Convergence of in $L_1$ and probability [closed]
Let $X_n$ be a sequence of random variables such that $X_n → 0$ (converges in probability. Then show that if $|X_n | ≤ A$, for some $A > 0$ and all $n$, then $E[|X_n |] → 0$ and $E[X_n ] → 0$.
probability-theory
asked Nov 21 '18 at 12:37
Christy
12110
closed as off-topic by saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus Nov 22 '18 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let $X_n$ be a sequence of random variables such that $X_n → 0$ (converges in probability. Then show that if $|X_n | ≤ A$, for some $A > 0$ and all $n$, then $E[|X_n |] → 0$ and $E[X_n ] → 0$.
probability-theory
asked Nov 21 '18 at 12:37
Christy
12110
closed as off-topic by saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus Nov 22 '18 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
Any thoughts on the problem? What have you tried?
– saz
Nov 21 '18 at 12:44
As $X_n$ converges in probability, $P(|X_n| > epsilon) ->0$. By Markov inequality, $P(|X_n| > epsilon) < E(X_n)/ epsilon$. Therefore, $E(|X_n|)>0$?
– Christy
Nov 21 '18 at 12:55
@saz please provide some direction.
– Christy
Nov 21 '18 at 13:14
add a comment |
Let $X_n$ be a sequence of random variables such that $X_n → 0$ (converges in probability. Then show that if $|X_n | ≤ A$, for some $A > 0$ and all $n$, then $E[|X_n |] → 0$ and $E[X_n ] → 0$.
probability-theory
asked Nov 21 '18 at 12:37
Christy
12110
Let $X_n$ be a sequence of random variables such that $X_n → 0$ (converges in probability. Then show that if $|X_n | ≤ A$, for some $A > 0$ and all $n$, then $E[|X_n |] → 0$ and $E[X_n ] → 0$.
probability-theory
probability-theory
asked Nov 21 '18 at 12:37
Christy
12110
asked Nov 21 '18 at 12:37
Christy
12110
asked Nov 21 '18 at 12:37
Christy
12110
asked Nov 21 '18 at 12:37
Christy
12110
asked Nov 21 '18 at 12:37
Christy
12110
12110
closed as off-topic by saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus Nov 22 '18 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus Nov 22 '18 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – saz, Davide Giraudo, Scientifica, Don Thousand, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
Any thoughts on the problem? What have you tried?
– saz
Nov 21 '18 at 12:44
As $X_n$ converges in probability, $P(|X_n| > epsilon) ->0$. By Markov inequality, $P(|X_n| > epsilon) < E(X_n)/ epsilon$. Therefore, $E(|X_n|)>0$?
– Christy
Nov 21 '18 at 12:55
@saz please provide some direction.
– Christy
Nov 21 '18 at 13:14
add a comment |
Any thoughts on the problem? What have you tried?
– saz
Nov 21 '18 at 12:44
As $X_n$ converges in probability, $P(|X_n| > epsilon) ->0$. By Markov inequality, $P(|X_n| > epsilon) < E(X_n)/ epsilon$. Therefore, $E(|X_n|)>0$?
– Christy
Nov 21 '18 at 12:55
@saz please provide some direction.
– Christy
Nov 21 '18 at 13:14
Any thoughts on the problem? What have you tried?
– saz
Nov 21 '18 at 12:44
Any thoughts on the problem? What have you tried?
– saz
Nov 21 '18 at 12:44
As $X_n$ converges in probability, $P(|X_n| > epsilon) ->0$. By Markov inequality, $P(|X_n| > epsilon) < E(X_n)/ epsilon$. Therefore, $E(|X_n|)>0$?
– Christy
Nov 21 '18 at 12:55
As $X_n$ converges in probability, $P(|X_n| > epsilon) ->0$. By Markov inequality, $P(|X_n| > epsilon) < E(X_n)/ epsilon$. Therefore, $E(|X_n|)>0$?
