Mixing
Prove that the maximum of $n$ independent standard normal random variables, is asymptotically equivalent to...
$begingroup$
Lets $(X_n)_{ninmathbb{N}}$ be an iid sequence of standard normal random variables. Define $$M_n=max_{1leq ileq n} X_i.$$
Prove that $$lim_{nrightarrowinfty} frac{M_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used the fact that $$left(frac{1}{x}-frac{1}{x^3}right)e^{-frac{x^2}{2}}leqmathbb P(X_n>x)leq frac{1}{x}e^{-frac{x^2}{2}},$$
and the Borel Cantelli lemmas to prove that
$$limsup_{nrightarrowinfty} frac{X_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used Davide Giraudo's comment to show $$limsup_n frac{M_n}{sqrt{2log n}}=1quad text{a.s.}$$
I have no idea how to compute the $liminf$. Borel-Cantelli give us tools to compute the $limsup$ of sets, I am unsure of how to argue almost sure convergence. Any help would be appreciated.
probability probability-theory probability-distributions probability-limit-theorems
edited Aug 6 '16 at 19:03
Michael Hardy
1
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
$endgroup$
add a comment |
$begingroup$
Lets $(X_n)_{ninmathbb{N}}$ be an iid sequence of standard normal random variables. Define $$M_n=max_{1leq ileq n} X_i.$$
Prove that $$lim_{nrightarrowinfty} frac{M_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used the fact that $$left(frac{1}{x}-frac{1}{x^3}right)e^{-frac{x^2}{2}}leqmathbb P(X_n>x)leq frac{1}{x}e^{-frac{x^2}{2}},$$
and the Borel Cantelli lemmas to prove that
$$limsup_{nrightarrowinfty} frac{X_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used Davide Giraudo's comment to show $$limsup_n frac{M_n}{sqrt{2log n}}=1quad text{a.s.}$$
I have no idea how to compute the $liminf$. Borel-Cantelli give us tools to compute the $limsup$ of sets, I am unsure of how to argue almost sure convergence. Any help would be appreciated.
probability probability-theory probability-distributions probability-limit-theorems
edited Aug 6 '16 at 19:03
Michael Hardy
1
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
$endgroup$
add a comment |
$begingroup$
Lets $(X_n)_{ninmathbb{N}}$ be an iid sequence of standard normal random variables. Define $$M_n=max_{1leq ileq n} X_i.$$
Prove that $$lim_{nrightarrowinfty} frac{M_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used the fact that $$left(frac{1}{x}-frac{1}{x^3}right)e^{-frac{x^2}{2}}leqmathbb P(X_n>x)leq frac{1}{x}e^{-frac{x^2}{2}},$$
and the Borel Cantelli lemmas to prove that
$$limsup_{nrightarrowinfty} frac{X_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used Davide Giraudo's comment to show $$limsup_n frac{M_n}{sqrt{2log n}}=1quad text{a.s.}$$
I have no idea how to compute the $liminf$. Borel-Cantelli give us tools to compute the $limsup$ of sets, I am unsure of how to argue almost sure convergence. Any help would be appreciated.
probability probability-theory probability-distributions probability-limit-theorems
edited Aug 6 '16 at 19:03
Michael Hardy
1
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
$endgroup$
Lets $(X_n)_{ninmathbb{N}}$ be an iid sequence of standard normal random variables. Define $$M_n=max_{1leq ileq n} X_i.$$
Prove that $$lim_{nrightarrowinfty} frac{M_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used the fact that $$left(frac{1}{x}-frac{1}{x^3}right)e^{-frac{x^2}{2}}leqmathbb P(X_n>x)leq frac{1}{x}e^{-frac{x^2}{2}},$$
and the Borel Cantelli lemmas to prove that
$$limsup_{nrightarrowinfty} frac{X_n}{sqrt{2log n}}=1quadtext{a.s.}$$
I used Davide Giraudo's comment to show $$limsup_n frac{M_n}{sqrt{2log n}}=1quad text{a.s.}$$
I have no idea how to compute the $liminf$. Borel-Cantelli give us tools to compute the $limsup$ of sets, I am unsure of how to argue almost sure convergence. Any help would be appreciated.
