$chi_n to chi$ in distribution, $chi sim mathrm {exp}(1)$ then $frac {1}{n log (1-frac {chi_{n}}{n})}to -frac...
Suppose $chi_n to chi, ~~ chi sim mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - chi_n/n$. Show that
begin{align*}
frac{1}{n log rho_n}
end{align*}
converges in distribution to $-frac {1}{chi}$.
For this question, I want to solve it by proving $frac{1}{n log rho_n}+frac {1}{chi_{n}}to0$ in probability, and therefore, $frac{1}{n log rho_n}to -frac {1}{chi}$ in distribution. This is because $-frac {1}{chi_{n}}to -frac {1}{chi}$ in distribution.
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac{1}{n log rho_n}+frac {1}{n(1-rho_{n})}\
&=frac {1}{n}(frac{1}{ log rho_n}+frac {1}{1-rho_{n}})\
&=frac {1}{n}left(frac {1}{log (1-frac {chi_{n}}{n})}+frac {n}{chi_{n}}right)
end{align*}
By Taylor expansion of $log (1-x)$ at $x=0$,
begin{align*}
log (1-x)=0-x+ldots
end{align*}
Therefore,
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac {1}{n}(-frac {n}{chi_{n}}+frac {n}{chi_{n}})\
&=0
end{align*}
The thing is I'm not sure that I can use the Taylor expansion here. I'm hoping for a better proof.
probability-theory probability-distributions
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
|
show 10 more comments
Suppose $chi_n to chi, ~~ chi sim mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - chi_n/n$. Show that
begin{align*}
frac{1}{n log rho_n}
end{align*}
converges in distribution to $-frac {1}{chi}$.
For this question, I want to solve it by proving $frac{1}{n log rho_n}+frac {1}{chi_{n}}to0$ in probability, and therefore, $frac{1}{n log rho_n}to -frac {1}{chi}$ in distribution. This is because $-frac {1}{chi_{n}}to -frac {1}{chi}$ in distribution.
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac{1}{n log rho_n}+frac {1}{n(1-rho_{n})}\
&=frac {1}{n}(frac{1}{ log rho_n}+frac {1}{1-rho_{n}})\
&=frac {1}{n}left(frac {1}{log (1-frac {chi_{n}}{n})}+frac {n}{chi_{n}}right)
end{align*}
By Taylor expansion of $log (1-x)$ at $x=0$,
begin{align*}
log (1-x)=0-x+ldots
end{align*}
Therefore,
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac {1}{n}(-frac {n}{chi_{n}}+frac {n}{chi_{n}})\
&=0
end{align*}
The thing is I'm not sure that I can use the Taylor expansion here. I'm hoping for a better proof.
probability-theory probability-distributions
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
What is the meaning of "$to0$" in your statement that $$frac{1}{n log rho_n}-frac {1}{chi_{n}}to0$$ which you say you want to prove?
– Did
Dec 9 '18 at 7:38
Later on, how do you explain the step $$frac{1}{ log rho_n}-frac {1}{1-rho_{n}}=-frac {2n}{chi_{n}} ?$$
– Did
Dec 9 '18 at 7:39
@Did $to 0$ means convergence in distribution to 0.
– Jiexiong687691
Dec 9 '18 at 8:31
@Did For that step, I've edited the question
– Jiexiong687691
Dec 9 '18 at 8:32
@Did I want to show that $frac {1}{nlog rho_{n}} to frac {1}{chi}$ in distribution. I want to use the theorem that suppose a sequence $(X_n)_{n geq 0}$ of random variables converge in distribution to a random variable $X$. Suppose further that another sequence $(Y_n)_{n geq 0}$ of random variables is such that $X_n - Y_n$ converges to zero in probability. Then $Y$ converges to $X$ in distribution.
– Jiexiong687691
Dec 9 '18 at 8:40
|
show 10 more comments
Suppose $chi_n to chi, ~~ chi sim mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - chi_n/n$. Show that
begin{align*}
frac{1}{n log rho_n}
end{align*}
converges in distribution to $-frac {1}{chi}$.