– Christy
Nov 21 '18 at 12:55
@saz please provide some direction.
– Christy
Nov 21 '18 at 13:14
@saz please provide some direction.
– Christy
Nov 21 '18 at 13:14
add a comment |
1 Answer
1
active
oldest
votes
Fix any $varepsilon>0$, and observe that
$$
mathbb{E}|X_n| = int_{|X_n| > varepsilon} |X_n| d mathbb{P} + int_{|X_n| leq varepsilon} |X_n| d mathbb{P} leq A mathbb{P}(|X_n| > varepsilon) + varepsilon.
$$
Now use the fact that $X_n to 0$ in probability, and that $varepsilon>0$ was arbitrary.
answered Nov 21 '18 at 20:19
Hayk
2,0721213
So $E|X_n| -> epsilon$ ?
– Christy
Nov 21 '18 at 23:42
1
@Christy, from the displayed inequality, applying limits on both sides we get $$ limsuplimits_{n toinfty } mathbb{E}|X_n| leq varepsilon (1) $$ thanks to the condition $mathbb{P}(|X_n| > varepsilon ) to 0$ (convergence of $X_n$ to $0$ in probability). Since $varepsilon>0$ is arbitrary, the inequality $(1)$ implies that the $lim mathbb{E}|X_n| $ exists and equals $0$.
– Hayk
Nov 22 '18 at 4:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Fix any $varepsilon>0$, and observe that
$$
mathbb{E}|X_n| = int_{|X_n| > varepsilon} |X_n| d mathbb{P} + int_{|X_n| leq varepsilon} |X_n| d mathbb{P} leq A mathbb{P}(|X_n| > varepsilon) + varepsilon.
$$
Now use the fact that $X_n to 0$ in probability, and that $varepsilon>0$ was arbitrary.
answered Nov 21 '18 at 20:19
Hayk
2,0721213
So $E|X_n| -> epsilon$ ?
– Christy
Nov 21 '18 at 23:42
1
@Christy, from the displayed inequality, applying limits on both sides we get $$ limsuplimits_{n toinfty } mathbb{E}|X_n| leq varepsilon (1) $$ thanks to the condition $mathbb{P}(|X_n| > varepsilon ) to 0$ (convergence of $X_n$ to $0$ in probability). Since $varepsilon>0$ is arbitrary, the inequality $(1)$ implies that the $lim mathbb{E}|X_n| $ exists and equals $0$.
– Hayk
Nov 22 '18 at 4:57
add a comment |
Fix any $varepsilon>0$, and observe that
$$
mathbb{E}|X_n| = int_{|X_n| > varepsilon} |X_n| d mathbb{P} + int_{|X_n| leq varepsilon} |X_n| d mathbb{P} leq A mathbb{P}(|X_n| > varepsilon) + varepsilon.
$$
Now use the fact that $X_n to 0$ in probability, and that $varepsilon>0$ was arbitrary.
answered Nov 21 '18 at 20:19
Hayk
2,0721213
So $E|X_n| -> epsilon$ ?
– Christy
Nov 21 '18 at 23:42
1
@Christy, from the displayed inequality, applying limits on both sides we get $$ limsuplimits_{n toinfty } mathbb{E}|X_n| leq varepsilon (1) $$ thanks to the condition $mathbb{P}(|X_n| > varepsilon ) to 0$ (convergence of $X_n$ to $0$ in probability). Since $varepsilon>0$ is arbitrary, the inequality $(1)$ implies that the $lim mathbb{E}|X_n| $ exists and equals $0$.