probability probability-theory probability-distributions probability-limit-theorems
probability probability-theory probability-distributions probability-limit-theorems
edited Aug 6 '16 at 19:03
Michael Hardy
1
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
edited Aug 6 '16 at 19:03
Michael Hardy
1
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
edited Aug 6 '16 at 19:03
Michael Hardy
1
edited Aug 6 '16 at 19:03
Michael Hardy
1
edited Aug 6 '16 at 19:03
Michael Hardy
1
1
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
asked Oct 7 '14 at 3:32
anonymousanonymous
1206
1206
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your estimate gives that for each positive $varepsilon$, we have
$$sum_imathbb P(M_{2^{i}}>(1+varepsilon)sqrt 2sqrt{i+1})<infty$$
(we use the fact that $mathbb P(M_n>x)leqslant nmathbb P(X_1gt x)$).
We thus deduce that $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1+varepsilon$ almost surely. Taking $varepsilon:=1/k$, we get $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1$ almost surely. To conclude, notice that if $2^ileqslant nlt 2^{i+1}$,
$$frac{M_n}{sqrt{2log n}}leqslant frac{M_{2^{i+1}}}{sqrt{2i}}.$$
For the $liminf$, define for a fixed positive $varepsilon$, $$A_n:=left{frac{M_n}{sqrt{2log n}}lt 1-varepsilonright}.$$
A use of the estimate on the tail of the normal distribution show that the series $sum_n mathbb P(A_n)$ is convergent, hence by the Borel-Cantelli lemma, we have $mathbb Pleft(limsup_n A_nright)=0$. This means that for almost every $omega$, we can find $N=N(omega)$ such that if $ngeqslant N(omega)$, then $$frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon.$$
This implies that $$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon quad mbox{a.e.}$$
Here again, taking $varepsilon:=1/k$, we obtain that
$$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1quad mbox{a.e.}$$
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
So I am mainly having trouble with justifying the first sum converges. I get $$mathbb{P}(M_{2^i}>sqrt{(1+epsilon)2(i+1)})=1-(1-mathbb{P}(X_{2^i}>sqrt{(1+epsilon)2(i+1)}))^{2^i} leq 1-(1-frac{1}{sqrt{(1+epsilon)2(i+1)}}e^{-(1+epsilon)(i+1)})^{2^i}$$
$endgroup$
– anonymous
Oct 7 '14 at 10:15
$begingroup$
I edited. Is it clearer?
$endgroup$
– Davide Giraudo
Oct 7 '14 at 10:18
$begingroup$
yes significantly. Thank you. So that shows that $limsup$ of $frac{M_n}{sqrt{2log n}}leq 1$, but not the $liminfgeq1$.
$endgroup$
– anonymous
Oct 7 '14 at 10:27
$begingroup$
It may be not the case that $liminfgeq 1$, however the limsup is greater than $1$ because $M_ngeq X_n$.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:10
$begingroup$
Sorry, I missed the fact that you have a limit and not only a limsup.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:22
|
show 4 more comments
$begingroup$
Not an answer, but a related comment that is too long for a comment box ...
This question made me curious to compare:
- the pdf of the sample maximum (given a sample of size $n$ drawn on a N(0,1) parent), say $f(x;n):$
$$f(x) = frac{2^{frac{1}{2}-n} n e^{-frac{x^2}{2}} left(1+text{erf}left(frac{x}{sqrt{2}}right)right)^{n-1}}{sqrt{pi }}$$
where erf(.) denotes the error function, to
- the asymptote proposed by the question $sqrt{2 log n}$
... when $n$ is very large (say $n$ = 1 million).