For this question, I want to solve it by proving $frac{1}{n log rho_n}+frac {1}{chi_{n}}to0$ in probability, and therefore, $frac{1}{n log rho_n}to -frac {1}{chi}$ in distribution. This is because $-frac {1}{chi_{n}}to -frac {1}{chi}$ in distribution.
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac{1}{n log rho_n}+frac {1}{n(1-rho_{n})}\
&=frac {1}{n}(frac{1}{ log rho_n}+frac {1}{1-rho_{n}})\
&=frac {1}{n}left(frac {1}{log (1-frac {chi_{n}}{n})}+frac {n}{chi_{n}}right)
end{align*}
By Taylor expansion of $log (1-x)$ at $x=0$,
begin{align*}
log (1-x)=0-x+ldots
end{align*}
Therefore,
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac {1}{n}(-frac {n}{chi_{n}}+frac {n}{chi_{n}})\
&=0
end{align*}
The thing is I'm not sure that I can use the Taylor expansion here. I'm hoping for a better proof.
probability-theory probability-distributions
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
Suppose $chi_n to chi, ~~ chi sim mathrm{exp}(1)$ in distribution, and set $rho_n = 1 - chi_n/n$. Show that
begin{align*}
frac{1}{n log rho_n}
end{align*}
converges in distribution to $-frac {1}{chi}$.
For this question, I want to solve it by proving $frac{1}{n log rho_n}+frac {1}{chi_{n}}to0$ in probability, and therefore, $frac{1}{n log rho_n}to -frac {1}{chi}$ in distribution. This is because $-frac {1}{chi_{n}}to -frac {1}{chi}$ in distribution.
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac{1}{n log rho_n}+frac {1}{n(1-rho_{n})}\
&=frac {1}{n}(frac{1}{ log rho_n}+frac {1}{1-rho_{n}})\
&=frac {1}{n}left(frac {1}{log (1-frac {chi_{n}}{n})}+frac {n}{chi_{n}}right)
end{align*}
By Taylor expansion of $log (1-x)$ at $x=0$,
begin{align*}
log (1-x)=0-x+ldots
end{align*}
Therefore,
begin{align*}
frac{1}{n log rho_n}+frac {1}{chi_{n}}&=frac {1}{n}(-frac {n}{chi_{n}}+frac {n}{chi_{n}})\
&=0
end{align*}
The thing is I'm not sure that I can use the Taylor expansion here. I'm hoping for a better proof.
probability-theory probability-distributions
probability-theory probability-distributions
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
edited Dec 13 '18 at 8:04
Jiexiong687691
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
asked Dec 9 '18 at 6:28
Jiexiong687691Jiexiong687691
705
705
What is the meaning of "$to0$" in your statement that $$frac{1}{n log rho_n}-frac {1}{chi_{n}}to0$$ which you say you want to prove?
– Did
Dec 9 '18 at 7:38
Later on, how do you explain the step $$frac{1}{ log rho_n}-frac {1}{1-rho_{n}}=-frac {2n}{chi_{n}} ?$$
– Did
Dec 9 '18 at 7:39
@Did $to 0$ means convergence in distribution to 0.
– Jiexiong687691
Dec 9 '18 at 8:31
@Did For that step, I've edited the question
– Jiexiong687691
Dec 9 '18 at 8:32
@Did I want to show that $frac {1}{nlog rho_{n}} to frac {1}{chi}$ in distribution. I want to use the theorem that suppose a sequence $(X_n)_{n geq 0}$ of random variables converge in distribution to a random variable $X$. Suppose further that another sequence $(Y_n)_{n geq 0}$ of random variables is such that $X_n - Y_n$ converges to zero in probability. Then $Y$ converges to $X$ in distribution.
– Jiexiong687691
Dec 9 '18 at 8:40
|
show 10 more comments
What is the meaning of "$to0$" in your statement that $$frac{1}{n log rho_n}-frac {1}{chi_{n}}to0$$ which you say you want to prove?
– Did
Dec 9 '18 at 7:38
Later on, how do you explain the step $$frac{1}{ log rho_n}-frac {1}{1-rho_{n}}=-frac {2n}{chi_{n}} ?$$
– Did
Dec 9 '18 at 7:39
@Did $to 0$ means convergence in distribution to 0.
