Mixing
Functions and sequence of random variables
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Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.
$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$
I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.
probability convergence
asked Jul 29 at 22:56
Sergio Andrade
143112
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up vote
0
down vote
favorite
Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.
$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$
I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.
probability convergence
asked Jul 29 at 22:56
Sergio Andrade
143112
1
Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.
$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$
I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.
probability convergence
asked Jul 29 at 22:56
Sergio Andrade
143112
Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.
$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$
I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.
probability convergence
asked Jul 29 at 22:56
Sergio Andrade
143112
asked Jul 29 at 22:56
Sergio Andrade
143112
asked Jul 29 at 22:56
Sergio Andrade
143112
asked Jul 29 at 22:56
Sergio Andrade
143112
143112
1
Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45
add a comment |Â
1
Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45
1
1
Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45
Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
add a comment |Â
up vote
2
down vote
accepted
Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
answered Jul 29 at 23:45
Kavi Rama Murthy
19.7k2829
19.7k2829
add a comment |Â
add a comment |Â
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1
Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45