The diagram compares:
The distribution does not, in fact, converge to the asymptote in any meaningful manner, even when $n$ is very large, such as $n$ = 1 million or 100 million or even a billion.
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
$endgroup$
$begingroup$
Are you using the right logarithm? (I.e. base $e$ and not base 10?)
$endgroup$
– Chill2Macht
Oct 1 '18 at 0:13
1
$begingroup$
Yes - $sqrt{2 log left(10^6right)}$ = 5.25 ... as shown in the diagram above (the red line). By contrast, with base 10, the same calculation would be equal to $2 sqrt{3}$ which is approx 3.46.
$endgroup$
– wolfies
Oct 1 '18 at 7:17
add a comment |
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2 Answers
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votes
2 Answers
2
active
oldest
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active
oldest
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$begingroup$
Your estimate gives that for each positive $varepsilon$, we have
$$sum_imathbb P(M_{2^{i}}>(1+varepsilon)sqrt 2sqrt{i+1})<infty$$
(we use the fact that $mathbb P(M_n>x)leqslant nmathbb P(X_1gt x)$).
We thus deduce that $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1+varepsilon$ almost surely. Taking $varepsilon:=1/k$, we get $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1$ almost surely. To conclude, notice that if $2^ileqslant nlt 2^{i+1}$,
$$frac{M_n}{sqrt{2log n}}leqslant frac{M_{2^{i+1}}}{sqrt{2i}}.$$
For the $liminf$, define for a fixed positive $varepsilon$, $$A_n:=left{frac{M_n}{sqrt{2log n}}lt 1-varepsilonright}.$$
A use of the estimate on the tail of the normal distribution show that the series $sum_n mathbb P(A_n)$ is convergent, hence by the Borel-Cantelli lemma, we have $mathbb Pleft(limsup_n A_nright)=0$. This means that for almost every $omega$, we can find $N=N(omega)$ such that if $ngeqslant N(omega)$, then $$frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon.$$
This implies that $$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon quad mbox{a.e.}$$
Here again, taking $varepsilon:=1/k$, we obtain that
$$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1quad mbox{a.e.}$$
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
So I am mainly having trouble with justifying the first sum converges. I get $$mathbb{P}(M_{2^i}>sqrt{(1+epsilon)2(i+1)})=1-(1-mathbb{P}(X_{2^i}>sqrt{(1+epsilon)2(i+1)}))^{2^i} leq 1-(1-frac{1}{sqrt{(1+epsilon)2(i+1)}}e^{-(1+epsilon)(i+1)})^{2^i}$$
$endgroup$
– anonymous
Oct 7 '14 at 10:15
$begingroup$
I edited. Is it clearer?
$endgroup$
– Davide Giraudo
Oct 7 '14 at 10:18
$begingroup$
yes significantly. Thank you. So that shows that $limsup$ of $frac{M_n}{sqrt{2log n}}leq 1$, but not the $liminfgeq1$.
$endgroup$
– anonymous
Oct 7 '14 at 10:27
$begingroup$
It may be not the case that $liminfgeq 1$, however the limsup is greater than $1$ because $M_ngeq X_n$.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:10
$begingroup$
Sorry, I missed the fact that you have a limit and not only a limsup.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:22
|
show 4 more comments
$begingroup$
Your estimate gives that for each positive $varepsilon$, we have
$$sum_imathbb P(M_{2^{i}}>(1+varepsilon)sqrt 2sqrt{i+1})<infty$$
(we use the fact that $mathbb P(M_n>x)leqslant nmathbb P(X_1gt x)$).
