Mixing
$X_nto0$ in probability and ${c_n}$ is a bdd seq of real numbers, then $c_nX_nto0$ in probability.
Suppose $X_nto0$ in probability and ${c_n}$ is a bounded sequence of real numbers. Prove that $c_nX_nto0$ in probability.
I believe I need to use a theorem, which states that:
If ${X_n}_{n=1}^infty$ and $X$ are random variables on a probability space, $(Omega,mathcal{F},P)$. Then, $X_nto X$ in probability as $ntoinfty$ iff every subsequence ${X_{n_{m}}}$ has a further subsequence ${X_{n_{m_{k}}}}$ such that $X_{n_{m_{k}}}to X$ a.s. as $ktoinfty$.
I am having a difficult time formulating my thoughts though. Any help would be appreciated.
probability analysis probability-theory measure-theory
asked Nov 20 '18 at 17:25
Dragonite
1,045420
add a comment |
Suppose $X_nto0$ in probability and ${c_n}$ is a bounded sequence of real numbers. Prove that $c_nX_nto0$ in probability.
I believe I need to use a theorem, which states that:
If ${X_n}_{n=1}^infty$ and $X$ are random variables on a probability space, $(Omega,mathcal{F},P)$. Then, $X_nto X$ in probability as $ntoinfty$ iff every subsequence ${X_{n_{m}}}$ has a further subsequence ${X_{n_{m_{k}}}}$ such that $X_{n_{m_{k}}}to X$ a.s. as $ktoinfty$.
I am having a difficult time formulating my thoughts though. Any help would be appreciated.
probability analysis probability-theory measure-theory
asked Nov 20 '18 at 17:25
Dragonite
1,045420
add a comment |
Suppose $X_nto0$ in probability and ${c_n}$ is a bounded sequence of real numbers. Prove that $c_nX_nto0$ in probability.
I believe I need to use a theorem, which states that:
If ${X_n}_{n=1}^infty$ and $X$ are random variables on a probability space, $(Omega,mathcal{F},P)$. Then, $X_nto X$ in probability as $ntoinfty$ iff every subsequence ${X_{n_{m}}}$ has a further subsequence ${X_{n_{m_{k}}}}$ such that $X_{n_{m_{k}}}to X$ a.s. as $ktoinfty$.
I am having a difficult time formulating my thoughts though. Any help would be appreciated.
probability analysis probability-theory measure-theory
asked Nov 20 '18 at 17:25
Dragonite
1,045420
Suppose $X_nto0$ in probability and ${c_n}$ is a bounded sequence of real numbers. Prove that $c_nX_nto0$ in probability.
I believe I need to use a theorem, which states that:
If ${X_n}_{n=1}^infty$ and $X$ are random variables on a probability space, $(Omega,mathcal{F},P)$. Then, $X_nto X$ in probability as $ntoinfty$ iff every subsequence ${X_{n_{m}}}$ has a further subsequence ${X_{n_{m_{k}}}}$ such that $X_{n_{m_{k}}}to X$ a.s. as $ktoinfty$.
I am having a difficult time formulating my thoughts though. Any help would be appreciated.
probability analysis probability-theory measure-theory
probability analysis probability-theory measure-theory
asked Nov 20 '18 at 17:25
Dragonite
1,045420
asked Nov 20 '18 at 17:25
Dragonite
1,045420
asked Nov 20 '18 at 17:25
Dragonite
1,045420
asked Nov 20 '18 at 17:25
Dragonite
1,045420
asked Nov 20 '18 at 17:25
Dragonite
1,045420
1,045420
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You can use that theorem to prove the result, but I think that makes it more complicated than it needs to be.
Hint: The typical condition of convergence in probability is:
$$X_n stackrel{p}{to} X iff forall epsilon > 0, lim_{n to infty} mathbb P(|X_n - X| > epsilon) = 0.$$
Here, that means that for any $epsilon > 0$, $lim_{n to infty} mathbb P(|X_n| > epsilon) = 0$. What can you say about $mathbb P(|c_n X_n| > epsilon)$?
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
$mathbb{P}(|c_n||X_n|>epsilon)leqmathbb{P}(|sup_{ninmathbb{N}}|c_n||X_n|>epsilon)$. Then, let $ntoinfty$, so that we get $sup_{ninmathbb{N}}|c_n|mathbb{P}(|X_n|>epsilon)=0$?
