Mixing
Slutsky's Theorem - sufficient conditions for product to converge to zero
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Consider two sequences of random variables, $,{a_n}$ and ${b_n}$ and suppose that $,a_nstackrel{p}{longrightarrow}0$. Provided that $,b_nstackrel{d}{longrightarrow}Z$ with $Zsim F$ for a given distribution $F$, I understood that Slutsky's theorem dictates that $$a_n,b_nstackrel{d}{longrightarrow}0,Z$$such that effectively
$$a_n,b_nstackrel{p}{longrightarrow}0.$$
However, when defining "constant" random variables $,a_n=1/n$ and $b_n=n$, it clearly follows that
$$a_n,b_nstackrel{p}{longrightarrow}1.$$
I'd very much appreciate, if someone could briefly delineate a set of high-level conditions on ${b_n}$ that would be sufficient for the result that$$a_n,b_nstackrel{p}{longrightarrow}0quadtext{provided that },,,a_nstackrel{p}{longrightarrow}0.$$
E.g. does it suffice to assume that $b_n$ is $O_p(1)$ - which would clearly be violated in my counter example.
Thank you very much for your time.
Best regards,
Jon
probability convergence
asked Jan 24 at 22:37
J.BeckJ.Beck
637
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add a comment |
$begingroup$
Consider two sequences of random variables, $,{a_n}$ and ${b_n}$ and suppose that $,a_nstackrel{p}{longrightarrow}0$. Provided that $,b_nstackrel{d}{longrightarrow}Z$ with $Zsim F$ for a given distribution $F$, I understood that Slutsky's theorem dictates that $$a_n,b_nstackrel{d}{longrightarrow}0,Z$$such that effectively
$$a_n,b_nstackrel{p}{longrightarrow}0.$$
However, when defining "constant" random variables $,a_n=1/n$ and $b_n=n$, it clearly follows that
$$a_n,b_nstackrel{p}{longrightarrow}1.$$
I'd very much appreciate, if someone could briefly delineate a set of high-level conditions on ${b_n}$ that would be sufficient for the result that$$a_n,b_nstackrel{p}{longrightarrow}0quadtext{provided that },,,a_nstackrel{p}{longrightarrow}0.$$
E.g. does it suffice to assume that $b_n$ is $O_p(1)$ - which would clearly be violated in my counter example.
Thank you very much for your time.
Best regards,
Jon
probability convergence
asked Jan 24 at 22:37
J.BeckJ.Beck
637
$endgroup$
$begingroup$
The reason why $a_nb_n to_p 0$ is not working when $b_n = n$ is that $(b_n)$ does not converge to real-valued distribution, i.e. $b_nto infty$.
$endgroup$
– Song
Jan 24 at 22:55
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Thank you very much for you response. Just to clarify; so if the limiting distribution of $b_n$ was defined on a support $Xisubseteqmathbb{R}$, that would be sufficient for $a_n,b_nto_p0$ in general?
$endgroup$
– J.Beck
Jan 24 at 23:01
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Yes, you are right! If the limiting distribution, say $b$ satisfies $Bbb P(b=pminfty)=0$, then Slutsky theorem holds
$endgroup$
– Song
Jan 24 at 23:02
$begingroup$
Alright, thank you so much. Sounds straightforward enough.
$endgroup$
– J.Beck
Jan 24 at 23:05
add a comment |
$begingroup$
Consider two sequences of random variables, $,{a_n}$ and ${b_n}$ and suppose that $,a_nstackrel{p}{longrightarrow}0$. Provided that $,b_nstackrel{d}{longrightarrow}Z$ with $Zsim F$ for a given distribution $F$, I understood that Slutsky's theorem dictates that $$a_n,b_nstackrel{d}{longrightarrow}0,Z$$such that effectively
$$a_n,b_nstackrel{p}{longrightarrow}0.$$
However, when defining "constant" random variables $,a_n=1/n$ and $b_n=n$, it clearly follows that
$$a_n,b_nstackrel{p}{longrightarrow}1.$$
I'd very much appreciate, if someone could briefly delineate a set of high-level conditions on ${b_n}$ that would be sufficient for the result that$$a_n,b_nstackrel{p}{longrightarrow}0quadtext{provided that },,,a_nstackrel{p}{longrightarrow}0.$$
E.g. does it suffice to assume that $b_n$ is $O_p(1)$ - which would clearly be violated in my counter example.
Thank you very much for your time.
Best regards,
Jon
probability convergence
asked Jan 24 at 22:37
J.BeckJ.Beck
637
$endgroup$
Consider two sequences of random variables, $,{a_n}$ and ${b_n}$ and suppose that $,a_nstackrel{p}{longrightarrow}0$. Provided that $,b_nstackrel{d}{longrightarrow}Z$ with $Zsim F$ for a given distribution $F$, I understood that Slutsky's theorem dictates that $$a_n,b_nstackrel{d}{longrightarrow}0,Z$$such that effectively
$$a_n,b_nstackrel{p}{longrightarrow}0.$$
However, when defining "constant" random variables $,a_n=1/n$ and $b_n=n$, it clearly follows that
$$a_n,b_nstackrel{p}{longrightarrow}1.$$
I'd very much appreciate, if someone could briefly delineate a set of high-level conditions on ${b_n}$ that would be sufficient for the result that$$a_n,b_nstackrel{p}{longrightarrow}0quadtext{provided that },,,a_nstackrel{p}{longrightarrow}0.$$
E.g. does it suffice to assume that $b_n$ is $O_p(1)$ - which would clearly be violated in my counter example.
