Does a continuous function of real numbers preserve continuity of random variables?
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In particular, if $X_1, X_2, ldots$ is a sequence of random variables such that $X_nto X$ as $ntoinfty$, does it follow that if $f$ is continuous function (over reals) that $lim_{ntoinfty} f(X_n) = f(X)$?
What if $X$ is constant, is this still true?
continuity random-variables
asked Nov 19 at 20:04
Anon
406313
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In particular, if $X_1, X_2, ldots$ is a sequence of random variables such that $X_nto X$ as $ntoinfty$, does it follow that if $f$ is continuous function (over reals) that $lim_{ntoinfty} f(X_n) = f(X)$?
What if $X$ is constant, is this still true?
continuity random-variables
asked Nov 19 at 20:04
Anon
406313
In what sense $X_nto X$?
– Dante Grevino
Nov 19 at 20:16
We have $P(|X_n - X| > epsilon) to 0$ as $nto infty$.
– Anon
Nov 19 at 20:18
OK, in probability sense.
– Dante Grevino
Nov 19 at 20:36
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In particular, if $X_1, X_2, ldots$ is a sequence of random variables such that $X_nto X$ as $ntoinfty$, does it follow that if $f$ is continuous function (over reals) that $lim_{ntoinfty} f(X_n) = f(X)$?
What if $X$ is constant, is this still true?
continuity random-variables
asked Nov 19 at 20:04
Anon
406313
In particular, if $X_1, X_2, ldots$ is a sequence of random variables such that $X_nto X$ as $ntoinfty$, does it follow that if $f$ is continuous function (over reals) that $lim_{ntoinfty} f(X_n) = f(X)$?
What if $X$ is constant, is this still true?
continuity random-variables
continuity random-variables
asked Nov 19 at 20:04
Anon
406313
asked Nov 19 at 20:04
Anon
406313
asked Nov 19 at 20:04
Anon
406313
asked Nov 19 at 20:04
Anon
406313
asked Nov 19 at 20:04
Anon
406313
406313
In what sense $X_nto X$?
– Dante Grevino
Nov 19 at 20:16
We have $P(|X_n - X| > epsilon) to 0$ as $nto infty$.
– Anon
Nov 19 at 20:18
OK, in probability sense.
– Dante Grevino
Nov 19 at 20:36
add a comment |
In what sense $X_nto X$?
– Dante Grevino
Nov 19 at 20:16
We have $P(|X_n - X| > epsilon) to 0$ as $nto infty$.
– Anon
Nov 19 at 20:18
OK, in probability sense.
– Dante Grevino
Nov 19 at 20:36
In what sense $X_nto X$?
– Dante Grevino
Nov 19 at 20:16
In what sense $X_nto X$?
– Dante Grevino
Nov 19 at 20:16
We have $P(|X_n - X| > epsilon) to 0$ as $nto infty$.
– Anon
Nov 19 at 20:18
We have $P(|X_n - X| > epsilon) to 0$ as $nto infty$.
– Anon
Nov 19 at 20:18
OK, in probability sense.
– Dante Grevino
Nov 19 at 20:36
OK, in probability sense.
– Dante Grevino
Nov 19 at 20:36
add a comment |
1 Answer
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Let $epsilon$ and $delta$ be positive real numbers. Define the set
$$
A_delta = {xin mathbb{R}mid exists yin mathbb{R} s.t. |x-y|<delta, |g(x)-g(y)|>epsilon}
$$
Note that $P(|g(X_n)-g(X)|>epsilon) leq P(|X_n-X|geqdelta)+P(Xin A_delta)$.
So, taking the superior limit as $ntoinfty$, we get
$$
overline{lim_{ntoinfty}}P(|g(X_n)-g(X)|>epsilon)leq P(Xin A_delta),
$$
for every $delta>0$. Because the continuity of $g$, we have that $lim_{deltato 0^+}P(Xin A_delta) =0$. And we are done.
answered Nov 19 at 20:35
Dante Grevino
8817
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
0
down vote
Let $epsilon$ and $delta$ be positive real numbers. Define the set
$$
A_delta = {xin mathbb{R}mid exists yin mathbb{R} s.t. |x-y|<delta, |g(x)-g(y)|>epsilon}
$$
Note that $P(|g(X_n)-g(X)|>epsilon) leq P(|X_n-X|geqdelta)+P(Xin A_delta)$.
So, taking the superior limit as $ntoinfty$, we get
$$
overline{lim_{ntoinfty}}P(|g(X_n)-g(X)|>epsilon)leq P(Xin A_delta),
$$
for every $delta>0$. Because the continuity of $g$, we have that $lim_{deltato 0^+}P(Xin A_delta) =0$. And we are done.
answered Nov 19 at 20:35
Dante Grevino
8817
add a comment |
up vote
0
down vote
Let $epsilon$ and $delta$ be positive real numbers. Define the set
$$
A_delta = {xin mathbb{R}mid exists yin mathbb{R} s.t. |x-y|<delta, |g(x)-g(y)|>epsilon}
$$
Note that $P(|g(X_n)-g(X)|>epsilon) leq P(|X_n-X|geqdelta)+P(Xin A_delta)$.
So, taking the superior limit as $ntoinfty$, we get
$$
overline{lim_{ntoinfty}}P(|g(X_n)-g(X)|>epsilon)leq P(Xin A_delta),
$$
for every $delta>0$. Because the continuity of $g$, we have that $lim_{deltato 0^+}P(Xin A_delta) =0$. And we are done.
answered Nov 19 at 20:35
Dante Grevino
8817
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $epsilon$ and $delta$ be positive real numbers. Define the set
$$
A_delta = {xin mathbb{R}mid exists yin mathbb{R} s.t. |x-y|<delta, |g(x)-g(y)|>epsilon}
$$
Note that $P(|g(X_n)-g(X)|>epsilon) leq P(|X_n-X|geqdelta)+P(Xin A_delta)$.
So, taking the superior limit as $ntoinfty$, we get
$$
overline{lim_{ntoinfty}}P(|g(X_n)-g(X)|>epsilon)leq P(Xin A_delta),
$$
for every $delta>0$. Because the continuity of $g$, we have that $lim_{deltato 0^+}P(Xin A_delta) =0$. And we are done.
answered Nov 19 at 20:35
Dante Grevino
8817
Let $epsilon$ and $delta$ be positive real numbers. Define the set
$$
A_delta = {xin mathbb{R}mid exists yin mathbb{R} s.t. |x-y|<delta, |g(x)-g(y)|>epsilon}
$$
Note that $P(|g(X_n)-g(X)|>epsilon) leq P(|X_n-X|geqdelta)+P(Xin A_delta)$.
So, taking the superior limit as $ntoinfty$, we get
$$
overline{lim_{ntoinfty}}P(|g(X_n)-g(X)|>epsilon)leq P(Xin A_delta),
$$
for every $delta>0$. Because the continuity of $g$, we have that $lim_{deltato 0^+}P(Xin A_delta) =0$. And we are done.
answered Nov 19 at 20:35
Dante Grevino
8817
answered Nov 19 at 20:35
Dante Grevino
8817
answered Nov 19 at 20:35
Dante Grevino
8817
answered Nov 19 at 20:35
Dante Grevino
8817
8817
add a comment |
add a comment |
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In what sense $X_nto X$?
– Dante Grevino
Nov 19 at 20:16
We have $P(|X_n - X| > epsilon) to 0$ as $nto infty$.
– Anon
Nov 19 at 20:18
OK, in probability sense.
– Dante Grevino
Nov 19 at 20:36