Converges to $0$ in probability












3












$begingroup$


Suppose $chi_n to chi:=mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - frac{chi_n}{n}$. Given that $frac{1}{n log rho_n} to -frac {1}{chi}$, show that
begin{align*}
frac{1}{nsqrt{log n}} frac{log chi_n}{log rho_n}
end{align*}
converges to zero in probability. My understanding is that it suffices to show $frac {log chi_{n}}{sqrt {log n}} to 0$ in probability. In other words, I want to show: $$forallepsilon >0:lim_{ntoinfty}mathbb {P}left(Bigg|frac {log chi_{n}}{sqrt {log n}}Bigg|>epsilonright)=0$$ But I found that it's a little bit hard to show this.






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You might justify your understanding by noting that for $epsilon>0$: $${|A_nB_n| > epsilon} subseteq {|A_n|>sqrt{epsilon}} cup {|B_n|>sqrt{epsilon}} $$
    $endgroup$
    – Michael
    Dec 13 '18 at 7:17








  • 1




    $begingroup$
    Do you mean that $chi$ is distributed exponentially with parameter $1$ or do you mean it is a deterministic random variable?
    $endgroup$
    – orange
    Dec 13 '18 at 8:18










  • $begingroup$
    Yes, $chi$ is defined as exp(1).
    $endgroup$
    – Jiexiong687691
    Dec 13 '18 at 8:23






  • 1




    $begingroup$
    If $X_n to^d X$ and $X$ is almost surely constant then $X_n to X$ in probability.
    $endgroup$
    – orange
    Dec 13 '18 at 8:57
















3












$begingroup$


Suppose $chi_n to chi:=mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - frac{chi_n}{n}$. Given that $frac{1}{n log rho_n} to -frac {1}{chi}$, show that
begin{align*}
frac{1}{nsqrt{log n}} frac{log chi_n}{log rho_n}
end{align*}
converges to zero in probability. My understanding is that it suffices to show $frac {log chi_{n}}{sqrt {log n}} to 0$ in probability. In other words, I want to show: $$forallepsilon >0:lim_{ntoinfty}mathbb {P}left(Bigg|frac {log chi_{n}}{sqrt {log n}}Bigg|>epsilonright)=0$$ But I found that it's a little bit hard to show this.






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You might justify your understanding by noting that for $epsilon>0$: $${|A_nB_n| > epsilon} subseteq {|A_n|>sqrt{epsilon}} cup {|B_n|>sqrt{epsilon}} $$
    $endgroup$
    – Michael
    Dec 13 '18 at 7:17








  • 1




    $begingroup$
    Do you mean that $chi$ is distributed exponentially with parameter $1$ or do you mean it is a deterministic random variable?
    $endgroup$
    – orange
    Dec 13 '18 at 8:18










  • $begingroup$
    Yes, $chi$ is defined as exp(1).
    $endgroup$
    – Jiexiong687691
    Dec 13 '18 at 8:23






  • 1




    $begingroup$
    If $X_n to^d X$ and $X$ is almost surely constant then $X_n to X$ in probability.
    $endgroup$
    – orange
    Dec 13 '18 at 8:57














3












3








3


1



$begingroup$


Suppose $chi_n to chi:=mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - frac{chi_n}{n}$. Given that $frac{1}{n log rho_n} to -frac {1}{chi}$, show that
begin{align*}
frac{1}{nsqrt{log n}} frac{log chi_n}{log rho_n}
end{align*}
converges to zero in probability. My understanding is that it suffices to show $frac {log chi_{n}}{sqrt {log n}} to 0$ in probability. In other words, I want to show: $$forallepsilon >0:lim_{ntoinfty}mathbb {P}left(Bigg|frac {log chi_{n}}{sqrt {log n}}Bigg|>epsilonright)=0$$ But I found that it's a little bit hard to show this.