– Jiexiong687691
Dec 9 '18 at 8:31
@Did For that step, I've edited the question
– Jiexiong687691
Dec 9 '18 at 8:32
@Did I want to show that $frac {1}{nlog rho_{n}} to frac {1}{chi}$ in distribution. I want to use the theorem that suppose a sequence $(X_n)_{n geq 0}$ of random variables converge in distribution to a random variable $X$. Suppose further that another sequence $(Y_n)_{n geq 0}$ of random variables is such that $X_n - Y_n$ converges to zero in probability. Then $Y$ converges to $X$ in distribution.
– Jiexiong687691
Dec 9 '18 at 8:40
What is the meaning of "$to0$" in your statement that $$frac{1}{n log rho_n}-frac {1}{chi_{n}}to0$$ which you say you want to prove?
– Did
Dec 9 '18 at 7:38
What is the meaning of "$to0$" in your statement that $$frac{1}{n log rho_n}-frac {1}{chi_{n}}to0$$ which you say you want to prove?
– Did
Dec 9 '18 at 7:38
Later on, how do you explain the step $$frac{1}{ log rho_n}-frac {1}{1-rho_{n}}=-frac {2n}{chi_{n}} ?$$
– Did
Dec 9 '18 at 7:39
Later on, how do you explain the step $$frac{1}{ log rho_n}-frac {1}{1-rho_{n}}=-frac {2n}{chi_{n}} ?$$
– Did
Dec 9 '18 at 7:39
@Did $to 0$ means convergence in distribution to 0.
– Jiexiong687691
Dec 9 '18 at 8:31
@Did $to 0$ means convergence in distribution to 0.
– Jiexiong687691
Dec 9 '18 at 8:31
@Did For that step, I've edited the question
– Jiexiong687691
Dec 9 '18 at 8:32
@Did For that step, I've edited the question
– Jiexiong687691
Dec 9 '18 at 8:32
@Did I want to show that $frac {1}{nlog rho_{n}} to frac {1}{chi}$ in distribution. I want to use the theorem that suppose a sequence $(X_n)_{n geq 0}$ of random variables converge in distribution to a random variable $X$. Suppose further that another sequence $(Y_n)_{n geq 0}$ of random variables is such that $X_n - Y_n$ converges to zero in probability. Then $Y$ converges to $X$ in distribution.
– Jiexiong687691
Dec 9 '18 at 8:40
@Did I want to show that $frac {1}{nlog rho_{n}} to frac {1}{chi}$ in distribution. I want to use the theorem that suppose a sequence $(X_n)_{n geq 0}$ of random variables converge in distribution to a random variable $X$. Suppose further that another sequence $(Y_n)_{n geq 0}$ of random variables is such that $X_n - Y_n$ converges to zero in probability. Then $Y$ converges to $X$ in distribution.
– Jiexiong687691
Dec 9 '18 at 8:40
|
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What is the meaning of "$to0$" in your statement that $$frac{1}{n log rho_n}-frac {1}{chi_{n}}to0$$ which you say you want to prove?
– Did
Dec 9 '18 at 7:38
Later on, how do you explain the step $$frac{1}{ log rho_n}-frac {1}{1-rho_{n}}=-frac {2n}{chi_{n}} ?$$
– Did
Dec 9 '18 at 7:39
@Did $to 0$ means convergence in distribution to 0.
– Jiexiong687691
Dec 9 '18 at 8:31
@Did For that step, I've edited the question
– Jiexiong687691
Dec 9 '18 at 8:32
@Did I want to show that $frac {1}{nlog rho_{n}} to frac {1}{chi}$ in distribution. I want to use the theorem that suppose a sequence $(X_n)_{n geq 0}$ of random variables converge in distribution to a random variable $X$. Suppose further that another sequence $(Y_n)_{n geq 0}$ of random variables is such that $X_n - Y_n$ converges to zero in probability. Then $Y$ converges to $X$ in distribution.
– Jiexiong687691
Dec 9 '18 at 8:40