Thank you very much for your time.
Best regards,
Jon
probability convergence
probability convergence
asked Jan 24 at 22:37
J.BeckJ.Beck
637
asked Jan 24 at 22:37
J.BeckJ.Beck
637
asked Jan 24 at 22:37
J.BeckJ.Beck
637
asked Jan 24 at 22:37
J.BeckJ.Beck
637
asked Jan 24 at 22:37
J.BeckJ.Beck
637
637
$begingroup$
The reason why $a_nb_n to_p 0$ is not working when $b_n = n$ is that $(b_n)$ does not converge to real-valued distribution, i.e. $b_nto infty$.
$endgroup$
– Song
Jan 24 at 22:55
$begingroup$
Thank you very much for you response. Just to clarify; so if the limiting distribution of $b_n$ was defined on a support $Xisubseteqmathbb{R}$, that would be sufficient for $a_n,b_nto_p0$ in general?
$endgroup$
– J.Beck
Jan 24 at 23:01
$begingroup$
Yes, you are right! If the limiting distribution, say $b$ satisfies $Bbb P(b=pminfty)=0$, then Slutsky theorem holds
$endgroup$
– Song
Jan 24 at 23:02
$begingroup$
Alright, thank you so much. Sounds straightforward enough.
$endgroup$
– J.Beck
Jan 24 at 23:05
add a comment |
$begingroup$
The reason why $a_nb_n to_p 0$ is not working when $b_n = n$ is that $(b_n)$ does not converge to real-valued distribution, i.e. $b_nto infty$.
$endgroup$
– Song
Jan 24 at 22:55
$begingroup$
Thank you very much for you response. Just to clarify; so if the limiting distribution of $b_n$ was defined on a support $Xisubseteqmathbb{R}$, that would be sufficient for $a_n,b_nto_p0$ in general?
$endgroup$
– J.Beck
Jan 24 at 23:01
$begingroup$
Yes, you are right! If the limiting distribution, say $b$ satisfies $Bbb P(b=pminfty)=0$, then Slutsky theorem holds
$endgroup$
– Song
Jan 24 at 23:02
$begingroup$
Alright, thank you so much. Sounds straightforward enough.
$endgroup$
– J.Beck
Jan 24 at 23:05
$begingroup$
The reason why $a_nb_n to_p 0$ is not working when $b_n = n$ is that $(b_n)$ does not converge to real-valued distribution, i.e. $b_nto infty$.
$endgroup$
– Song
Jan 24 at 22:55
$begingroup$
The reason why $a_nb_n to_p 0$ is not working when $b_n = n$ is that $(b_n)$ does not converge to real-valued distribution, i.e. $b_nto infty$.
$endgroup$
– Song
Jan 24 at 22:55
$begingroup$
Thank you very much for you response. Just to clarify; so if the limiting distribution of $b_n$ was defined on a support $Xisubseteqmathbb{R}$, that would be sufficient for $a_n,b_nto_p0$ in general?
$endgroup$
– J.Beck
Jan 24 at 23:01
$begingroup$
Thank you very much for you response. Just to clarify; so if the limiting distribution of $b_n$ was defined on a support $Xisubseteqmathbb{R}$, that would be sufficient for $a_n,b_nto_p0$ in general?
$endgroup$
– J.Beck
Jan 24 at 23:01
$begingroup$
Yes, you are right! If the limiting distribution, say $b$ satisfies $Bbb P(b=pminfty)=0$, then Slutsky theorem holds
$endgroup$
– Song
Jan 24 at 23:02
$begingroup$
Yes, you are right! If the limiting distribution, say $b$ satisfies $Bbb P(b=pminfty)=0$, then Slutsky theorem holds
$endgroup$
– Song
Jan 24 at 23:02
$begingroup$
Alright, thank you so much. Sounds straightforward enough.
$endgroup$
– J.Beck
Jan 24 at 23:05
$begingroup$
Alright, thank you so much. Sounds straightforward enough.
$endgroup$
– J.Beck
Jan 24 at 23:05
add a comment |
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$begingroup$
The reason why $a_nb_n to_p 0$ is not working when $b_n = n$ is that $(b_n)$ does not converge to real-valued distribution, i.e. $b_nto infty$.
$endgroup$
– Song
Jan 24 at 22:55
$begingroup$
Thank you very much for you response. Just to clarify; so if the limiting distribution of $b_n$ was defined on a support $Xisubseteqmathbb{R}$, that would be sufficient for $a_n,b_nto_p0$ in general?
$endgroup$
– J.Beck
Jan 24 at 23:01
$begingroup$
Yes, you are right! If the limiting distribution, say $b$ satisfies $Bbb P(b=pminfty)=0$, then Slutsky theorem holds
$endgroup$
– Song
Jan 24 at 23:02
$begingroup$
Alright, thank you so much. Sounds straightforward enough.
$endgroup$
– J.Beck
Jan 24 at 23:05