Mixing
If two different statistics converge to a same function, does it mean both of them converge to a same...
$begingroup$
Let $A(x)$ be a (deterministic) function, and let $A_n(x)$ and $A_n^*(x)$ be two different estimators of $A(x)$. Define $B_n=sup|A_n-A|$ and $B_n^*=sup|A_n^*-A|$. If for a sequence $(c_n)$ and a distribution function $F$
- $B_n=o_pleft(frac{1}{c_n}right)$
- $limPr(c_n B_nleq x)=F(x)$
- $sup|A_n-A_n^*|=o_pleft(frac{1}{c_n}right)$
does it mean $limPr(c_n B_n^*leq x)=F(x)$?
My answer is "yes", but my proof is kind of strange. Since we have
begin{eqnarray*}
B_n^*leqsup|A_n^*-A_n|+sup|A_n-A|=o_pleft(frac{1}{c_n}right),
end{eqnarray*}
then it is abvious that $|B_n-B_n^*|=o_pleft(frac{1}{c_n}right)$. From this, if $n$ is large enough, then
begin{eqnarray*}
Pr(c_n B_n^*leq x)=Pr(c_n B_nleq x)rightarrow F(x).
end{eqnarray*}
I think, my intuition is right, but the proof is too heuristic and not firm. Do you guys have any idea what is my mistake and how to fix it? Thank you so much
probability sequences-and-series statistics convergence random-variables
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
$endgroup$
add a comment |
$begingroup$
Let $A(x)$ be a (deterministic) function, and let $A_n(x)$ and $A_n^*(x)$ be two different estimators of $A(x)$. Define $B_n=sup|A_n-A|$ and $B_n^*=sup|A_n^*-A|$. If for a sequence $(c_n)$ and a distribution function $F$
- $B_n=o_pleft(frac{1}{c_n}right)$
- $limPr(c_n B_nleq x)=F(x)$
- $sup|A_n-A_n^*|=o_pleft(frac{1}{c_n}right)$
does it mean $limPr(c_n B_n^*leq x)=F(x)$?
My answer is "yes", but my proof is kind of strange. Since we have
begin{eqnarray*}
B_n^*leqsup|A_n^*-A_n|+sup|A_n-A|=o_pleft(frac{1}{c_n}right),
end{eqnarray*}
then it is abvious that $|B_n-B_n^*|=o_pleft(frac{1}{c_n}right)$. From this, if $n$ is large enough, then
begin{eqnarray*}
Pr(c_n B_n^*leq x)=Pr(c_n B_nleq x)rightarrow F(x).
end{eqnarray*}
I think, my intuition is right, but the proof is too heuristic and not firm. Do you guys have any idea what is my mistake and how to fix it? Thank you so much
probability sequences-and-series statistics convergence random-variables
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
$endgroup$
add a comment |
$begingroup$
Let $A(x)$ be a (deterministic) function, and let $A_n(x)$ and $A_n^*(x)$ be two different estimators of $A(x)$. Define $B_n=sup|A_n-A|$ and $B_n^*=sup|A_n^*-A|$. If for a sequence $(c_n)$ and a distribution function $F$
- $B_n=o_pleft(frac{1}{c_n}right)$
- $limPr(c_n B_nleq x)=F(x)$
- $sup|A_n-A_n^*|=o_pleft(frac{1}{c_n}right)$
does it mean $limPr(c_n B_n^*leq x)=F(x)$?
