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Question about the converge of a random sequence involving a Gaussian random variable
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Consider a random variable $X$ that is Gaussian with mean $mu$ and variance $sigma^2$. I am trying to understand whether the following statement holds:
beginequation
fracXn xrightarrow[nrightarrowinfty]texta.s. 0.
endequation
I feel that this statement should not hold for almost sure convergence but it should hold for in probability convergence. I used Markov inequality and could not prove the result for almost sure convergence, because an assumption of the form $mathbbE [ |X|] < infty$ was needed. Any help would be really appreciated! Thank you for your time!
probability convergence
asked Jul 22 at 22:24
SpawnKilleR
320111
 |Â
show 1 more comment
up vote
0
down vote
favorite
Consider a random variable $X$ that is Gaussian with mean $mu$ and variance $sigma^2$. I am trying to understand whether the following statement holds:
beginequation
fracXn xrightarrow[nrightarrowinfty]texta.s. 0.
endequation
I feel that this statement should not hold for almost sure convergence but it should hold for in probability convergence. I used Markov inequality and could not prove the result for almost sure convergence, because an assumption of the form $mathbbE [ |X|] < infty$ was needed. Any help would be really appreciated! Thank you for your time!
probability convergence
asked Jul 22 at 22:24
SpawnKilleR
320111
1
$mathbbE [ |X|] < infty$ is true for a Gaussian distributed random variable. I think it is bounded above by $|mu|+sqrtfrac2pisigma$ or more simply by $|mu|+sigma$
– Henry
Jul 22 at 23:00
I think you are right! Thanks for the input!
– SpawnKilleR
Jul 22 at 23:26
3
I am not sure that I understand the question. For each $omega in Omega$, $X(omega)$ is a real number which does not depend on $n$. Therefore $X(omega)/n to 0$ as $ntoinfty$. Do you mean that $X_n$ is a sequence of Gaussian random variables and you want to investigate $X_n/n$?
– Chris Janjigian
Jul 22 at 23:36
No it is $fracXn$. That's what I also thought but since Gaussian random variables are unbounded I wasn't sure if this is a rigorous argument.
– SpawnKilleR
Jul 22 at 23:38
1
Yep that argument is rigorous, you do not need it to be bounded. The point is that the result holds $omega$ by $omega$.
– Chris Janjigian
Jul 23 at 0:47
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a random variable $X$ that is Gaussian with mean $mu$ and variance $sigma^2$. I am trying to understand whether the following statement holds:
beginequation
fracXn xrightarrow[nrightarrowinfty]texta.s. 0.
endequation
I feel that this statement should not hold for almost sure convergence but it should hold for in probability convergence. I used Markov inequality and could not prove the result for almost sure convergence, because an assumption of the form $mathbbE [ |X|] < infty$ was needed. Any help would be really appreciated! Thank you for your time!
probability convergence
asked Jul 22 at 22:24
SpawnKilleR
320111
Consider a random variable $X$ that is Gaussian with mean $mu$ and variance $sigma^2$. I am trying to understand whether the following statement holds:
beginequation
fracXn xrightarrow[nrightarrowinfty]texta.s. 0.
endequation
I feel that this statement should not hold for almost sure convergence but it should hold for in probability convergence. I used Markov inequality and could not prove the result for almost sure convergence, because an assumption of the form $mathbbE [ |X|] < infty$ was needed. Any help would be really appreciated! Thank you for your time!
probability convergence
asked Jul 22 at 22:24
SpawnKilleR
320111
asked Jul 22 at 22:24
SpawnKilleR
320111
asked Jul 22 at 22:24
SpawnKilleR
320111
asked Jul 22 at 22:24
SpawnKilleR
320111
320111
1
$mathbbE [ |X|] < infty$ is true for a Gaussian distributed random variable. I think it is bounded above by $|mu|+sqrtfrac2pisigma$ or more simply by $|mu|+sigma$
– Henry
Jul 22 at 23:00
I think you are right! Thanks for the input!
– SpawnKilleR
Jul 22 at 23:26
3
I am not sure that I understand the question. For each $omega in Omega$, $X(omega)$ is a real number which does not depend on $n$. Therefore $X(omega)/n to 0$ as $ntoinfty$. Do you mean that $X_n$ is a sequence of Gaussian random variables and you want to investigate $X_n/n$?
– Chris Janjigian
Jul 22 at 23:36
No it is $fracXn$. That's what I also thought but since Gaussian random variables are unbounded I wasn't sure if this is a rigorous argument.
