Showing this sum converges (sum of normal tail probabilities)?












0












$begingroup$


Let $S_n sim N(0,n)$ so that $S_n$ is the partial sum process of standard normal random variables.



My objective is to prove:



$$sum_{n=1}^infty P(S_n geq (1+epsilon)sqrt{2 n log log n }) < infty$$



for all $epsilon > 0$.



I cannot use the law of iterated logarithm, as the reason I want to prove the above inequality is to prove a weaker form of iterated logarithm ($leq 1$ instead of $=1$).



My attempt has been to bound the tail probabilities using known inequalities for standard normal. Firstly, we rewrite it as



$$sum_{n=1}^infty P(S_n/sqrt{n} geq (1+epsilon)sqrt{2 log log n }) < infty$$



$$sum_{n=1}^infty (1-Phi((1+epsilon)sqrt{2 log log n })) < infty$$



Inequalities I know of:



$$1-Phi(x) leq x^{-1} e^{-x^2/2}/sqrt{2pi}$$



$$1-Phi(x) leq frac{1}{2} e^{-x^2/2}$$



But neither is enough to bound the sum by a convergent sequence. Alternatively, I also know that



$1-Phi(x) sim phi(x)$ but that doesn't seem to help






probability convergence self-learning







share|cite|improve this question











$endgroup$












  • $begingroup$
    In the first inequality for $1-Phi(x)$ you have stated there is also an inequality in the opposite direction with a different constant. It appears therefore that the result you are trying to prove is actually false.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 5:44
















0












$begingroup$


Let $S_n sim N(0,n)$ so that $S_n$ is the partial sum process of standard normal random variables.



My objective is to prove:



$$sum_{n=1}^infty P(S_n geq (1+epsilon)sqrt{2 n log log n }) < infty$$



for all $epsilon > 0$.



I cannot use the law of iterated logarithm, as the reason I want to prove the above inequality is to prove a weaker form of iterated logarithm ($leq 1$ instead of $=1$).



My attempt has been to bound the tail probabilities using known inequalities for standard normal. Firstly, we rewrite it as



$$sum_{n=1}^infty P(S_n/sqrt{n} geq (1+epsilon)sqrt{2 log log n }) < infty$$



$$sum_{n=1}^infty (1-Phi((1+epsilon)sqrt{2 log log n })) < infty$$



Inequalities I know of:



$$1-Phi(x) leq x^{-1} e^{-x^2/2}/sqrt{2pi}$$



$$1-Phi(x) leq frac{1}{2} e^{-x^2/2}$$



But neither is enough to bound the sum by a convergent sequence. Alternatively, I also know that



$1-Phi(x) sim phi(x)$ but that doesn't seem to help






probability convergence self-learning







share|cite|improve this question











$endgroup$












  • $begingroup$
    In the first inequality for $1-Phi(x)$ you have stated there is also an inequality in the opposite direction with a different constant. It appears therefore that the result you are trying to prove is actually false.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 5:44














0












0








0





$begingroup$


Let $S_n sim N(0,n)$ so that $S_n$ is the partial sum process of standard normal random variables.



My objective is to prove:



$$sum_{n=1}^infty P(S_n geq (1+epsilon)sqrt{2 n log log n }) < infty$$



for all $epsilon > 0$.



I cannot use the law of iterated logarithm, as the reason I want to prove the above inequality is to prove a weaker form of iterated logarithm ($leq 1$ instead of $=1$).



My attempt has been to bound the tail probabilities using known inequalities for standard normal. Firstly, we rewrite it as



$$sum_{n=1}^infty P(S_n/sqrt{n} geq (1+epsilon)sqrt{2 log log n }) < infty$$



$$sum_{n=1}^infty (1-Phi((1+epsilon)sqrt{2 log log n })) < infty$$



Inequalities I know of:



$$1-Phi(x) leq x^{-1} e^{-x^2/2}/sqrt{2pi}$$



$$1-Phi(x) leq frac{1}{2} e^{-x^2/2}$$



But neither is enough to bound the sum by a convergent sequence. Alternatively, I also know that



$1-Phi(x) sim phi(x)$ but that doesn't seem to help






probability convergence self-learning







share|cite|improve this question











$endgroup$




Let $S_n sim N(0,n)$ so that $S_n$ is the partial sum process of standard normal random variables.



My objective is to prove:



$$sum_{n=1}^infty P(S_n geq (1+epsilon)sqrt{2 n log log n }) < infty$$



for all $epsilon > 0$.



I cannot use the law of iterated logarithm, as the reason I want to prove the above inequality is to prove a weaker form of iterated logarithm ($leq 1$ instead of $=1$).