My answer is "yes", but my proof is kind of strange. Since we have
begin{eqnarray*}
B_n^*leqsup|A_n^*-A_n|+sup|A_n-A|=o_pleft(frac{1}{c_n}right),
end{eqnarray*}
then it is abvious that $|B_n-B_n^*|=o_pleft(frac{1}{c_n}right)$. From this, if $n$ is large enough, then
begin{eqnarray*}
Pr(c_n B_n^*leq x)=Pr(c_n B_nleq x)rightarrow F(x).
end{eqnarray*}
I think, my intuition is right, but the proof is too heuristic and not firm. Do you guys have any idea what is my mistake and how to fix it? Thank you so much
probability sequences-and-series statistics convergence random-variables
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
$endgroup$
Let $A(x)$ be a (deterministic) function, and let $A_n(x)$ and $A_n^*(x)$ be two different estimators of $A(x)$. Define $B_n=sup|A_n-A|$ and $B_n^*=sup|A_n^*-A|$. If for a sequence $(c_n)$ and a distribution function $F$
- $B_n=o_pleft(frac{1}{c_n}right)$
- $limPr(c_n B_nleq x)=F(x)$
- $sup|A_n-A_n^*|=o_pleft(frac{1}{c_n}right)$
does it mean $limPr(c_n B_n^*leq x)=F(x)$?
My answer is "yes", but my proof is kind of strange. Since we have
begin{eqnarray*}
B_n^*leqsup|A_n^*-A_n|+sup|A_n-A|=o_pleft(frac{1}{c_n}right),
end{eqnarray*}
then it is abvious that $|B_n-B_n^*|=o_pleft(frac{1}{c_n}right)$. From this, if $n$ is large enough, then
begin{eqnarray*}
Pr(c_n B_n^*leq x)=Pr(c_n B_nleq x)rightarrow F(x).
end{eqnarray*}
I think, my intuition is right, but the proof is too heuristic and not firm. Do you guys have any idea what is my mistake and how to fix it? Thank you so much
probability sequences-and-series statistics convergence random-variables
probability sequences-and-series statistics convergence random-variables
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
asked Jan 23 at 7:27
Rizky Reza FujisakiRizky Reza Fujisaki
44429
44429
add a comment |
add a comment |
1 Answer
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$begingroup$
If $|B_n-B_n^{*}| = O_p left( frac{1}{c_n} right)$ then $|c_nB_n-c_nB_n^{*}| = O_p left( 1 right)$. Since $|c_nB_n-c_nB_n^{*}|$ does not converge in probability to zero for large $n$, you cannot conclude that $Pr(c_nB_n^{*}leq x)$ converges towards $Pr(c_nB_nleq x)$.
Wikipedia defines the notation $o_p$ as follows:
$X_n=o_p(a_n) text{ if } lim_{nrightarrow infty}mathrm{Pr}left(left|frac{X_n}{a_n}right|geq epsilonright)=0$.
With this notation it can be deduced that $F(x)=0$ when $x<0$, that $F(x)=1$ when $x>0$ and that $mathrm{Pr}(c_nB_n*leq x)=F(x)$ when $x ne 0$. However there exists a counterexample in the case when $x=0$.
Let $A(x)=0$, $A_n(x)=1$, $A_n^* (x)=0$ and $c_n=frac{1}{n}$. It can be verified that all three conditions hold. Note that $c_nB_n=frac{1}{n}$ and hence that $F(0)=lim_{nrightarrowinfty}mathrm{Pr}(frac{1}{n}leq 0)=0$. On the other hand, $mathrm{Pr}(c_nB_n^* leq 0)=1$ since $B_n^*=0$. It follows that $lim_{nrightarrowinfty}mathrm{Pr}(c_nB_n^*leq 0)=1 neq F(0)$.
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
$begingroup$
Thank you so much for answering, but, it is not big-O, it's $|B_n-B_n^*|=o_p(c_n^{-1})$, little-o, so $|c_n B_n-c_n B_n^*|=o_p(1)$, and it will converge to $0$ in probability, so is my argument valid?
$endgroup$
– Rizky Reza Fujisaki
Jan 24 at 2:21
add a comment |
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$begingroup$
If $|B_n-B_n^{*}| = O_p left( frac{1}{c_n} right)$ then $|c_nB_n-c_nB_n^{*}| = O_p left( 1 right)$. Since $|c_nB_n-c_nB_n^{*}|$ does not converge in probability to zero for large $n$, you cannot conclude that $Pr(c_nB_n^{*}leq x)$ converges towards $Pr(c_nB_nleq x)$.
