Proof using chebyshev's inequality [closed]
$begingroup$
Consider the sequence of random non-negative values $X_1...X_n$ where
$E[X_n], Var[X_n] > 0$ for all $n in mathbb{N}$ and:
$$lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0$$
Prove:
$$lim_{nrightarrow infty} P(X_n >0) = 1$$
How would I prove this using Chebyshev's inequality?
probability probability-theory discrete-mathematics
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
$endgroup$
closed as off-topic by Dando18, Shailesh, Davide Giraudo, user10354138, Kevin Dec 13 '18 at 14:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Shailesh, Davide Giraudo, user10354138, Kevin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the sequence of random non-negative values $X_1...X_n$ where
$E[X_n], Var[X_n] > 0$ for all $n in mathbb{N}$ and:
$$lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0$$
Prove:
$$lim_{nrightarrow infty} P(X_n >0) = 1$$
How would I prove this using Chebyshev's inequality?
probability probability-theory discrete-mathematics
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
$endgroup$
closed as off-topic by Dando18, Shailesh, Davide Giraudo, user10354138, Kevin Dec 13 '18 at 14:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Shailesh, Davide Giraudo, user10354138, Kevin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the sequence of random non-negative values $X_1...X_n$ where
$E[X_n], Var[X_n] > 0$ for all $n in mathbb{N}$ and:
$$lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0$$
Prove:
$$lim_{nrightarrow infty} P(X_n >0) = 1$$
How would I prove this using Chebyshev's inequality?
probability probability-theory discrete-mathematics
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
$endgroup$
Consider the sequence of random non-negative values $X_1...X_n$ where
$E[X_n], Var[X_n] > 0$ for all $n in mathbb{N}$ and:
$$lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0$$
Prove:
$$lim_{nrightarrow infty} P(X_n >0) = 1$$
How would I prove this using Chebyshev's inequality?
probability probability-theory discrete-mathematics
probability probability-theory discrete-mathematics
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
asked Dec 13 '18 at 4:24
J. LastinJ. Lastin
936
936
closed as off-topic by Dando18, Shailesh, Davide Giraudo, user10354138, Kevin Dec 13 '18 at 14:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Shailesh, Davide Giraudo, user10354138, Kevin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Dando18, Shailesh, Davide Giraudo, user10354138, Kevin Dec 13 '18 at 14:31
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Shailesh, Davide Giraudo, user10354138, Kevin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Chebyshev's inequality is in the right direction, but before trying to apply it without a plan, take a look at how assumption looks like: it says that
$$
lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0.
$$ If one looks at the assumption carefully, one may notice that $frac{1}{E[X_n]}$ plays a role of normalizing constant. So, by letting $$X_n' :=frac{X_n}{E[X_n]},$$ one gets
$$
E[X_n'] = 1,quad lim_{nrightarrow infty} Var[X'_n] = 0.
$$ The problem gets simpler now. Chebyshev's inequality implies that $$
P(X_n>0) = P(X'_n>0)geq P(|X'_n -1|leq frac{1}{2}) geq 1- 4Var[X'_n]to 1,
$$
(in fact, $X'_n to_p 1$ in probability) and hence that $lim_{ntoinfty}P(X_n>0)=1$ as desired.
answered Dec 13 '18 at 4:55
SongSong
13.1k632
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Chebyshev's inequality is in the right direction, but before trying to apply it without a plan, take a look at how assumption looks like: it says that
$$
lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0.
$$ If one looks at the assumption carefully, one may notice that $frac{1}{E[X_n]}$ plays a role of normalizing constant. So, by letting $$X_n' :=frac{X_n}{E[X_n]},$$ one gets
$$
E[X_n'] = 1,quad lim_{nrightarrow infty} Var[X'_n] = 0.
$$ The problem gets simpler now. Chebyshev's inequality implies that $$
P(X_n>0) = P(X'_n>0)geq P(|X'_n -1|leq frac{1}{2}) geq 1- 4Var[X'_n]to 1,
$$
(in fact, $X'_n to_p 1$ in probability) and hence that $lim_{ntoinfty}P(X_n>0)=1$ as desired.
answered Dec 13 '18 at 4:55
SongSong
13.1k632
$endgroup$
add a comment |
$begingroup$
Chebyshev's inequality is in the right direction, but before trying to apply it without a plan, take a look at how assumption looks like: it says that
$$
lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0.
$$ If one looks at the assumption carefully, one may notice that $frac{1}{E[X_n]}$ plays a role of normalizing constant. So, by letting $$X_n' :=frac{X_n}{E[X_n]},$$ one gets
$$
E[X_n'] = 1,quad lim_{nrightarrow infty} Var[X'_n] = 0.
$$ The problem gets simpler now. Chebyshev's inequality implies that $$
P(X_n>0) = P(X'_n>0)geq P(|X'_n -1|leq frac{1}{2}) geq 1- 4Var[X'_n]to 1,
$$
(in fact, $X'_n to_p 1$ in probability) and hence that $lim_{ntoinfty}P(X_n>0)=1$ as desired.
answered Dec 13 '18 at 4:55
SongSong
13.1k632
$endgroup$
add a comment |
$begingroup$
Chebyshev's inequality is in the right direction, but before trying to apply it without a plan, take a look at how assumption looks like: it says that
$$
lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0.
$$ If one looks at the assumption carefully, one may notice that $frac{1}{E[X_n]}$ plays a role of normalizing constant. So, by letting $$X_n' :=frac{X_n}{E[X_n]},$$ one gets
$$
E[X_n'] = 1,quad lim_{nrightarrow infty} Var[X'_n] = 0.
$$ The problem gets simpler now. Chebyshev's inequality implies that $$
P(X_n>0) = P(X'_n>0)geq P(|X'_n -1|leq frac{1}{2}) geq 1- 4Var[X'_n]to 1,
$$
(in fact, $X'_n to_p 1$ in probability) and hence that $lim_{ntoinfty}P(X_n>0)=1$ as desired.
answered Dec 13 '18 at 4:55
SongSong
13.1k632
$endgroup$
Chebyshev's inequality is in the right direction, but before trying to apply it without a plan, take a look at how assumption looks like: it says that
$$
lim_{nrightarrow infty} frac{Var[X_n]}{E[X_n]^2} = 0.
$$ If one looks at the assumption carefully, one may notice that $frac{1}{E[X_n]}$ plays a role of normalizing constant. So, by letting $$X_n' :=frac{X_n}{E[X_n]},$$ one gets
$$
E[X_n'] = 1,quad lim_{nrightarrow infty} Var[X'_n] = 0.
$$ The problem gets simpler now. Chebyshev's inequality implies that $$
P(X_n>0) = P(X'_n>0)geq P(|X'_n -1|leq frac{1}{2}) geq 1- 4Var[X'_n]to 1,
$$
(in fact, $X'_n to_p 1$ in probability) and hence that $lim_{ntoinfty}P(X_n>0)=1$ as desired.
answered Dec 13 '18 at 4:55
SongSong
13.1k632
answered Dec 13 '18 at 4:55
SongSong
13.1k632
answered Dec 13 '18 at 4:55
SongSong
13.1k632
answered Dec 13 '18 at 4:55
SongSong
13.1k632
13.1k632
add a comment |
add a comment |
