner. It is based on the following notion of asymptotic tightness:
Definition: A sequence of random variables $X_1, X_2, ldots$ taking values in a metric space with metric $d$ is asymptotically tight if for every $epsilon > 0$ there exists a compact $K_0$ such that for every $delta > 0$ we have $varliminf_n to infty P(X_n in K_0^delta) ge 1 - epsilon$ where $K_0^delta = x: d(x, K_0) < delta$ is the $delta$-enlargement of $K_0$.
For convenience, the definition of uniform tightness is
Definition: A sequence of random variables $X_1, X_2, ldots$ is uniformly tight if for every $epsilon > 0$ there exists a compact $K$ such that $P(X_n in K) ge 1 - epsilon$ for every $K$.
The problem is:
Claim: A sequence $X_n$ is uniformly tight if and only if it is asymptotically tight and each $X_n$ is tight.
I cannot seem to prove the non-trivial direction of this claim. The following hint is given:
Hint: Fix $epsilon > 0$. Take a compact $K_0$ with $varliminf P(X_n in K_0^delta) ge 1 - epsilon$ for every $delta$. Choose $n_1 < n_2 < cdots $ such that $P(X_n in K_0^1/m) ge 1 - 2 epsilon$ for $n ge n_m$. For $n_m < n le n_m+1$ choose a compact $K_n$ with $P(X_n in K_0^1/m - K_n) < epsilon$. Now $K = bigcup_n = 0 ^ infty K_n$ is compact and $P(X_n in K) ge 1 - 3 epsilon$.
Now, I understand everything in the hint except for the claim that $K = bigcup_n = 0 ^ infty K_n$ is compact. I assume I'm missing something in how $K_n$ is constructed that makes the countable union compact, since obviously a countable union of compact sets isn't necessarily compact. I also can't seem to find a proof of this claim anywhere other than this book, except a proof in the book Introduction to Empirical Processes and Semiparametric Inference which apparently just copies this hint and leaves the claim that $K$ is compact as "an exercise left to the reader."
EDIT: I think I have a sketch now for compactness, so if anyone could verify it I would be appreciative. The $K_n$ can be chosen so that $K_0 subseteq K_n subseteq K_0^1/m$ for $n_m < n le n_m+1$. Given an open cover $U_alpha$ of $K$, find a finite subcover of $K_0$, $U_1, ldots, U_K$. Next, find $m$ such that $K_0^1/m subseteq bigcup_k = 1 ^K U_k$ (exists because $K_0$ is compact and $bigcup_k = 1 ^ K U_k$ is open). Hence $U_1, ldots, U_K$ is an open cover for $K_n$ for all $n > n_m$. Now, just use compactness of $K_1, ldots, K_n_m$ extend the open cover $K_n_m + 1, K_n_m + 2, ldots$ to one which covers all $K_n$.
probability measure-theory probability-theory
I'm stumped on the following question, which is problem 1.3.9 in the book Weak Convergence and Empirical Proceses by van der Vaart and Wellner. It is based on the following notion of asymptotic tightness:
Definition: A sequence of random variables $X_1, X_2, ldots$ taking values in a metric space with metric $d$ is asymptotically tight if for every $epsilon > 0$ there exists a compact $K_0$ such that for every $delta > 0$ we have $varliminf_n to infty P(X_n in K_0^delta) ge 1 - epsilon$ where $K_0^delta = x: d(x, K_0) < delta$ is the $delta$-enlargement of $K_0$.
For convenience, the definition of uniform tightness is
Definition: A sequence of random variables $X_1, X_2, ldots$ is uniformly tight if for every $epsilon > 0$ there exists a compact $K$ such that $P(X_n in K) ge 1 - epsilon$ for every $K$.
The problem is:
Claim: A sequence $X_n$ is uniformly tight if and only if it is asymptotically tight and each $X_n$ is tight.
