Convergence results for conditional random random variables
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Suppose that $X_n$ is a sequence of random variables such that $X_n rightarrow X$ either almost surely or in $L^p$ or both.
Now consider the sequence of conditional random variables $f(X_n) | X_n in B$ for some measurable set $B$ and some nice function $f$ (we can assume it's Lipschitz to make everything easy). Do I still preserve convergence properties for this conditional random variable, i.e. does $f(X_n) | X_n in B rightarrow f(X) | X in B$ in some sense (a.s. or in $L^p)$?
I'm stuck verifying this by using the definition of almost sure convergence or $L^p$ convergence.
For example, regarding almost sure convergence, I have to check that $P(lim_nrightarrow infty f(X_n)|X_n in B = f(X)|Xin B)$ which looks strange from a notation perspective because of the conditional. Or should I resort to using conditional expectation?
real-analysis probability probability-theory convergence random-variables
asked Aug 17 at 5:09
Tomas Jorovic
1,56721325
add a comment |Â
up vote
0
down vote
favorite
Suppose that $X_n$ is a sequence of random variables such that $X_n rightarrow X$ either almost surely or in $L^p$ or both.
Now consider the sequence of conditional random variables $f(X_n) | X_n in B$ for some measurable set $B$ and some nice function $f$ (we can assume it's Lipschitz to make everything easy). Do I still preserve convergence properties for this conditional random variable, i.e. does $f(X_n) | X_n in B rightarrow f(X) | X in B$ in some sense (a.s. or in $L^p)$?
I'm stuck verifying this by using the definition of almost sure convergence or $L^p$ convergence.
For example, regarding almost sure convergence, I have to check that $P(lim_nrightarrow infty f(X_n)|X_n in B = f(X)|Xin B)$ which looks strange from a notation perspective because of the conditional. Or should I resort to using conditional expectation?
real-analysis probability probability-theory convergence random-variables
asked Aug 17 at 5:09
Tomas Jorovic
1,56721325
2
You cannot define $f(X_n)|X_n$ as random variables. You use conditioning only for distributions and expectations. So almost sure convergence does not even make sense here.
– Kavi Rama Murthy
Aug 17 at 5:38
Perhaps you are really trying to show $E[f(X_n)|X_n in B] rightarrow E[f(X)|Xin B]$? Of course this is not always true, take $B= mathbbR$, $X=0$, $f(x)=x$, then the statement is equivalent to showing $E[X_n]rightarrow 0$ whenever $X_nrightarrow 0$ with prob 1, which is not always true.
– Michael
Aug 17 at 6:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $X_n$ is a sequence of random variables such that $X_n rightarrow X$ either almost surely or in $L^p$ or both.
Now consider the sequence of conditional random variables $f(X_n) | X_n in B$ for some measurable set $B$ and some nice function $f$ (we can assume it's Lipschitz to make everything easy). Do I still preserve convergence properties for this conditional random variable, i.e. does $f(X_n) | X_n in B rightarrow f(X) | X in B$ in some sense (a.s. or in $L^p)$?
I'm stuck verifying this by using the definition of almost sure convergence or $L^p$ convergence.
For example, regarding almost sure convergence, I have to check that $P(lim_nrightarrow infty f(X_n)|X_n in B = f(X)|Xin B)$ which looks strange from a notation perspective because of the conditional. Or should I resort to using conditional expectation?
real-analysis probability probability-theory convergence random-variables
asked Aug 17 at 5:09
Tomas Jorovic
1,56721325
Suppose that $X_n$ is a sequence of random variables such that $X_n rightarrow X$ either almost surely or in $L^p$ or both.
Now consider the sequence of conditional random variables $f(X_n) | X_n in B$ for some measurable set $B$ and some nice function $f$ (we can assume it's Lipschitz to make everything easy). Do I still preserve convergence properties for this conditional random variable, i.e. does $f(X_n) | X_n in B rightarrow f(X) | X in B$ in some sense (a.s. or in $L^p)$?
I'm stuck verifying this by using the definition of almost sure convergence or $L^p$ convergence.
For example, regarding almost sure convergence, I have to check that $P(lim_nrightarrow infty f(X_n)|X_n in B = f(X)|Xin B)$ which looks strange from a notation perspective because of the conditional. Or should I resort to using conditional expectation?
real-analysis probability probability-theory convergence random-variables
asked Aug 17 at 5:09
Tomas Jorovic
1,56721325
asked Aug 17 at 5:09
Tomas Jorovic
1,56721325
asked Aug 17 at 5:09
Tomas Jorovic
1,56721325
asked Aug 17 at 5:09
Tomas Jorovic
1,56721325
1,56721325
2
You cannot define $f(X_n)|X_n$ as random variables. You use conditioning only for distributions and expectations. So almost sure convergence does not even make sense here.
– Kavi Rama Murthy
Aug 17 at 5:38
Perhaps you are really trying to show $E[f(X_n)|X_n in B] rightarrow E[f(X)|Xin B]$? Of course this is not always true, take $B= mathbbR$, $X=0$, $f(x)=x$, then the statement is equivalent to showing $E[X_n]rightarrow 0$ whenever $X_nrightarrow 0$ with prob 1, which is not always true.
– Michael
Aug 17 at 6:27
add a comment |Â
2
You cannot define $f(X_n)|X_n$ as random variables. You use conditioning only for distributions and expectations. So almost sure convergence does not even make sense here.
– Kavi Rama Murthy
Aug 17 at 5:38
Perhaps you are really trying to show $E[f(X_n)|X_n in B] rightarrow E[f(X)|Xin B]$? Of course this is not always true, take $B= mathbbR$, $X=0$, $f(x)=x$, then the statement is equivalent to showing $E[X_n]rightarrow 0$ whenever $X_nrightarrow 0$ with prob 1, which is not always true.
– Michael
Aug 17 at 6:27
2
2
You cannot define $f(X_n)|X_n$ as random variables. You use conditioning only for distributions and expectations. So almost sure convergence does not even make sense here.
– Kavi Rama Murthy
Aug 17 at 5:38
You cannot define $f(X_n)|X_n$ as random variables. You use conditioning only for distributions and expectations. So almost sure convergence does not even make sense here.
– Kavi Rama Murthy
Aug 17 at 5:38
Perhaps you are really trying to show $E[f(X_n)|X_n in B] rightarrow E[f(X)|Xin B]$? Of course this is not always true, take $B= mathbbR$, $X=0$, $f(x)=x$, then the statement is equivalent to showing $E[X_n]rightarrow 0$ whenever $X_nrightarrow 0$ with prob 1, which is not always true.
– Michael
Aug 17 at 6:27
Perhaps you are really trying to show $E[f(X_n)|X_n in B] rightarrow E[f(X)|Xin B]$? Of course this is not always true, take $B= mathbbR$, $X=0$, $f(x)=x$, then the statement is equivalent to showing $E[X_n]rightarrow 0$ whenever $X_nrightarrow 0$ with prob 1, which is not always true.
– Michael
Aug 17 at 6:27
add a comment |Â
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2
You cannot define $f(X_n)|X_n$ as random variables. You use conditioning only for distributions and expectations. So almost sure convergence does not even make sense here.
– Kavi Rama Murthy
Aug 17 at 5:38
Perhaps you are really trying to show $E[f(X_n)|X_n in B] rightarrow E[f(X)|Xin B]$? Of course this is not always true, take $B= mathbbR$, $X=0$, $f(x)=x$, then the statement is equivalent to showing $E[X_n]rightarrow 0$ whenever $X_nrightarrow 0$ with prob 1, which is not always true.
– Michael
Aug 17 at 6:27