Wikipedia defines the notation $o_p$ as follows:
$X_n=o_p(a_n) text{ if } lim_{nrightarrow infty}mathrm{Pr}left(left|frac{X_n}{a_n}right|geq epsilonright)=0$.
With this notation it can be deduced that $F(x)=0$ when $x<0$, that $F(x)=1$ when $x>0$ and that $mathrm{Pr}(c_nB_n*leq x)=F(x)$ when $x ne 0$. However there exists a counterexample in the case when $x=0$.
Let $A(x)=0$, $A_n(x)=1$, $A_n^* (x)=0$ and $c_n=frac{1}{n}$. It can be verified that all three conditions hold. Note that $c_nB_n=frac{1}{n}$ and hence that $F(0)=lim_{nrightarrowinfty}mathrm{Pr}(frac{1}{n}leq 0)=0$. On the other hand, $mathrm{Pr}(c_nB_n^* leq 0)=1$ since $B_n^*=0$. It follows that $lim_{nrightarrowinfty}mathrm{Pr}(c_nB_n^*leq 0)=1 neq F(0)$.
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
$begingroup$
Thank you so much for answering, but, it is not big-O, it's $|B_n-B_n^*|=o_p(c_n^{-1})$, little-o, so $|c_n B_n-c_n B_n^*|=o_p(1)$, and it will converge to $0$ in probability, so is my argument valid?
$endgroup$
– Rizky Reza Fujisaki
Jan 24 at 2:21
add a comment |
$begingroup$
If $|B_n-B_n^{*}| = O_p left( frac{1}{c_n} right)$ then $|c_nB_n-c_nB_n^{*}| = O_p left( 1 right)$. Since $|c_nB_n-c_nB_n^{*}|$ does not converge in probability to zero for large $n$, you cannot conclude that $Pr(c_nB_n^{*}leq x)$ converges towards $Pr(c_nB_nleq x)$.
Wikipedia defines the notation $o_p$ as follows:
$X_n=o_p(a_n) text{ if } lim_{nrightarrow infty}mathrm{Pr}left(left|frac{X_n}{a_n}right|geq epsilonright)=0$.
With this notation it can be deduced that $F(x)=0$ when $x<0$, that $F(x)=1$ when $x>0$ and that $mathrm{Pr}(c_nB_n*leq x)=F(x)$ when $x ne 0$. However there exists a counterexample in the case when $x=0$.
Let $A(x)=0$, $A_n(x)=1$, $A_n^* (x)=0$ and $c_n=frac{1}{n}$. It can be verified that all three conditions hold. Note that $c_nB_n=frac{1}{n}$ and hence that $F(0)=lim_{nrightarrowinfty}mathrm{Pr}(frac{1}{n}leq 0)=0$. On the other hand, $mathrm{Pr}(c_nB_n^* leq 0)=1$ since $B_n^*=0$. It follows that $lim_{nrightarrowinfty}mathrm{Pr}(c_nB_n^*leq 0)=1 neq F(0)$.
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
$begingroup$
Thank you so much for answering, but, it is not big-O, it's $|B_n-B_n^*|=o_p(c_n^{-1})$, little-o, so $|c_n B_n-c_n B_n^*|=o_p(1)$, and it will converge to $0$ in probability, so is my argument valid?
$endgroup$
– Rizky Reza Fujisaki
Jan 24 at 2:21
add a comment |
$begingroup$
If $|B_n-B_n^{*}| = O_p left( frac{1}{c_n} right)$ then $|c_nB_n-c_nB_n^{*}| = O_p left( 1 right)$. Since $|c_nB_n-c_nB_n^{*}|$ does not converge in probability to zero for large $n$, you cannot conclude that $Pr(c_nB_n^{*}leq x)$ converges towards $Pr(c_nB_nleq x)$.
Wikipedia defines the notation $o_p$ as follows:
$X_n=o_p(a_n) text{ if } lim_{nrightarrow infty}mathrm{Pr}left(left|frac{X_n}{a_n}right|geq epsilonright)=0$.