We thus deduce that $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1+varepsilon$ almost surely. Taking $varepsilon:=1/k$, we get $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1$ almost surely. To conclude, notice that if $2^ileqslant nlt 2^{i+1}$,
$$frac{M_n}{sqrt{2log n}}leqslant frac{M_{2^{i+1}}}{sqrt{2i}}.$$
For the $liminf$, define for a fixed positive $varepsilon$, $$A_n:=left{frac{M_n}{sqrt{2log n}}lt 1-varepsilonright}.$$
A use of the estimate on the tail of the normal distribution show that the series $sum_n mathbb P(A_n)$ is convergent, hence by the Borel-Cantelli lemma, we have $mathbb Pleft(limsup_n A_nright)=0$. This means that for almost every $omega$, we can find $N=N(omega)$ such that if $ngeqslant N(omega)$, then $$frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon.$$
This implies that $$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon quad mbox{a.e.}$$
Here again, taking $varepsilon:=1/k$, we obtain that
$$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1quad mbox{a.e.}$$
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
So I am mainly having trouble with justifying the first sum converges. I get $$mathbb{P}(M_{2^i}>sqrt{(1+epsilon)2(i+1)})=1-(1-mathbb{P}(X_{2^i}>sqrt{(1+epsilon)2(i+1)}))^{2^i} leq 1-(1-frac{1}{sqrt{(1+epsilon)2(i+1)}}e^{-(1+epsilon)(i+1)})^{2^i}$$
$endgroup$
– anonymous
Oct 7 '14 at 10:15
$begingroup$
I edited. Is it clearer?
$endgroup$
– Davide Giraudo
Oct 7 '14 at 10:18
$begingroup$
yes significantly. Thank you. So that shows that $limsup$ of $frac{M_n}{sqrt{2log n}}leq 1$, but not the $liminfgeq1$.
$endgroup$
– anonymous
Oct 7 '14 at 10:27
$begingroup$
It may be not the case that $liminfgeq 1$, however the limsup is greater than $1$ because $M_ngeq X_n$.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:10
$begingroup$
Sorry, I missed the fact that you have a limit and not only a limsup.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:22
|
show 4 more comments
$begingroup$
Your estimate gives that for each positive $varepsilon$, we have
$$sum_imathbb P(M_{2^{i}}>(1+varepsilon)sqrt 2sqrt{i+1})<infty$$
(we use the fact that $mathbb P(M_n>x)leqslant nmathbb P(X_1gt x)$).
We thus deduce that $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1+varepsilon$ almost surely. Taking $varepsilon:=1/k$, we get $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1$ almost surely. To conclude, notice that if $2^ileqslant nlt 2^{i+1}$,
$$frac{M_n}{sqrt{2log n}}leqslant frac{M_{2^{i+1}}}{sqrt{2i}}.$$
For the $liminf$, define for a fixed positive $varepsilon$, $$A_n:=left{frac{M_n}{sqrt{2log n}}lt 1-varepsilonright}.$$
A use of the estimate on the tail of the normal distribution show that the series $sum_n mathbb P(A_n)$ is convergent, hence by the Borel-Cantelli lemma, we have $mathbb Pleft(limsup_n A_nright)=0$. This means that for almost every $omega$, we can find $N=N(omega)$ such that if $ngeqslant N(omega)$, then $$frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon.$$
This implies that $$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon quad mbox{a.e.}$$
Here again, taking $varepsilon:=1/k$, we obtain that
$$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1quad mbox{a.e.}$$
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
Your estimate gives that for each positive $varepsilon$, we have
$$sum_imathbb P(M_{2^{i}}>(1+varepsilon)sqrt 2sqrt{i+1})<infty$$
(we use the fact that $mathbb P(M_n>x)leqslant nmathbb P(X_1gt x)$).