– Dragonite
Nov 20 '18 at 18:37
Not quite -- there's no mechanism that lets you "factor" either the $sup$ or the $c_n$ terms out of the probability operator like that. What happens if you divide the $c_n$ terms across the inequality instead?
– Aaron Montgomery
Nov 20 '18 at 18:40
Can we do $mathbb{P}(|c_n||X_n|>epsilon)=mathbb{P}(|X_n|>epsilon/|c_n|)$ and then send the $epsilon/|c_n|$ to 0 using boundedness of $c_n$?
– Dragonite
Nov 20 '18 at 18:42
Close -- but you don't have to "send the $epsilon / |c_n|$ to 0," and you don't even really know that you can -- observe that the $|c_n|$ terms could be close to (or equal to) 0. Use the bound instead: $|c_n X_n| < M |X_n|$, so $mathbb P(|c_n X_n| > epsilon) leq mathbb P(M |X_n| > epsilon)$ (why)?, and work with the right-hand expression instead.
– Aaron Montgomery
Nov 20 '18 at 18:47
1
Okay, I see now. Thank you!
– Dragonite
Nov 20 '18 at 18:51
add a comment |
Abridged solution using the theorem you wrote. There is $|c_n| leq alpha$ for all $n,$ then the relation $X_{n_{m_k}} to 0$ (being equivalent to $|X_{n_{m_k}}| to 0$), implies $0 leq |c_{n_{m_k}} X_{n_{m_k}}| leq alpha |X_{n_{m_k}}| to 0,$ hence the result. Q.E.D.
answered Nov 20 '18 at 17:52
Will M.
2,387314
add a comment |
Note that
$$
C|X_n|ge |c_n X_n|
$$
for some $C>0$ (if $c_n$ is identically zero the result is trivial) since the sequence $(c_n)$ is bounded. In particular, given $epsilon>0$,
$$
P(|c_nX_n|>epsilon)leq P(C|X_n|>epsilon)to 0
$$
since one event is a subset of the other and $X_nto 0$ in probability.
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
add a comment |
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3 Answers
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You can use that theorem to prove the result, but I think that makes it more complicated than it needs to be.
Hint: The typical condition of convergence in probability is:
$$X_n stackrel{p}{to} X iff forall epsilon > 0, lim_{n to infty} mathbb P(|X_n - X| > epsilon) = 0.$$
Here, that means that for any $epsilon > 0$, $lim_{n to infty} mathbb P(|X_n| > epsilon) = 0$. What can you say about $mathbb P(|c_n X_n| > epsilon)$?
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
$mathbb{P}(|c_n||X_n|>epsilon)leqmathbb{P}(|sup_{ninmathbb{N}}|c_n||X_n|>epsilon)$. Then, let $ntoinfty$, so that we get $sup_{ninmathbb{N}}|c_n|mathbb{P}(|X_n|>epsilon)=0$?
– Dragonite
Nov 20 '18 at 18:37
Not quite -- there's no mechanism that lets you "factor" either the $sup$ or the $c_n$ terms out of the probability operator like that. What happens if you divide the $c_n$ terms across the inequality instead?
– Aaron Montgomery
Nov 20 '18 at 18:40
Can we do $mathbb{P}(|c_n||X_n|>epsilon)=mathbb{P}(|X_n|>epsilon/|c_n|)$ and then send the $epsilon/|c_n|$ to 0 using boundedness of $c_n$?
– Dragonite
Nov 20 '18 at 18:42
Close -- but you don't have to "send the $epsilon / |c_n|$ to 0," and you don't even really know that you can -- observe that the $|c_n|$ terms could be close to (or equal to) 0. Use the bound instead: $|c_n X_n| < M |X_n|$, so $mathbb P(|c_n X_n| > epsilon) leq mathbb P(M |X_n| > epsilon)$ (why)?, and work with the right-hand expression instead.
– Aaron Montgomery
Nov 20 '18 at 18:47
1
Okay, I see now. Thank you!
– Dragonite
Nov 20 '18 at 18:51
add a comment |
You can use that theorem to prove the result, but I think that makes it more complicated than it needs to be.