– SpawnKilleR
Jul 22 at 23:38
1
Yep that argument is rigorous, you do not need it to be bounded. The point is that the result holds $omega$ by $omega$.
– Chris Janjigian
Jul 23 at 0:47
 |Â
show 1 more comment
1
$mathbbE [ |X|] < infty$ is true for a Gaussian distributed random variable. I think it is bounded above by $|mu|+sqrtfrac2pisigma$ or more simply by $|mu|+sigma$
– Henry
Jul 22 at 23:00
I think you are right! Thanks for the input!
– SpawnKilleR
Jul 22 at 23:26
3
I am not sure that I understand the question. For each $omega in Omega$, $X(omega)$ is a real number which does not depend on $n$. Therefore $X(omega)/n to 0$ as $ntoinfty$. Do you mean that $X_n$ is a sequence of Gaussian random variables and you want to investigate $X_n/n$?
– Chris Janjigian
Jul 22 at 23:36
No it is $fracXn$. That's what I also thought but since Gaussian random variables are unbounded I wasn't sure if this is a rigorous argument.
– SpawnKilleR
Jul 22 at 23:38
1
Yep that argument is rigorous, you do not need it to be bounded. The point is that the result holds $omega$ by $omega$.
– Chris Janjigian
Jul 23 at 0:47
1
1
$mathbbE [ |X|] < infty$ is true for a Gaussian distributed random variable. I think it is bounded above by $|mu|+sqrtfrac2pisigma$ or more simply by $|mu|+sigma$
– Henry
Jul 22 at 23:00
$mathbbE [ |X|] < infty$ is true for a Gaussian distributed random variable. I think it is bounded above by $|mu|+sqrtfrac2pisigma$ or more simply by $|mu|+sigma$
– Henry
Jul 22 at 23:00
I think you are right! Thanks for the input!
– SpawnKilleR
Jul 22 at 23:26
I think you are right! Thanks for the input!
– SpawnKilleR
Jul 22 at 23:26
3
3
I am not sure that I understand the question. For each $omega in Omega$, $X(omega)$ is a real number which does not depend on $n$. Therefore $X(omega)/n to 0$ as $ntoinfty$. Do you mean that $X_n$ is a sequence of Gaussian random variables and you want to investigate $X_n/n$?
– Chris Janjigian
Jul 22 at 23:36
I am not sure that I understand the question. For each $omega in Omega$, $X(omega)$ is a real number which does not depend on $n$. Therefore $X(omega)/n to 0$ as $ntoinfty$. Do you mean that $X_n$ is a sequence of Gaussian random variables and you want to investigate $X_n/n$?
– Chris Janjigian
Jul 22 at 23:36
No it is $fracXn$. That's what I also thought but since Gaussian random variables are unbounded I wasn't sure if this is a rigorous argument.
– SpawnKilleR
Jul 22 at 23:38
No it is $fracXn$. That's what I also thought but since Gaussian random variables are unbounded I wasn't sure if this is a rigorous argument.
– SpawnKilleR
Jul 22 at 23:38
1
1
Yep that argument is rigorous, you do not need it to be bounded. The point is that the result holds $omega$ by $omega$.
– Chris Janjigian
Jul 23 at 0:47
Yep that argument is rigorous, you do not need it to be bounded. The point is that the result holds $omega$ by $omega$.
– Chris Janjigian
Jul 23 at 0:47
 |Â
show 1 more comment
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1
$mathbbE [ |X|] < infty$ is true for a Gaussian distributed random variable. I think it is bounded above by $|mu|+sqrtfrac2pisigma$ or more simply by $|mu|+sigma$
– Henry
Jul 22 at 23:00
I think you are right! Thanks for the input!
– SpawnKilleR
Jul 22 at 23:26
3
I am not sure that I understand the question. For each $omega in Omega$, $X(omega)$ is a real number which does not depend on $n$. Therefore $X(omega)/n to 0$ as $ntoinfty$. Do you mean that $X_n$ is a sequence of Gaussian random variables and you want to investigate $X_n/n$?
– Chris Janjigian
Jul 22 at 23:36
No it is $fracXn$. That's what I also thought but since Gaussian random variables are unbounded I wasn't sure if this is a rigorous argument.
– SpawnKilleR
Jul 22 at 23:38
1
Yep that argument is rigorous, you do not need it to be bounded. The point is that the result holds $omega$ by $omega$.
– Chris Janjigian
Jul 23 at 0:47