probability







share|cite|improve this question











$endgroup$




Suppose $chi_n to chi:=mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - frac{chi_n}{n}$. Given that $frac{1}{n log rho_n} to -frac {1}{chi}$, show that
begin{align*}
frac{1}{nsqrt{log n}} frac{log chi_n}{log rho_n}
end{align*}
converges to zero in probability. My understanding is that it suffices to show $frac {log chi_{n}}{sqrt {log n}} to 0$ in probability. In other words, I want to show: $$forallepsilon >0:lim_{ntoinfty}mathbb {P}left(Bigg|frac {log chi_{n}}{sqrt {log n}}Bigg|>epsilonright)=0$$ But I found that it's a little bit hard to show this.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 9:25









orange

650215




650215










asked Dec 13 '18 at 1:54









Jiexiong687691Jiexiong687691

825




825








  • 1




    $begingroup$
    You might justify your understanding by noting that for $epsilon>0$: $${|A_nB_n| > epsilon} subseteq {|A_n|>sqrt{epsilon}} cup {|B_n|>sqrt{epsilon}} $$
    $endgroup$
    – Michael
    Dec 13 '18 at 7:17








  • 1




    $begingroup$
    Do you mean that $chi$ is distributed exponentially with parameter $1$ or do you mean it is a deterministic random variable?
    $endgroup$
    – orange
    Dec 13 '18 at 8:18










  • $begingroup$
    Yes, $chi$ is defined as exp(1).
    $endgroup$
    – Jiexiong687691
    Dec 13 '18 at 8:23






  • 1




    $begingroup$
    If $X_n to^d X$ and $X$ is almost surely constant then $X_n to X$ in probability.
    $endgroup$
    – orange
    Dec 13 '18 at 8:57














  • 1




    $begingroup$
    You might justify your understanding by noting that for $epsilon>0$: $${|A_nB_n| > epsilon} subseteq {|A_n|>sqrt{epsilon}} cup {|B_n|>sqrt{epsilon}} $$
    $endgroup$
    – Michael
    Dec 13 '18 at 7:17








  • 1




    $begingroup$
    Do you mean that $chi$ is distributed exponentially with parameter $1$ or do you mean it is a deterministic random variable?
    $endgroup$
    – orange
    Dec 13 '18 at 8:18










  • $begingroup$
    Yes, $chi$ is defined as exp(1).
    $endgroup$
    – Jiexiong687691
    Dec 13 '18 at 8:23






  • 1




    $begingroup$
    If $X_n to^d X$ and $X$ is almost surely constant then $X_n to X$ in probability.
    $endgroup$
    – orange
    Dec 13 '18 at 8:57








1




1




$begingroup$
You might justify your understanding by noting that for $epsilon>0$: $${|A_nB_n| > epsilon} subseteq {|A_n|>sqrt{epsilon}} cup {|B_n|>sqrt{epsilon}} $$
$endgroup$
– Michael
Dec 13 '18 at 7:17






$begingroup$
You might justify your understanding by noting that for $epsilon>0$: $${|A_nB_n| > epsilon} subseteq {|A_n|>sqrt{epsilon}} cup {|B_n|>sqrt{epsilon}} $$
$endgroup$
– Michael
Dec 13 '18 at 7:17






1




1




$begingroup$
Do you mean that $chi$ is distributed exponentially with parameter $1$ or do you mean it is a deterministic random variable?
$endgroup$
– orange
Dec 13 '18 at 8:18




$begingroup$
Do you mean that $chi$ is distributed exponentially with parameter $1$ or do you mean it is a deterministic random variable?
$endgroup$
– orange
Dec 13 '18 at 8:18












$begingroup$
Yes, $chi$ is defined as exp(1).
$endgroup$
– Jiexiong687691
Dec 13 '18 at 8:23




$begingroup$
Yes, $chi$ is defined as exp(1).
$endgroup$
– Jiexiong687691
Dec 13 '18 at 8:23




1




1




$begingroup$
If $X_n to^d X$ and $X$ is almost surely constant then $X_n to X$ in probability.
$endgroup$
– orange
Dec 13 '18 at 8:57




$begingroup$
If $X_n to^d X$ and $X$ is almost surely constant then $X_n to X$ in probability.
$endgroup$
– orange
Dec 13 '18 at 8:57










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