Hint: The typical condition of convergence in probability is:
$$X_n stackrel{p}{to} X iff forall epsilon > 0, lim_{n to infty} mathbb P(|X_n - X| > epsilon) = 0.$$
Here, that means that for any $epsilon > 0$, $lim_{n to infty} mathbb P(|X_n| > epsilon) = 0$. What can you say about $mathbb P(|c_n X_n| > epsilon)$?
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
$mathbb{P}(|c_n||X_n|>epsilon)leqmathbb{P}(|sup_{ninmathbb{N}}|c_n||X_n|>epsilon)$. Then, let $ntoinfty$, so that we get $sup_{ninmathbb{N}}|c_n|mathbb{P}(|X_n|>epsilon)=0$?
– Dragonite
Nov 20 '18 at 18:37
Not quite -- there's no mechanism that lets you "factor" either the $sup$ or the $c_n$ terms out of the probability operator like that. What happens if you divide the $c_n$ terms across the inequality instead?
– Aaron Montgomery
Nov 20 '18 at 18:40
Can we do $mathbb{P}(|c_n||X_n|>epsilon)=mathbb{P}(|X_n|>epsilon/|c_n|)$ and then send the $epsilon/|c_n|$ to 0 using boundedness of $c_n$?
– Dragonite
Nov 20 '18 at 18:42
Close -- but you don't have to "send the $epsilon / |c_n|$ to 0," and you don't even really know that you can -- observe that the $|c_n|$ terms could be close to (or equal to) 0. Use the bound instead: $|c_n X_n| < M |X_n|$, so $mathbb P(|c_n X_n| > epsilon) leq mathbb P(M |X_n| > epsilon)$ (why)?, and work with the right-hand expression instead.
– Aaron Montgomery
Nov 20 '18 at 18:47
1
Okay, I see now. Thank you!
– Dragonite
Nov 20 '18 at 18:51
add a comment |
You can use that theorem to prove the result, but I think that makes it more complicated than it needs to be.
Hint: The typical condition of convergence in probability is:
$$X_n stackrel{p}{to} X iff forall epsilon > 0, lim_{n to infty} mathbb P(|X_n - X| > epsilon) = 0.$$
Here, that means that for any $epsilon > 0$, $lim_{n to infty} mathbb P(|X_n| > epsilon) = 0$. What can you say about $mathbb P(|c_n X_n| > epsilon)$?
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
You can use that theorem to prove the result, but I think that makes it more complicated than it needs to be.
Hint: The typical condition of convergence in probability is:
$$X_n stackrel{p}{to} X iff forall epsilon > 0, lim_{n to infty} mathbb P(|X_n - X| > epsilon) = 0.$$
Here, that means that for any $epsilon > 0$, $lim_{n to infty} mathbb P(|X_n| > epsilon) = 0$. What can you say about $mathbb P(|c_n X_n| > epsilon)$?
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
answered Nov 20 '18 at 17:48
Aaron Montgomery
4,712523
4,712523
$mathbb{P}(|c_n||X_n|>epsilon)leqmathbb{P}(|sup_{ninmathbb{N}}|c_n||X_n|>epsilon)$. Then, let $ntoinfty$, so that we get $sup_{ninmathbb{N}}|c_n|mathbb{P}(|X_n|>epsilon)=0$?
– Dragonite
Nov 20 '18 at 18:37
Not quite -- there's no mechanism that lets you "factor" either the $sup$ or the $c_n$ terms out of the probability operator like that. What happens if you divide the $c_n$ terms across the inequality instead?
– Aaron Montgomery
Nov 20 '18 at 18:40
Can we do $mathbb{P}(|c_n||X_n|>epsilon)=mathbb{P}(|X_n|>epsilon/|c_n|)$ and then send the $epsilon/|c_n|$ to 0 using boundedness of $c_n$?
– Dragonite
Nov 20 '18 at 18:42
Close -- but you don't have to "send the $epsilon / |c_n|$ to 0," and you don't even really know that you can -- observe that the $|c_n|$ terms could be close to (or equal to) 0. Use the bound instead: $|c_n X_n| < M |X_n|$, so $mathbb P(|c_n X_n| > epsilon) leq mathbb P(M |X_n| > epsilon)$ (why)?, and work with the right-hand expression instead.