With this notation it can be deduced that $F(x)=0$ when $x<0$, that $F(x)=1$ when $x>0$ and that $mathrm{Pr}(c_nB_n*leq x)=F(x)$ when $x ne 0$. However there exists a counterexample in the case when $x=0$.
Let $A(x)=0$, $A_n(x)=1$, $A_n^* (x)=0$ and $c_n=frac{1}{n}$. It can be verified that all three conditions hold. Note that $c_nB_n=frac{1}{n}$ and hence that $F(0)=lim_{nrightarrowinfty}mathrm{Pr}(frac{1}{n}leq 0)=0$. On the other hand, $mathrm{Pr}(c_nB_n^* leq 0)=1$ since $B_n^*=0$. It follows that $lim_{nrightarrowinfty}mathrm{Pr}(c_nB_n^*leq 0)=1 neq F(0)$.
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
$endgroup$
If $|B_n-B_n^{*}| = O_p left( frac{1}{c_n} right)$ then $|c_nB_n-c_nB_n^{*}| = O_p left( 1 right)$. Since $|c_nB_n-c_nB_n^{*}|$ does not converge in probability to zero for large $n$, you cannot conclude that $Pr(c_nB_n^{*}leq x)$ converges towards $Pr(c_nB_nleq x)$.
Wikipedia defines the notation $o_p$ as follows:
$X_n=o_p(a_n) text{ if } lim_{nrightarrow infty}mathrm{Pr}left(left|frac{X_n}{a_n}right|geq epsilonright)=0$.
With this notation it can be deduced that $F(x)=0$ when $x<0$, that $F(x)=1$ when $x>0$ and that $mathrm{Pr}(c_nB_n*leq x)=F(x)$ when $x ne 0$. However there exists a counterexample in the case when $x=0$.
Let $A(x)=0$, $A_n(x)=1$, $A_n^* (x)=0$ and $c_n=frac{1}{n}$. It can be verified that all three conditions hold. Note that $c_nB_n=frac{1}{n}$ and hence that $F(0)=lim_{nrightarrowinfty}mathrm{Pr}(frac{1}{n}leq 0)=0$. On the other hand, $mathrm{Pr}(c_nB_n^* leq 0)=1$ since $B_n^*=0$. It follows that $lim_{nrightarrowinfty}mathrm{Pr}(c_nB_n^*leq 0)=1 neq F(0)$.
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
edited Jan 24 at 17:18
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
answered Jan 23 at 8:38
Angela RichardsonAngela Richardson
5,40411734
5,40411734
$begingroup$
Thank you so much for answering, but, it is not big-O, it's $|B_n-B_n^*|=o_p(c_n^{-1})$, little-o, so $|c_n B_n-c_n B_n^*|=o_p(1)$, and it will converge to $0$ in probability, so is my argument valid?
$endgroup$
– Rizky Reza Fujisaki
Jan 24 at 2:21
add a comment |
$begingroup$
Thank you so much for answering, but, it is not big-O, it's $|B_n-B_n^*|=o_p(c_n^{-1})$, little-o, so $|c_n B_n-c_n B_n^*|=o_p(1)$, and it will converge to $0$ in probability, so is my argument valid?
$endgroup$
– Rizky Reza Fujisaki
Jan 24 at 2:21
$begingroup$
Thank you so much for answering, but, it is not big-O, it's $|B_n-B_n^*|=o_p(c_n^{-1})$, little-o, so $|c_n B_n-c_n B_n^*|=o_p(1)$, and it will converge to $0$ in probability, so is my argument valid?
$endgroup$
– Rizky Reza Fujisaki
Jan 24 at 2:21
$begingroup$
Thank you so much for answering, but, it is not big-O, it's $|B_n-B_n^*|=o_p(c_n^{-1})$, little-o, so $|c_n B_n-c_n B_n^*|=o_p(1)$, and it will converge to $0$ in probability, so is my argument valid?
$endgroup$
– Rizky Reza Fujisaki
Jan 24 at 2:21
add a comment |
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