We thus deduce that $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1+varepsilon$ almost surely. Taking $varepsilon:=1/k$, we get $limsup_iM_{2^{i}}/(sqrt{2(i+1)})leqslant 1$ almost surely. To conclude, notice that if $2^ileqslant nlt 2^{i+1}$,
$$frac{M_n}{sqrt{2log n}}leqslant frac{M_{2^{i+1}}}{sqrt{2i}}.$$
For the $liminf$, define for a fixed positive $varepsilon$, $$A_n:=left{frac{M_n}{sqrt{2log n}}lt 1-varepsilonright}.$$
A use of the estimate on the tail of the normal distribution show that the series $sum_n mathbb P(A_n)$ is convergent, hence by the Borel-Cantelli lemma, we have $mathbb Pleft(limsup_n A_nright)=0$. This means that for almost every $omega$, we can find $N=N(omega)$ such that if $ngeqslant N(omega)$, then $$frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon.$$
This implies that $$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1-varepsilon quad mbox{a.e.}$$
Here again, taking $varepsilon:=1/k$, we obtain that
$$liminf_{nto infty}frac{M_n(omega)}{sqrt{2log n}}geqslant 1quad mbox{a.e.}$$
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
edited Oct 8 '14 at 13:36
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
answered Oct 7 '14 at 9:29
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
$begingroup$
So I am mainly having trouble with justifying the first sum converges. I get $$mathbb{P}(M_{2^i}>sqrt{(1+epsilon)2(i+1)})=1-(1-mathbb{P}(X_{2^i}>sqrt{(1+epsilon)2(i+1)}))^{2^i} leq 1-(1-frac{1}{sqrt{(1+epsilon)2(i+1)}}e^{-(1+epsilon)(i+1)})^{2^i}$$
$endgroup$
– anonymous
Oct 7 '14 at 10:15
$begingroup$
I edited. Is it clearer?
$endgroup$
– Davide Giraudo
Oct 7 '14 at 10:18
$begingroup$
yes significantly. Thank you. So that shows that $limsup$ of $frac{M_n}{sqrt{2log n}}leq 1$, but not the $liminfgeq1$.
$endgroup$
– anonymous
Oct 7 '14 at 10:27
$begingroup$
It may be not the case that $liminfgeq 1$, however the limsup is greater than $1$ because $M_ngeq X_n$.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:10
$begingroup$
Sorry, I missed the fact that you have a limit and not only a limsup.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:22
|
show 4 more comments
$begingroup$
So I am mainly having trouble with justifying the first sum converges. I get $$mathbb{P}(M_{2^i}>sqrt{(1+epsilon)2(i+1)})=1-(1-mathbb{P}(X_{2^i}>sqrt{(1+epsilon)2(i+1)}))^{2^i} leq 1-(1-frac{1}{sqrt{(1+epsilon)2(i+1)}}e^{-(1+epsilon)(i+1)})^{2^i}$$
$endgroup$
– anonymous
Oct 7 '14 at 10:15
$begingroup$
I edited. Is it clearer?
$endgroup$
– Davide Giraudo
Oct 7 '14 at 10:18
$begingroup$
yes significantly. Thank you. So that shows that $limsup$ of $frac{M_n}{sqrt{2log n}}leq 1$, but not the $liminfgeq1$.
$endgroup$
– anonymous
Oct 7 '14 at 10:27
$begingroup$
It may be not the case that $liminfgeq 1$, however the limsup is greater than $1$ because $M_ngeq X_n$.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:10
$begingroup$
Sorry, I missed the fact that you have a limit and not only a limsup.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:22
$begingroup$
So I am mainly having trouble with justifying the first sum converges. I get $$mathbb{P}(M_{2^i}>sqrt{(1+epsilon)2(i+1)})=1-(1-mathbb{P}(X_{2^i}>sqrt{(1+epsilon)2(i+1)}))^{2^i} leq 1-(1-frac{1}{sqrt{(1+epsilon)2(i+1)}}e^{-(1+epsilon)(i+1)})^{2^i}$$
$endgroup$
– anonymous
Oct 7 '14 at 10:15
$begingroup$
So I am mainly having trouble with justifying the first sum converges. I get $$mathbb{P}(M_{2^i}>sqrt{(1+epsilon)2(i+1)})=1-(1-mathbb{P}(X_{2^i}>sqrt{(1+epsilon)2(i+1)}))^{2^i} leq 1-(1-frac{1}{sqrt{(1+epsilon)2(i+1)}}e^{-(1+epsilon)(i+1)})^{2^i}$$
$endgroup$
– anonymous
Oct 7 '14 at 10:15
$begingroup$
I edited. Is it clearer?
