Limes superior of $X_n>alpha ln(n)$
$begingroup$
Given the random variable $X_n$, to be exponential distributed with parameter $1$, we define
$$A_n:=left{X_n>alpha ln(n)right}.$$
We've shown in the classes that for $alpha leq 1$, $A_n$ occurs infinitely times, and for $alpha>1$ finitely times. Now calculating the limes superior $overline{lim}_{nrightarrowinfty}frac{X_n}{ln(n)}$, we have said that $frac{X_n}{ln(n)}geq 1$ is true for infinitely many $n$.
Why is that true? Is this fact connected with the case $alpha leq 1$?
probability-theory convergence limsup-and-liminf exponential-distribution
edited Dec 10 '18 at 20:50
user376343
3,3532825
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
$endgroup$
add a comment |
$begingroup$
Given the random variable $X_n$, to be exponential distributed with parameter $1$, we define
$$A_n:=left{X_n>alpha ln(n)right}.$$
We've shown in the classes that for $alpha leq 1$, $A_n$ occurs infinitely times, and for $alpha>1$ finitely times. Now calculating the limes superior $overline{lim}_{nrightarrowinfty}frac{X_n}{ln(n)}$, we have said that $frac{X_n}{ln(n)}geq 1$ is true for infinitely many $n$.
Why is that true? Is this fact connected with the case $alpha leq 1$?
probability-theory convergence limsup-and-liminf exponential-distribution
edited Dec 10 '18 at 20:50
user376343
3,3532825
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
$endgroup$
add a comment |
$begingroup$
Given the random variable $X_n$, to be exponential distributed with parameter $1$, we define
$$A_n:=left{X_n>alpha ln(n)right}.$$
We've shown in the classes that for $alpha leq 1$, $A_n$ occurs infinitely times, and for $alpha>1$ finitely times. Now calculating the limes superior $overline{lim}_{nrightarrowinfty}frac{X_n}{ln(n)}$, we have said that $frac{X_n}{ln(n)}geq 1$ is true for infinitely many $n$.
Why is that true? Is this fact connected with the case $alpha leq 1$?
probability-theory convergence limsup-and-liminf exponential-distribution
edited Dec 10 '18 at 20:50
user376343
3,3532825
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
$endgroup$
Given the random variable $X_n$, to be exponential distributed with parameter $1$, we define
$$A_n:=left{X_n>alpha ln(n)right}.$$
We've shown in the classes that for $alpha leq 1$, $A_n$ occurs infinitely times, and for $alpha>1$ finitely times. Now calculating the limes superior $overline{lim}_{nrightarrowinfty}frac{X_n}{ln(n)}$, we have said that $frac{X_n}{ln(n)}geq 1$ is true for infinitely many $n$.
Why is that true? Is this fact connected with the case $alpha leq 1$?
probability-theory convergence limsup-and-liminf exponential-distribution
probability-theory convergence limsup-and-liminf exponential-distribution
edited Dec 10 '18 at 20:50
user376343
3,3532825
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
edited Dec 10 '18 at 20:50
user376343
3,3532825
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
edited Dec 10 '18 at 20:50
user376343
3,3532825
edited Dec 10 '18 at 20:50
user376343
3,3532825
edited Dec 10 '18 at 20:50
user376343
3,3532825
3,3532825
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
asked Dec 3 '18 at 9:52
ThesinusThesinus
247210
247210
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
$P(X_n >alpha log, n) =int_{alpha log, n}^{infty} e^{-x}, dx=frac 1 {n^{alpha}}$. Assuming independence of $X_n$'s we can conclude from Borel Cantelli Lemma and convergence of $sum frac 1 {n^{alpha}}$ that $X_n leq alpha log, n$ for all $n$ sufficently large with probability $1$ if $alpha >1$. If $alpha leq 1$ the same argument shows that $X_n >alpha log, n$ for inifintely many $n$ with probability $1$. This implies that $lim sup frac {X_n} {log, n} geq 1$ with probability $1$.
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$P(X_n >alpha log, n) =int_{alpha log, n}^{infty} e^{-x}, dx=frac 1 {n^{alpha}}$. Assuming independence of $X_n$'s we can conclude from Borel Cantelli Lemma and convergence of $sum frac 1 {n^{alpha}}$ that $X_n leq alpha log, n$ for all $n$ sufficently large with probability $1$ if $alpha >1$. If $alpha leq 1$ the same argument shows that $X_n >alpha log, n$ for inifintely many $n$ with probability $1$. This implies that $lim sup frac {X_n} {log, n} geq 1$ with probability $1$.
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
add a comment |
$begingroup$
$P(X_n >alpha log, n) =int_{alpha log, n}^{infty} e^{-x}, dx=frac 1 {n^{alpha}}$. Assuming independence of $X_n$'s we can conclude from Borel Cantelli Lemma and convergence of $sum frac 1 {n^{alpha}}$ that $X_n leq alpha log, n$ for all $n$ sufficently large with probability $1$ if $alpha >1$. If $alpha leq 1$ the same argument shows that $X_n >alpha log, n$ for inifintely many $n$ with probability $1$. This implies that $lim sup frac {X_n} {log, n} geq 1$ with probability $1$.
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
add a comment |
$begingroup$
$P(X_n >alpha log, n) =int_{alpha log, n}^{infty} e^{-x}, dx=frac 1 {n^{alpha}}$. Assuming independence of $X_n$'s we can conclude from Borel Cantelli Lemma and convergence of $sum frac 1 {n^{alpha}}$ that $X_n leq alpha log, n$ for all $n$ sufficently large with probability $1$ if $alpha >1$. If $alpha leq 1$ the same argument shows that $X_n >alpha log, n$ for inifintely many $n$ with probability $1$. This implies that $lim sup frac {X_n} {log, n} geq 1$ with probability $1$.
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
$P(X_n >alpha log, n) =int_{alpha log, n}^{infty} e^{-x}, dx=frac 1 {n^{alpha}}$. Assuming independence of $X_n$'s we can conclude from Borel Cantelli Lemma and convergence of $sum frac 1 {n^{alpha}}$ that $X_n leq alpha log, n$ for all $n$ sufficently large with probability $1$ if $alpha >1$. If $alpha leq 1$ the same argument shows that $X_n >alpha log, n$ for inifintely many $n$ with probability $1$. This implies that $lim sup frac {X_n} {log, n} geq 1$ with probability $1$.
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
edited Dec 3 '18 at 10:06
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Dec 3 '18 at 9:59
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
53.8k32055
add a comment |
add a comment |
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