– Aaron Montgomery
Nov 20 '18 at 18:47
1
Okay, I see now. Thank you!
– Dragonite
Nov 20 '18 at 18:51
add a comment |
$mathbb{P}(|c_n||X_n|>epsilon)leqmathbb{P}(|sup_{ninmathbb{N}}|c_n||X_n|>epsilon)$. Then, let $ntoinfty$, so that we get $sup_{ninmathbb{N}}|c_n|mathbb{P}(|X_n|>epsilon)=0$?
– Dragonite
Nov 20 '18 at 18:37
Not quite -- there's no mechanism that lets you "factor" either the $sup$ or the $c_n$ terms out of the probability operator like that. What happens if you divide the $c_n$ terms across the inequality instead?
– Aaron Montgomery
Nov 20 '18 at 18:40
Can we do $mathbb{P}(|c_n||X_n|>epsilon)=mathbb{P}(|X_n|>epsilon/|c_n|)$ and then send the $epsilon/|c_n|$ to 0 using boundedness of $c_n$?
– Dragonite
Nov 20 '18 at 18:42
Close -- but you don't have to "send the $epsilon / |c_n|$ to 0," and you don't even really know that you can -- observe that the $|c_n|$ terms could be close to (or equal to) 0. Use the bound instead: $|c_n X_n| < M |X_n|$, so $mathbb P(|c_n X_n| > epsilon) leq mathbb P(M |X_n| > epsilon)$ (why)?, and work with the right-hand expression instead.
– Aaron Montgomery
Nov 20 '18 at 18:47
1
Okay, I see now. Thank you!
– Dragonite
Nov 20 '18 at 18:51
$mathbb{P}(|c_n||X_n|>epsilon)leqmathbb{P}(|sup_{ninmathbb{N}}|c_n||X_n|>epsilon)$. Then, let $ntoinfty$, so that we get $sup_{ninmathbb{N}}|c_n|mathbb{P}(|X_n|>epsilon)=0$?
– Dragonite
Nov 20 '18 at 18:37
$mathbb{P}(|c_n||X_n|>epsilon)leqmathbb{P}(|sup_{ninmathbb{N}}|c_n||X_n|>epsilon)$. Then, let $ntoinfty$, so that we get $sup_{ninmathbb{N}}|c_n|mathbb{P}(|X_n|>epsilon)=0$?
– Dragonite
Nov 20 '18 at 18:37
Not quite -- there's no mechanism that lets you "factor" either the $sup$ or the $c_n$ terms out of the probability operator like that. What happens if you divide the $c_n$ terms across the inequality instead?
– Aaron Montgomery
Nov 20 '18 at 18:40
Not quite -- there's no mechanism that lets you "factor" either the $sup$ or the $c_n$ terms out of the probability operator like that. What happens if you divide the $c_n$ terms across the inequality instead?
– Aaron Montgomery
Nov 20 '18 at 18:40
Can we do $mathbb{P}(|c_n||X_n|>epsilon)=mathbb{P}(|X_n|>epsilon/|c_n|)$ and then send the $epsilon/|c_n|$ to 0 using boundedness of $c_n$?
– Dragonite
Nov 20 '18 at 18:42
Can we do $mathbb{P}(|c_n||X_n|>epsilon)=mathbb{P}(|X_n|>epsilon/|c_n|)$ and then send the $epsilon/|c_n|$ to 0 using boundedness of $c_n$?
– Dragonite
Nov 20 '18 at 18:42
Close -- but you don't have to "send the $epsilon / |c_n|$ to 0," and you don't even really know that you can -- observe that the $|c_n|$ terms could be close to (or equal to) 0. Use the bound instead: $|c_n X_n| < M |X_n|$, so $mathbb P(|c_n X_n| > epsilon) leq mathbb P(M |X_n| > epsilon)$ (why)?, and work with the right-hand expression instead.
– Aaron Montgomery
Nov 20 '18 at 18:47
Close -- but you don't have to "send the $epsilon / |c_n|$ to 0," and you don't even really know that you can -- observe that the $|c_n|$ terms could be close to (or equal to) 0. Use the bound instead: $|c_n X_n| < M |X_n|$, so $mathbb P(|c_n X_n| > epsilon) leq mathbb P(M |X_n| > epsilon)$ (why)?, and work with the right-hand expression instead.
