Optimal rate of growth of i.i.d. Gaussians?
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Suppose I have a countable collection ${N_k}_{k=1}^infty$ of independent $mathcal{N}(0.1)$ random variables and I want to estimate their rate of growth in the sense that I want to find a function $f:mathcal{N}tomathcal{R}$ such that
$$sup_k frac{vert N_kvert}{f(k)} <infty text{ a.s.}$$
By means of the basic inequality
$$mathbb{P} (vert N_kvert geq c)leq 2e^{-c^2/2}$$
and Borel-Cantelli lemma, I can obtain that this works for $f(k)=sqrt{2log(1+k)}$ (the constant 2 is there only because it simplifies calculations) and that such an $f$ is sufficiently optimal in the sense that $alpha=1/2$ is the threshold parameter for which this works with $f$ of the form $(log(1+k))^alpha$. The problem is that, if I define the variable
$$Y:= sup_k frac{vert N_kvert}{sqrt{2log(1+k)}}$$
then apparently $Y$ is heavy tailed distributed, in the sense that the estimate I'm able to obtain again by the above inequality is
$$mathbb{P}(Ygeq y)leq frac{C}{1+y^2}$$
for a suitable constant $C$. So the first thing I'm asking is whether this estimate is quite sharp, and $Y$ is indeed heavy-tailed, or there are better estimates which lead to $Y$ having some finite moments. The second question is, if this is not the case, if one defines instead
$$Y:= sup_k frac{vert N_kvert}{g(k)sqrt{2log(1+k)}}$$
for a suitable $g$ such that $g(k)toinfty$ as $ktoinfty$, if there is some kind of optimal $g$ that grows as slowly as possible but such that $Y$ now admits moments of all orders. My intuition tells me that this should work for instance with $g(k)=sqrt{log(log(1+k))}$ but I'm not able to perform nice calculations and I don't know if there is something even better, like $g$ even slower such that $Y$ still admits all moments.
probability-theory probability-distributions
asked Nov 15 at 10:55
Lucio
477215
add a comment |
up vote
7
down vote
favorite
Suppose I have a countable collection ${N_k}_{k=1}^infty$ of independent $mathcal{N}(0.1)$ random variables and I want to estimate their rate of growth in the sense that I want to find a function $f:mathcal{N}tomathcal{R}$ such that
$$sup_k frac{vert N_kvert}{f(k)} <infty text{ a.s.}$$
By means of the basic inequality
$$mathbb{P} (vert N_kvert geq c)leq 2e^{-c^2/2}$$
and Borel-Cantelli lemma, I can obtain that this works for $f(k)=sqrt{2log(1+k)}$ (the constant 2 is there only because it simplifies calculations) and that such an $f$ is sufficiently optimal in the sense that $alpha=1/2$ is the threshold parameter for which this works with $f$ of the form $(log(1+k))^alpha$. The problem is that, if I define the variable
$$Y:= sup_k frac{vert N_kvert}{sqrt{2log(1+k)}}$$
then apparently $Y$ is heavy tailed distributed, in the sense that the estimate I'm able to obtain again by the above inequality is
$$mathbb{P}(Ygeq y)leq frac{C}{1+y^2}$$
for a suitable constant $C$. So the first thing I'm asking is whether this estimate is quite sharp, and $Y$ is indeed heavy-tailed, or there are better estimates which lead to $Y$ having some finite moments. The second question is, if this is not the case, if one defines instead
$$Y:= sup_k frac{vert N_kvert}{g(k)sqrt{2log(1+k)}}$$
for a suitable $g$ such that $g(k)toinfty$ as $ktoinfty$, if there is some kind of optimal $g$ that grows as slowly as possible but such that $Y$ now admits moments of all orders. My intuition tells me that this should work for instance with $g(k)=sqrt{log(log(1+k))}$ but I'm not able to perform nice calculations and I don't know if there is something even better, like $g$ even slower such that $Y$ still admits all moments.
probability-theory probability-distributions
asked Nov 15 at 10:55
Lucio
477215
This might be related to the law of the iterated logarithm. Indeed, if we set $N_k=B_k/sqrt{k}$ where $B$ is a Brownian motion, then we have $limsup N_k/sqrt{2*loglog k}=1$ a.s. Of course this is not exactly the same as your setup because the $N_k$ are not independent, but the covariance $Cov(N_n,N_m)=sqrt{min(n,m)/max(n,m)}$ does tend to zero, so I suspect something could be still be said in your case.
– Mike Hawk
Nov 23 at 22:04
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Suppose I have a countable collection ${N_k}_{k=1}^infty$ of independent $mathcal{N}(0.1)$ random variables and I want to estimate their rate of growth in the sense that I want to find a function $f:mathcal{N}tomathcal{R}$ such that
$$sup_k frac{vert N_kvert}{f(k)} <infty text{ a.s.}$$
By means of the basic inequality
$$mathbb{P} (vert N_kvert geq c)leq 2e^{-c^2/2}$$
and Borel-Cantelli lemma, I can obtain that this works for $f(k)=sqrt{2log(1+k)}$ (the constant 2 is there only because it simplifies calculations) and that such an $f$ is sufficiently optimal in the sense that $alpha=1/2$ is the threshold parameter for which this works with $f$ of the form $(log(1+k))^alpha$. The problem is that, if I define the variable
$$Y:= sup_k frac{vert N_kvert}{sqrt{2log(1+k)}}$$
then apparently $Y$ is heavy tailed distributed, in the sense that the estimate I'm able to obtain again by the above inequality is
$$mathbb{P}(Ygeq y)leq frac{C}{1+y^2}$$
for a suitable constant $C$. So the first thing I'm asking is whether this estimate is quite sharp, and $Y$ is indeed heavy-tailed, or there are better estimates which lead to $Y$ having some finite moments. The second question is, if this is not the case, if one defines instead
$$Y:= sup_k frac{vert N_kvert}{g(k)sqrt{2log(1+k)}}$$
for a suitable $g$ such that $g(k)toinfty$ as $ktoinfty$, if there is some kind of optimal $g$ that grows as slowly as possible but such that $Y$ now admits moments of all orders. My intuition tells me that this should work for instance with $g(k)=sqrt{log(log(1+k))}$ but I'm not able to perform nice calculations and I don't know if there is something even better, like $g$ even slower such that $Y$ still admits all moments.