$endgroup$
– Davide Giraudo
Oct 7 '14 at 10:18
$begingroup$
I edited. Is it clearer?
$endgroup$
– Davide Giraudo
Oct 7 '14 at 10:18
$begingroup$
yes significantly. Thank you. So that shows that $limsup$ of $frac{M_n}{sqrt{2log n}}leq 1$, but not the $liminfgeq1$.
$endgroup$
– anonymous
Oct 7 '14 at 10:27
$begingroup$
yes significantly. Thank you. So that shows that $limsup$ of $frac{M_n}{sqrt{2log n}}leq 1$, but not the $liminfgeq1$.
$endgroup$
– anonymous
Oct 7 '14 at 10:27
$begingroup$
It may be not the case that $liminfgeq 1$, however the limsup is greater than $1$ because $M_ngeq X_n$.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:10
$begingroup$
It may be not the case that $liminfgeq 1$, however the limsup is greater than $1$ because $M_ngeq X_n$.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:10
$begingroup$
Sorry, I missed the fact that you have a limit and not only a limsup.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:22
$begingroup$
Sorry, I missed the fact that you have a limit and not only a limsup.
$endgroup$
– Davide Giraudo
Oct 7 '14 at 11:22
|
show 4 more comments
$begingroup$
Not an answer, but a related comment that is too long for a comment box ...
This question made me curious to compare:
- the pdf of the sample maximum (given a sample of size $n$ drawn on a N(0,1) parent), say $f(x;n):$
$$f(x) = frac{2^{frac{1}{2}-n} n e^{-frac{x^2}{2}} left(1+text{erf}left(frac{x}{sqrt{2}}right)right)^{n-1}}{sqrt{pi }}$$
where erf(.) denotes the error function, to
- the asymptote proposed by the question $sqrt{2 log n}$
... when $n$ is very large (say $n$ = 1 million).
The diagram compares:
The distribution does not, in fact, converge to the asymptote in any meaningful manner, even when $n$ is very large, such as $n$ = 1 million or 100 million or even a billion.
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
$endgroup$
$begingroup$
Are you using the right logarithm? (I.e. base $e$ and not base 10?)
$endgroup$
– Chill2Macht
Oct 1 '18 at 0:13
1
$begingroup$
Yes - $sqrt{2 log left(10^6right)}$ = 5.25 ... as shown in the diagram above (the red line). By contrast, with base 10, the same calculation would be equal to $2 sqrt{3}$ which is approx 3.46.
$endgroup$
– wolfies
Oct 1 '18 at 7:17
add a comment |
$begingroup$
Not an answer, but a related comment that is too long for a comment box ...
This question made me curious to compare:
- the pdf of the sample maximum (given a sample of size $n$ drawn on a N(0,1) parent), say $f(x;n):$
$$f(x) = frac{2^{frac{1}{2}-n} n e^{-frac{x^2}{2}} left(1+text{erf}left(frac{x}{sqrt{2}}right)right)^{n-1}}{sqrt{pi }}$$
where erf(.) denotes the error function, to
- the asymptote proposed by the question $sqrt{2 log n}$
... when $n$ is very large (say $n$ = 1 million).
The diagram compares:
The distribution does not, in fact, converge to the asymptote in any meaningful manner, even when $n$ is very large, such as $n$ = 1 million or 100 million or even a billion.
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
$endgroup$
$begingroup$
Are you using the right logarithm? (I.e. base $e$ and not base 10?)
$endgroup$
– Chill2Macht
Oct 1 '18 at 0:13
1
$begingroup$
Yes - $sqrt{2 log left(10^6right)}$ = 5.25 ... as shown in the diagram above (the red line). By contrast, with base 10, the same calculation would be equal to $2 sqrt{3}$ which is approx 3.46.
