Converges to $0$ in probability
$begingroup$
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
edited Dec 13 '18 at 9:25
orange
650215
asked Dec 13 '18 at 1:54
Jiexiong687691Jiexiong687691
825
$endgroup$
add a comment |
$begingroup$
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
edited Dec 13 '18 at 9:25
orange
650215
asked Dec 13 '18 at 1:54
Jiexiong687691Jiexiong687691
825
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 13 '18 at 9:25
orange
650215
asked Dec 13 '18 at 1:54
Jiexiong687691Jiexiong687691
825
$endgroup$
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
edited Dec 13 '18 at 9:25
orange
650215
asked Dec 13 '18 at 1:54
Jiexiong687691Jiexiong687691
825
edited Dec 13 '18 at 9:25
orange
650215
asked Dec 13 '18 at 1:54
Jiexiong687691Jiexiong687691
825
edited Dec 13 '18 at 9:25
orange
650215
edited Dec 13 '18 at 9:25
orange
650215
edited Dec 13 '18 at 9:25
orange
650215
650215
asked Dec 13 '18 at 1:54
Jiexiong687691Jiexiong687691
825
asked Dec 13 '18 at 1:54
Jiexiong687691Jiexiong687691
825
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
add a comment |
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
add a comment |
0
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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