My attempt has been to bound the tail probabilities using known inequalities for standard normal. Firstly, we rewrite it as



$$sum_{n=1}^infty P(S_n/sqrt{n} geq (1+epsilon)sqrt{2 log log n }) < infty$$



$$sum_{n=1}^infty (1-Phi((1+epsilon)sqrt{2 log log n })) < infty$$



Inequalities I know of:



$$1-Phi(x) leq x^{-1} e^{-x^2/2}/sqrt{2pi}$$



$$1-Phi(x) leq frac{1}{2} e^{-x^2/2}$$



But neither is enough to bound the sum by a convergent sequence. Alternatively, I also know that



$1-Phi(x) sim phi(x)$ but that doesn't seem to help






probability convergence self-learning




probability convergence self-learning






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 6:42







Xiaomi

















asked Dec 3 '18 at 5:11









XiaomiXiaomi

1,057115




1,057115












  • $begingroup$
    In the first inequality for $1-Phi(x)$ you have stated there is also an inequality in the opposite direction with a different constant. It appears therefore that the result you are trying to prove is actually false.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 5:44


















  • $begingroup$
    In the first inequality for $1-Phi(x)$ you have stated there is also an inequality in the opposite direction with a different constant. It appears therefore that the result you are trying to prove is actually false.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 5:44
















$begingroup$
In the first inequality for $1-Phi(x)$ you have stated there is also an inequality in the opposite direction with a different constant. It appears therefore that the result you are trying to prove is actually false.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 5:44




$begingroup$
In the first inequality for $1-Phi(x)$ you have stated there is also an inequality in the opposite direction with a different constant. It appears therefore that the result you are trying to prove is actually false.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 5:44










1 Answer
1






active

oldest

votes


















2












$begingroup$

(Too long for a comment)



This cannot be shown in the way you're trying. Indeed, the best possible this way is $Omega(sqrt{log n})$. This is because a commensurate lower bound to the gaussian CDF exists:



$$1 - Phi(x) ge frac{1}{sqrt{2pi}} frac{x}{x^2 + 1} e^{-x^2/2}$$



(to show this, integrate by parts one more time in the proof of the first upper bound you state).



One reason you are failing is the following: with the proof strategy, the following statement would also follow: Consider independent Gaussians $Z_n sim mathcal{N}(0,1).$ Then ${Z_n ge (1+epsilon)sqrt{2log log n} }$ occurs only finitely often a.s. It is intuitive that this is false (for a proof, use the lower bound above and B-C). The LIL truly uses the fact that one is dealing with a sum of Gaussians, by exploting the correlations between $S_n$ and $S_{(1+c)n}$ for small $c$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes sense. But that is strange, as this is an old exam question, and I don't think it would be that difficult.
    $endgroup$
    – Xiaomi
    Dec 3 '18 at 6:41










  • $begingroup$
    Well, maybe, since it's an exam question and since the LIL is pretty standard material for a grad prob. class, it's possible that they expect one to reproduce the upper bound proof from this?
    $endgroup$
    – stochasticboy321
    Dec 3 '18 at 21:24











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2












$begingroup$

(Too long for a comment)



This cannot be shown in the way you're trying. Indeed, the best possible this way is $Omega(sqrt{log n})$. This is because a commensurate lower bound to the gaussian CDF exists:



$$1 - Phi(x) ge frac{1}{sqrt{2pi}} frac{x}{x^2 + 1} e^{-x^2/2}$$



(to show this, integrate by parts one more time in the proof of the first upper bound you state).



One reason you are failing is the following: with the proof strategy, the following statement would also follow: Consider independent Gaussians $Z_n sim mathcal{N}(0,1).$ Then ${Z_n ge (1+epsilon)sqrt{2log log n} }$ occurs only finitely often a.s. It is intuitive that this is false (for a proof, use the lower bound above and B-C). The LIL truly uses the fact that one is dealing with a sum of Gaussians, by exploting the correlations between $S_n$ and $S_{(1+c)n}$ for small $c$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes sense. But that is strange, as this is an old exam question, and I don't think it would be that difficult.
    $endgroup$
    – Xiaomi
    Dec 3 '18 at 6:41










  • $begingroup$
    Well, maybe, since it's an exam question and since the LIL is pretty standard material for a grad prob. class, it's possible that they expect one to reproduce the upper bound proof from this?
    $endgroup$
    – stochasticboy321
    Dec 3 '18 at 21:24
















2












$begingroup$

(Too long for a comment)



This cannot be shown in the way you're trying. Indeed, the best possible this way is $Omega(sqrt{log n})$. This is because a commensurate lower bound to the gaussian CDF exists:



$$1 - Phi(x) ge frac{1}{sqrt{2pi}} frac{x}{x^2 + 1} e^{-x^2/2}$$



(to show this, integrate by parts one more time in the proof of the first upper bound you state).



