Mixing
convergence in probability to $0$ implies a.s convergence if the sequence is monotonic
$begingroup$
claim : let ${X_n, , n geq 1}$ be a monotonic sequence of random variables
$X_nrightarrow0quad text{in probability} implies X_nrightarrow 0quad text{a.s}$
proof attempt :
we have for every $epsilon > 0$ $$lim_{n to +infty}mathbb{P}(|X_n| geq epsilon) = 0$$
which intuitively means to me that : (since it holds for every $epsilon, , $ and $epsilon > 0$ and $|X_n |geq 0$)
$$lim_{n to +infty}mathbb{P}(|X_n| > 0) =lim_{n to +infty}mathbb{P}(|X_n| neq 0) = 0$$
I'm not sure concerning this step.... anyways
by the monotone convergence theorem :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| neq 0) = 0$$
then :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| = 0) = mathbb{P}(omega midlim_{n to +infty} X_n(w) = 0) = 1$$
thus the almost sure convergence, I need some feedback concerning this attempt at proof, thanks !
probability-theory proof-verification convergence
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
claim : let ${X_n, , n geq 1}$ be a monotonic sequence of random variables
$X_nrightarrow0quad text{in probability} implies X_nrightarrow 0quad text{a.s}$
proof attempt :
we have for every $epsilon > 0$ $$lim_{n to +infty}mathbb{P}(|X_n| geq epsilon) = 0$$
which intuitively means to me that : (since it holds for every $epsilon, , $ and $epsilon > 0$ and $|X_n |geq 0$)
$$lim_{n to +infty}mathbb{P}(|X_n| > 0) =lim_{n to +infty}mathbb{P}(|X_n| neq 0) = 0$$
I'm not sure concerning this step.... anyways
by the monotone convergence theorem :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| neq 0) = 0$$
then :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| = 0) = mathbb{P}(omega midlim_{n to +infty} X_n(w) = 0) = 1$$
thus the almost sure convergence, I need some feedback concerning this attempt at proof, thanks !
probability-theory proof-verification convergence
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
$endgroup$
1
$begingroup$
"I'm not sure concerning this step" Indeed, you do not justify it, and it happens to be wrong, as very simple examples show. To prove the result, note that if $(X_n)$ is monotonous, $X_nto X$ almost surely hence $X_nto X$ in probability. By the almost sure uniqueness of the limit in probability, $X=0$ almost surely, end of the proof.
$endgroup$
– Did
Jan 18 at 14:04
add a comment |
$begingroup$
claim : let ${X_n, , n geq 1}$ be a monotonic sequence of random variables
$X_nrightarrow0quad text{in probability} implies X_nrightarrow 0quad text{a.s}$
proof attempt :
we have for every $epsilon > 0$ $$lim_{n to +infty}mathbb{P}(|X_n| geq epsilon) = 0$$
which intuitively means to me that : (since it holds for every $epsilon, , $ and $epsilon > 0$ and $|X_n |geq 0$)
$$lim_{n to +infty}mathbb{P}(|X_n| > 0) =lim_{n to +infty}mathbb{P}(|X_n| neq 0) = 0$$
I'm not sure concerning this step.... anyways
by the monotone convergence theorem :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| neq 0) = 0$$
then :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| = 0) = mathbb{P}(omega midlim_{n to +infty} X_n(w) = 0) = 1$$
thus the almost sure convergence, I need some feedback concerning this attempt at proof, thanks !
probability-theory proof-verification convergence
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
$endgroup$
claim : let ${X_n, , n geq 1}$ be a monotonic sequence of random variables
$X_nrightarrow0quad text{in probability} implies X_nrightarrow 0quad text{a.s}$
proof attempt :
we have for every $epsilon > 0$ $$lim_{n to +infty}mathbb{P}(|X_n| geq epsilon) = 0$$
which intuitively means to me that : (since it holds for every $epsilon, , $ and $epsilon > 0$ and $|X_n |geq 0$)
$$lim_{n to +infty}mathbb{P}(|X_n| > 0) =lim_{n to +infty}mathbb{P}(|X_n| neq 0) = 0$$
I'm not sure concerning this step.... anyways
by the monotone convergence theorem :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| neq 0) = 0$$
then :
$$mathbb{P}(omega midlim_{n to +infty}|X_n(w)| = 0) = mathbb{P}(omega midlim_{n to +infty} X_n(w) = 0) = 1$$
thus the almost sure convergence, I need some feedback concerning this attempt at proof, thanks !
probability-theory proof-verification convergence
probability-theory proof-verification convergence
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
asked Jan 18 at 13:50
rapidracimrapidracim
1,7191419
1,7191419
1
$begingroup$
"I'm not sure concerning this step" Indeed, you do not justify it, and it happens to be wrong, as very simple examples show. To prove the result, note that if $(X_n)$ is monotonous, $X_nto X$ almost surely hence $X_nto X$ in probability. By the almost sure uniqueness of the limit in probability, $X=0$ almost surely, end of the proof.
$endgroup$
– Did
Jan 18 at 14:04
add a comment |
1
$begingroup$
"I'm not sure concerning this step" Indeed, you do not justify it, and it happens to be wrong, as very simple examples show. To prove the result, note that if $(X_n)$ is monotonous, $X_nto X$ almost surely hence $X_nto X$ in probability. By the almost sure uniqueness of the limit in probability, $X=0$ almost surely, end of the proof.
$endgroup$
– Did
Jan 18 at 14:04
1
1
$begingroup$
"I'm not sure concerning this step" Indeed, you do not justify it, and it happens to be wrong, as very simple examples show. To prove the result, note that if $(X_n)$ is monotonous, $X_nto X$ almost surely hence $X_nto X$ in probability. By the almost sure uniqueness of the limit in probability, $X=0$ almost surely, end of the proof.
$endgroup$
– Did
Jan 18 at 14:04
$begingroup$
"I'm not sure concerning this step" Indeed, you do not justify it, and it happens to be wrong, as very simple examples show. To prove the result, note that if $(X_n)$ is monotonous, $X_nto X$ almost surely hence $X_nto X$ in probability. By the almost sure uniqueness of the limit in probability, $X=0$ almost surely, end of the proof.
$endgroup$
– Did
Jan 18 at 14:04
add a comment |
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$begingroup$
"I'm not sure concerning this step" Indeed, you do not justify it, and it happens to be wrong, as very simple examples show. To prove the result, note that if $(X_n)$ is monotonous, $X_nto X$ almost surely hence $X_nto X$ in probability. By the almost sure uniqueness of the limit in probability, $X=0$ almost surely, end of the proof.
$endgroup$
– Did
Jan 18 at 14:04