I cannot seem to prove the non-trivial direction of this claim. The following hint is given:
Hint: Fix $epsilon > 0$. Take a compact $K_0$ with $varliminf P(X_n in K_0^delta) ge 1 - epsilon$ for every $delta$. Choose $n_1 < n_2 < cdots $ such that $P(X_n in K_0^1/m) ge 1 - 2 epsilon$ for $n ge n_m$. For $n_m < n le n_m+1$ choose a compact $K_n$ with $P(X_n in K_0^1/m - K_n) < epsilon$. Now $K = bigcup_n = 0 ^ infty K_n$ is compact and $P(X_n in K) ge 1 - 3 epsilon$.
Now, I understand everything in the hint except for the claim that $K = bigcup_n = 0 ^ infty K_n$ is compact. I assume I'm missing something in how $K_n$ is constructed that makes the countable union compact, since obviously a countable union of compact sets isn't necessarily compact. I also can't seem to find a proof of this claim anywhere other than this book, except a proof in the book Introduction to Empirical Processes and Semiparametric Inference which apparently just copies this hint and leaves the claim that $K$ is compact as "an exercise left to the reader."
EDIT: I think I have a sketch now for compactness, so if anyone could verify it I would be appreciative. The $K_n$ can be chosen so that $K_0 subseteq K_n subseteq K_0^1/m$ for $n_m < n le n_m+1$. Given an open cover $U_alpha$ of $K$, find a finite subcover of $K_0$, $U_1, ldots, U_K$. Next, find $m$ such that $K_0^1/m subseteq bigcup_k = 1 ^K U_k$ (exists because $K_0$ is compact and $bigcup_k = 1 ^ K U_k$ is open). Hence $U_1, ldots, U_K$ is an open cover for $K_n$ for all $n > n_m$. Now, just use compactness of $K_1, ldots, K_n_m$ extend the open cover $K_n_m + 1, K_n_m + 2, ldots$ to one which covers all $K_n$.
probability measure-theory probability-theory
edited Oct 12 '14 at 6:53
asked Oct 12 '14 at 6:26
guy
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2,41311626
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1 Answer
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A simpler way to show the non-trivial direction of the posted claim is the follwoing.
Since the sequence of random variables is asymptotically tight, we have that there is some $n^*$ such that, for all $n>n^*$, $sup_kge nP(X_nnotin K)<epsilon$. Uniform tightness thus follows by combining this result with the fact that the collection $(X_1,...,X_n^*)$ is finite and the fact that every finite collection of tight random variables is also tight.
Thanks for the answer. It has been a long time so I am a little bit hazy, but how do you know there exists an $n^*$ with this property? If I remember correctly, this property is stronger than asymptotic tightness; for asymptotic tightness we can only guarantee an $n^star$ such that $sup_k ge n P(X_n notin K^delta) le epsilon$ for a $delta$-enlargement of $K$, and I vaguely recall that Van der Vaart and Wellner explicitly state that you cannot prove equivalence between these properties.
â guy
Aug 17 at 2:08
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
A simpler way to show the non-trivial direction of the posted claim is the follwoing.
Since the sequence of random variables is asymptotically tight, we have that there is some $n^*$ such that, for all $n>n^*$, $sup_kge nP(X_nnotin K)<epsilon$. Uniform tightness thus follows by combining this result with the fact that the collection $(X_1,...,X_n^*)$ is finite and the fact that every finite collection of tight random variables is also tight.
Thanks for the answer. It has been a long time so I am a little bit hazy, but how do you know there exists an $n^*$ with this property? If I remember correctly, this property is stronger than asymptotic tightness; for asymptotic tightness we can only guarantee an $n^star$ such that $sup_k ge n P(X_n notin K^delta) le epsilon$ for a $delta$-enlargement of $K$, and I vaguely recall that Van der Vaart and Wellner explicitly state that you cannot prove equivalence between these properties.