probability-theory probability-distributions
asked Nov 15 at 10:55
Lucio
477215
Suppose I have a countable collection ${N_k}_{k=1}^infty$ of independent $mathcal{N}(0.1)$ random variables and I want to estimate their rate of growth in the sense that I want to find a function $f:mathcal{N}tomathcal{R}$ such that
$$sup_k frac{vert N_kvert}{f(k)} <infty text{ a.s.}$$
By means of the basic inequality
$$mathbb{P} (vert N_kvert geq c)leq 2e^{-c^2/2}$$
and Borel-Cantelli lemma, I can obtain that this works for $f(k)=sqrt{2log(1+k)}$ (the constant 2 is there only because it simplifies calculations) and that such an $f$ is sufficiently optimal in the sense that $alpha=1/2$ is the threshold parameter for which this works with $f$ of the form $(log(1+k))^alpha$. The problem is that, if I define the variable
$$Y:= sup_k frac{vert N_kvert}{sqrt{2log(1+k)}}$$
then apparently $Y$ is heavy tailed distributed, in the sense that the estimate I'm able to obtain again by the above inequality is
$$mathbb{P}(Ygeq y)leq frac{C}{1+y^2}$$
for a suitable constant $C$. So the first thing I'm asking is whether this estimate is quite sharp, and $Y$ is indeed heavy-tailed, or there are better estimates which lead to $Y$ having some finite moments. The second question is, if this is not the case, if one defines instead
$$Y:= sup_k frac{vert N_kvert}{g(k)sqrt{2log(1+k)}}$$
for a suitable $g$ such that $g(k)toinfty$ as $ktoinfty$, if there is some kind of optimal $g$ that grows as slowly as possible but such that $Y$ now admits moments of all orders. My intuition tells me that this should work for instance with $g(k)=sqrt{log(log(1+k))}$ but I'm not able to perform nice calculations and I don't know if there is something even better, like $g$ even slower such that $Y$ still admits all moments.
probability-theory probability-distributions
probability-theory probability-distributions
asked Nov 15 at 10:55
Lucio
477215
asked Nov 15 at 10:55
Lucio
477215
asked Nov 15 at 10:55
Lucio
477215
asked Nov 15 at 10:55
Lucio
477215
asked Nov 15 at 10:55
Lucio
477215
477215
This might be related to the law of the iterated logarithm. Indeed, if we set $N_k=B_k/sqrt{k}$ where $B$ is a Brownian motion, then we have $limsup N_k/sqrt{2*loglog k}=1$ a.s. Of course this is not exactly the same as your setup because the $N_k$ are not independent, but the covariance $Cov(N_n,N_m)=sqrt{min(n,m)/max(n,m)}$ does tend to zero, so I suspect something could be still be said in your case.
– Mike Hawk
Nov 23 at 22:04
add a comment |
This might be related to the law of the iterated logarithm. Indeed, if we set $N_k=B_k/sqrt{k}$ where $B$ is a Brownian motion, then we have $limsup N_k/sqrt{2*loglog k}=1$ a.s. Of course this is not exactly the same as your setup because the $N_k$ are not independent, but the covariance $Cov(N_n,N_m)=sqrt{min(n,m)/max(n,m)}$ does tend to zero, so I suspect something could be still be said in your case.
– Mike Hawk
Nov 23 at 22:04
This might be related to the law of the iterated logarithm. Indeed, if we set $N_k=B_k/sqrt{k}$ where $B$ is a Brownian motion, then we have $limsup N_k/sqrt{2*loglog k}=1$ a.s. Of course this is not exactly the same as your setup because the $N_k$ are not independent, but the covariance $Cov(N_n,N_m)=sqrt{min(n,m)/max(n,m)}$ does tend to zero, so I suspect something could be still be said in your case.
– Mike Hawk
Nov 23 at 22:04
This might be related to the law of the iterated logarithm. Indeed, if we set $N_k=B_k/sqrt{k}$ where $B$ is a Brownian motion, then we have $limsup N_k/sqrt{2*loglog k}=1$ a.s. Of course this is not exactly the same as your setup because the $N_k$ are not independent, but the covariance $Cov(N_n,N_m)=sqrt{min(n,m)/max(n,m)}$ does tend to zero, so I suspect something could be still be said in your case.
– Mike Hawk
Nov 23 at 22:04
add a comment |
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This might be related to the law of the iterated logarithm. Indeed, if we set $N_k=B_k/sqrt{k}$ where $B$ is a Brownian motion, then we have $limsup N_k/sqrt{2*loglog k}=1$ a.s. Of course this is not exactly the same as your setup because the $N_k$ are not independent, but the covariance $Cov(N_n,N_m)=sqrt{min(n,m)/max(n,m)}$ does tend to zero, so I suspect something could be still be said in your case.
– Mike Hawk
Nov 23 at 22:04