– Hayk
Nov 22 '18 at 4:57
add a comment |
Fix any $varepsilon>0$, and observe that
$$
mathbb{E}|X_n| = int_{|X_n| > varepsilon} |X_n| d mathbb{P} + int_{|X_n| leq varepsilon} |X_n| d mathbb{P} leq A mathbb{P}(|X_n| > varepsilon) + varepsilon.
$$
Now use the fact that $X_n to 0$ in probability, and that $varepsilon>0$ was arbitrary.
answered Nov 21 '18 at 20:19
Hayk
2,0721213
Fix any $varepsilon>0$, and observe that
$$
mathbb{E}|X_n| = int_{|X_n| > varepsilon} |X_n| d mathbb{P} + int_{|X_n| leq varepsilon} |X_n| d mathbb{P} leq A mathbb{P}(|X_n| > varepsilon) + varepsilon.
$$
Now use the fact that $X_n to 0$ in probability, and that $varepsilon>0$ was arbitrary.
answered Nov 21 '18 at 20:19
Hayk
2,0721213
answered Nov 21 '18 at 20:19
Hayk
2,0721213
answered Nov 21 '18 at 20:19
Hayk
2,0721213
answered Nov 21 '18 at 20:19
Hayk
2,0721213
2,0721213
So $E|X_n| -> epsilon$ ?
– Christy
Nov 21 '18 at 23:42
1
@Christy, from the displayed inequality, applying limits on both sides we get $$ limsuplimits_{n toinfty } mathbb{E}|X_n| leq varepsilon (1) $$ thanks to the condition $mathbb{P}(|X_n| > varepsilon ) to 0$ (convergence of $X_n$ to $0$ in probability). Since $varepsilon>0$ is arbitrary, the inequality $(1)$ implies that the $lim mathbb{E}|X_n| $ exists and equals $0$.
– Hayk
Nov 22 '18 at 4:57
add a comment |
So $E|X_n| -> epsilon$ ?
– Christy
Nov 21 '18 at 23:42
1
@Christy, from the displayed inequality, applying limits on both sides we get $$ limsuplimits_{n toinfty } mathbb{E}|X_n| leq varepsilon (1) $$ thanks to the condition $mathbb{P}(|X_n| > varepsilon ) to 0$ (convergence of $X_n$ to $0$ in probability). Since $varepsilon>0$ is arbitrary, the inequality $(1)$ implies that the $lim mathbb{E}|X_n| $ exists and equals $0$.
– Hayk
Nov 22 '18 at 4:57
So $E|X_n| -> epsilon$ ?
– Christy
Nov 21 '18 at 23:42
So $E|X_n| -> epsilon$ ?
– Christy
Nov 21 '18 at 23:42
1
1
@Christy, from the displayed inequality, applying limits on both sides we get $$ limsuplimits_{n toinfty } mathbb{E}|X_n| leq varepsilon (1) $$ thanks to the condition $mathbb{P}(|X_n| > varepsilon ) to 0$ (convergence of $X_n$ to $0$ in probability). Since $varepsilon>0$ is arbitrary, the inequality $(1)$ implies that the $lim mathbb{E}|X_n| $ exists and equals $0$.
– Hayk
Nov 22 '18 at 4:57
@Christy, from the displayed inequality, applying limits on both sides we get $$ limsuplimits_{n toinfty } mathbb{E}|X_n| leq varepsilon (1) $$ thanks to the condition $mathbb{P}(|X_n| > varepsilon ) to 0$ (convergence of $X_n$ to $0$ in probability). Since $varepsilon>0$ is arbitrary, the inequality $(1)$ implies that the $lim mathbb{E}|X_n| $ exists and equals $0$.
– Hayk
Nov 22 '18 at 4:57
add a comment |
Any thoughts on the problem? What have you tried?
– saz
Nov 21 '18 at 12:44
As $X_n$ converges in probability, $P(|X_n| > epsilon) ->0$. By Markov inequality, $P(|X_n| > epsilon) < E(X_n)/ epsilon$. Therefore, $E(|X_n|)>0$?
– Christy
Nov 21 '18 at 12:55
@saz please provide some direction.
– Christy
Nov 21 '18 at 13:14