One reason you are failing is the following: with the proof strategy, the following statement would also follow: Consider independent Gaussians $Z_n sim mathcal{N}(0,1).$ Then ${Z_n ge (1+epsilon)sqrt{2log log n} }$ occurs only finitely often a.s. It is intuitive that this is false (for a proof, use the lower bound above and B-C). The LIL truly uses the fact that one is dealing with a sum of Gaussians, by exploting the correlations between $S_n$ and $S_{(1+c)n}$ for small $c$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes sense. But that is strange, as this is an old exam question, and I don't think it would be that difficult.
    $endgroup$
    – Xiaomi
    Dec 3 '18 at 6:41










  • $begingroup$
    Well, maybe, since it's an exam question and since the LIL is pretty standard material for a grad prob. class, it's possible that they expect one to reproduce the upper bound proof from this?
    $endgroup$
    – stochasticboy321
    Dec 3 '18 at 21:24














2












2








2





$begingroup$

(Too long for a comment)



This cannot be shown in the way you're trying. Indeed, the best possible this way is $Omega(sqrt{log n})$. This is because a commensurate lower bound to the gaussian CDF exists:



$$1 - Phi(x) ge frac{1}{sqrt{2pi}} frac{x}{x^2 + 1} e^{-x^2/2}$$



(to show this, integrate by parts one more time in the proof of the first upper bound you state).



One reason you are failing is the following: with the proof strategy, the following statement would also follow: Consider independent Gaussians $Z_n sim mathcal{N}(0,1).$ Then ${Z_n ge (1+epsilon)sqrt{2log log n} }$ occurs only finitely often a.s. It is intuitive that this is false (for a proof, use the lower bound above and B-C). The LIL truly uses the fact that one is dealing with a sum of Gaussians, by exploting the correlations between $S_n$ and $S_{(1+c)n}$ for small $c$.






share|cite|improve this answer









$endgroup$



(Too long for a comment)



This cannot be shown in the way you're trying. Indeed, the best possible this way is $Omega(sqrt{log n})$. This is because a commensurate lower bound to the gaussian CDF exists:



$$1 - Phi(x) ge frac{1}{sqrt{2pi}} frac{x}{x^2 + 1} e^{-x^2/2}$$



(to show this, integrate by parts one more time in the proof of the first upper bound you state).



One reason you are failing is the following: with the proof strategy, the following statement would also follow: Consider independent Gaussians $Z_n sim mathcal{N}(0,1).$ Then ${Z_n ge (1+epsilon)sqrt{2log log n} }$ occurs only finitely often a.s. It is intuitive that this is false (for a proof, use the lower bound above and B-C). The LIL truly uses the fact that one is dealing with a sum of Gaussians, by exploting the correlations between $S_n$ and $S_{(1+c)n}$ for small $c$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 '18 at 5:44









stochasticboy321stochasticboy321

2,522617




2,522617












  • $begingroup$
    That makes sense. But that is strange, as this is an old exam question, and I don't think it would be that difficult.
    $endgroup$
    – Xiaomi
    Dec 3 '18 at 6:41










  • $begingroup$
    Well, maybe, since it's an exam question and since the LIL is pretty standard material for a grad prob. class, it's possible that they expect one to reproduce the upper bound proof from this?
    $endgroup$
    – stochasticboy321
    Dec 3 '18 at 21:24


















  • $begingroup$
    That makes sense. But that is strange, as this is an old exam question, and I don't think it would be that difficult.
    $endgroup$
    – Xiaomi
    Dec 3 '18 at 6:41










  • $begingroup$
    Well, maybe, since it's an exam question and since the LIL is pretty standard material for a grad prob. class, it's possible that they expect one to reproduce the upper bound proof from this?
    $endgroup$
    – stochasticboy321
    Dec 3 '18 at 21:24
















$begingroup$
That makes sense. But that is strange, as this is an old exam question, and I don't think it would be that difficult.
$endgroup$
– Xiaomi
Dec 3 '18 at 6:41




$begingroup$
That makes sense. But that is strange, as this is an old exam question, and I don't think it would be that difficult.
$endgroup$
– Xiaomi
Dec 3 '18 at 6:41












$begingroup$
Well, maybe, since it's an exam question and since the LIL is pretty standard material for a grad prob. class, it's possible that they expect one to reproduce the upper bound proof from this?
$endgroup$
– stochasticboy321
Dec 3 '18 at 21:24




$begingroup$
Well, maybe, since it's an exam question and since the LIL is pretty standard material for a grad prob. class, it's possible that they expect one to reproduce the upper bound proof from this?
$endgroup$
– stochasticboy321
Dec 3 '18 at 21:24


















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