â guy
Aug 17 at 2:08
add a comment |Â
up vote
1
down vote
A simpler way to show the non-trivial direction of the posted claim is the follwoing.
Since the sequence of random variables is asymptotically tight, we have that there is some $n^*$ such that, for all $n>n^*$, $sup_kge nP(X_nnotin K)<epsilon$. Uniform tightness thus follows by combining this result with the fact that the collection $(X_1,...,X_n^*)$ is finite and the fact that every finite collection of tight random variables is also tight.
Thanks for the answer. It has been a long time so I am a little bit hazy, but how do you know there exists an $n^*$ with this property? If I remember correctly, this property is stronger than asymptotic tightness; for asymptotic tightness we can only guarantee an $n^star$ such that $sup_k ge n P(X_n notin K^delta) le epsilon$ for a $delta$-enlargement of $K$, and I vaguely recall that Van der Vaart and Wellner explicitly state that you cannot prove equivalence between these properties.
â guy
Aug 17 at 2:08
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A simpler way to show the non-trivial direction of the posted claim is the follwoing.
Since the sequence of random variables is asymptotically tight, we have that there is some $n^*$ such that, for all $n>n^*$, $sup_kge nP(X_nnotin K)<epsilon$. Uniform tightness thus follows by combining this result with the fact that the collection $(X_1,...,X_n^*)$ is finite and the fact that every finite collection of tight random variables is also tight.
A simpler way to show the non-trivial direction of the posted claim is the follwoing.
Since the sequence of random variables is asymptotically tight, we have that there is some $n^*$ such that, for all $n>n^*$, $sup_kge nP(X_nnotin K)<epsilon$. Uniform tightness thus follows by combining this result with the fact that the collection $(X_1,...,X_n^*)$ is finite and the fact that every finite collection of tight random variables is also tight.
answered Aug 17 at 0:45
Chen
112
112
Thanks for the answer. It has been a long time so I am a little bit hazy, but how do you know there exists an $n^*$ with this property? If I remember correctly, this property is stronger than asymptotic tightness; for asymptotic tightness we can only guarantee an $n^star$ such that $sup_k ge n P(X_n notin K^delta) le epsilon$ for a $delta$-enlargement of $K$, and I vaguely recall that Van der Vaart and Wellner explicitly state that you cannot prove equivalence between these properties.
â guy
Aug 17 at 2:08
add a comment |Â
Thanks for the answer. It has been a long time so I am a little bit hazy, but how do you know there exists an $n^*$ with this property? If I remember correctly, this property is stronger than asymptotic tightness; for asymptotic tightness we can only guarantee an $n^star$ such that $sup_k ge n P(X_n notin K^delta) le epsilon$ for a $delta$-enlargement of $K$, and I vaguely recall that Van der Vaart and Wellner explicitly state that you cannot prove equivalence between these properties.
â guy
Aug 17 at 2:08
Thanks for the answer. It has been a long time so I am a little bit hazy, but how do you know there exists an $n^*$ with this property? If I remember correctly, this property is stronger than asymptotic tightness; for asymptotic tightness we can only guarantee an $n^star$ such that $sup_k ge n P(X_n notin K^delta) le epsilon$ for a $delta$-enlargement of $K$, and I vaguely recall that Van der Vaart and Wellner explicitly state that you cannot prove equivalence between these properties.
â guy
Aug 17 at 2:08
Thanks for the answer. It has been a long time so I am a little bit hazy, but how do you know there exists an $n^*$ with this property? If I remember correctly, this property is stronger than asymptotic tightness; for asymptotic tightness we can only guarantee an $n^star$ such that $sup_k ge n P(X_n notin K^delta) le epsilon$ for a $delta$-enlargement of $K$, and I vaguely recall that Van der Vaart and Wellner explicitly state that you cannot prove equivalence between these properties.
â guy
Aug 17 at 2:08
add a comment |Â
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