– Aaron Montgomery
Nov 20 '18 at 18:47
1
1
Okay, I see now. Thank you!
– Dragonite
Nov 20 '18 at 18:51
Okay, I see now. Thank you!
– Dragonite
Nov 20 '18 at 18:51
add a comment |
Abridged solution using the theorem you wrote. There is $|c_n| leq alpha$ for all $n,$ then the relation $X_{n_{m_k}} to 0$ (being equivalent to $|X_{n_{m_k}}| to 0$), implies $0 leq |c_{n_{m_k}} X_{n_{m_k}}| leq alpha |X_{n_{m_k}}| to 0,$ hence the result. Q.E.D.
answered Nov 20 '18 at 17:52
Will M.
2,387314
add a comment |
Abridged solution using the theorem you wrote. There is $|c_n| leq alpha$ for all $n,$ then the relation $X_{n_{m_k}} to 0$ (being equivalent to $|X_{n_{m_k}}| to 0$), implies $0 leq |c_{n_{m_k}} X_{n_{m_k}}| leq alpha |X_{n_{m_k}}| to 0,$ hence the result. Q.E.D.
answered Nov 20 '18 at 17:52
Will M.
2,387314
add a comment |
Abridged solution using the theorem you wrote. There is $|c_n| leq alpha$ for all $n,$ then the relation $X_{n_{m_k}} to 0$ (being equivalent to $|X_{n_{m_k}}| to 0$), implies $0 leq |c_{n_{m_k}} X_{n_{m_k}}| leq alpha |X_{n_{m_k}}| to 0,$ hence the result. Q.E.D.
answered Nov 20 '18 at 17:52
Will M.
2,387314
Abridged solution using the theorem you wrote. There is $|c_n| leq alpha$ for all $n,$ then the relation $X_{n_{m_k}} to 0$ (being equivalent to $|X_{n_{m_k}}| to 0$), implies $0 leq |c_{n_{m_k}} X_{n_{m_k}}| leq alpha |X_{n_{m_k}}| to 0,$ hence the result. Q.E.D.
answered Nov 20 '18 at 17:52
Will M.
2,387314
answered Nov 20 '18 at 17:52
Will M.
2,387314
answered Nov 20 '18 at 17:52
Will M.
2,387314
answered Nov 20 '18 at 17:52
Will M.
2,387314
2,387314
add a comment |
add a comment |
Note that
$$
C|X_n|ge |c_n X_n|
$$
for some $C>0$ (if $c_n$ is identically zero the result is trivial) since the sequence $(c_n)$ is bounded. In particular, given $epsilon>0$,
$$
P(|c_nX_n|>epsilon)leq P(C|X_n|>epsilon)to 0
$$
since one event is a subset of the other and $X_nto 0$ in probability.
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
add a comment |
Note that
$$
C|X_n|ge |c_n X_n|
$$
for some $C>0$ (if $c_n$ is identically zero the result is trivial) since the sequence $(c_n)$ is bounded. In particular, given $epsilon>0$,
$$
P(|c_nX_n|>epsilon)leq P(C|X_n|>epsilon)to 0
$$
since one event is a subset of the other and $X_nto 0$ in probability.
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
add a comment |
Note that
$$
C|X_n|ge |c_n X_n|
$$
for some $C>0$ (if $c_n$ is identically zero the result is trivial) since the sequence $(c_n)$ is bounded. In particular, given $epsilon>0$,
$$
P(|c_nX_n|>epsilon)leq P(C|X_n|>epsilon)to 0
$$
since one event is a subset of the other and $X_nto 0$ in probability.
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
Note that
$$
C|X_n|ge |c_n X_n|
$$
for some $C>0$ (if $c_n$ is identically zero the result is trivial) since the sequence $(c_n)$ is bounded. In particular, given $epsilon>0$,
$$
P(|c_nX_n|>epsilon)leq P(C|X_n|>epsilon)to 0
$$
since one event is a subset of the other and $X_nto 0$ in probability.
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
answered Nov 20 '18 at 17:58
Foobaz John
21.3k41251
21.3k41251
add a comment |
add a comment |
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