$endgroup$
– wolfies
Oct 1 '18 at 7:17
add a comment |
$begingroup$
Not an answer, but a related comment that is too long for a comment box ...
This question made me curious to compare:
- the pdf of the sample maximum (given a sample of size $n$ drawn on a N(0,1) parent), say $f(x;n):$
$$f(x) = frac{2^{frac{1}{2}-n} n e^{-frac{x^2}{2}} left(1+text{erf}left(frac{x}{sqrt{2}}right)right)^{n-1}}{sqrt{pi }}$$
where erf(.) denotes the error function, to
- the asymptote proposed by the question $sqrt{2 log n}$
... when $n$ is very large (say $n$ = 1 million).
The diagram compares:
The distribution does not, in fact, converge to the asymptote in any meaningful manner, even when $n$ is very large, such as $n$ = 1 million or 100 million or even a billion.
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
$endgroup$
Not an answer, but a related comment that is too long for a comment box ...
This question made me curious to compare:
- the pdf of the sample maximum (given a sample of size $n$ drawn on a N(0,1) parent), say $f(x;n):$
$$f(x) = frac{2^{frac{1}{2}-n} n e^{-frac{x^2}{2}} left(1+text{erf}left(frac{x}{sqrt{2}}right)right)^{n-1}}{sqrt{pi }}$$
where erf(.) denotes the error function, to
- the asymptote proposed by the question $sqrt{2 log n}$
... when $n$ is very large (say $n$ = 1 million).
The diagram compares:
The distribution does not, in fact, converge to the asymptote in any meaningful manner, even when $n$ is very large, such as $n$ = 1 million or 100 million or even a billion.
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
edited Oct 1 '18 at 7:20
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
answered Apr 2 '16 at 14:14
wolfieswolfies
4,2492923
4,2492923
$begingroup$
Are you using the right logarithm? (I.e. base $e$ and not base 10?)
$endgroup$
– Chill2Macht
Oct 1 '18 at 0:13
1
$begingroup$
Yes - $sqrt{2 log left(10^6right)}$ = 5.25 ... as shown in the diagram above (the red line). By contrast, with base 10, the same calculation would be equal to $2 sqrt{3}$ which is approx 3.46.
$endgroup$
– wolfies
Oct 1 '18 at 7:17
add a comment |
$begingroup$
Are you using the right logarithm? (I.e. base $e$ and not base 10?)
$endgroup$
– Chill2Macht
Oct 1 '18 at 0:13
1
$begingroup$
Yes - $sqrt{2 log left(10^6right)}$ = 5.25 ... as shown in the diagram above (the red line). By contrast, with base 10, the same calculation would be equal to $2 sqrt{3}$ which is approx 3.46.
$endgroup$
– wolfies
Oct 1 '18 at 7:17
$begingroup$
Are you using the right logarithm? (I.e. base $e$ and not base 10?)
$endgroup$
– Chill2Macht
Oct 1 '18 at 0:13
$begingroup$
Are you using the right logarithm? (I.e. base $e$ and not base 10?)
$endgroup$
– Chill2Macht
Oct 1 '18 at 0:13
1
1
$begingroup$
Yes - $sqrt{2 log left(10^6right)}$ = 5.25 ... as shown in the diagram above (the red line). By contrast, with base 10, the same calculation would be equal to $2 sqrt{3}$ which is approx 3.46.
$endgroup$
– wolfies
Oct 1 '18 at 7:17
$begingroup$
Yes - $sqrt{2 log left(10^6right)}$ = 5.25 ... as shown in the diagram above (the red line). By contrast, with base 10, the same calculation would be equal to $2 sqrt{3}$ which is approx 3.46.
$endgroup$
– wolfies
Oct 1 '18 at 7:17
add a comment |
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