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Deciding whether a statistic is larger than something (Chernoff bound? Algorithm?)
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I am given two statistics $X,Y$ such that $E(X)+E(Y) = lambda$ for some unknown $lambda in [epsilon/2,frac32epsilon]$ that we ultimately want to determine. Furthermore, $X$ is Bernoulli with $E(X^n)=V^n-1 E(X)$ and $E(X) leq V/2$ for a constant $V$, and $Var(X)=Var(Y)$. I am interested in knowing the order of $V-lambda$.
Letting $widetildeX = frac1nsum_i=0^n X_i$ where $X_i$ are $i.i.d.$ copies of $X$, I know that if I normalize the statistic and apply a Chernoff bound, I can get
$$Pr(V-2widetildeX geq (1+delta)(V-2 widetildeX)) < e^fracdelta^2 n2(1-frac2E(X)V).$$
Thus, I need to simply take $n$ to scale inversely with $(1-frac2 E(x)V)$ to obtain a bound that scales independently of $E(X)/V$. I'd like to do something similar for the statistic $Y$, but a measurement of one statistic prevents me from measuring the other, so I can either measure only $X$ or only $Y$. $Y$ is not Bernoulli and I'm looking to determine $lambda$. Applying, say, Hoeffding's inequality yields poor scaling in $E(Y)$, however, since I seemingly need to take quadratically more samples. Now, I do know that $lambda in [epsilon/2,3/2epsilon]$ for some very small $epsilon$. It would be good enough if I could determine whether $lambda$ is bigger than or smaller than $epsilon$, which (maybe) allows me to apply the Chernoff bound to $Y$ as well, but I was wondering if there is a more efficient/useful method given that I know what I'm looking for.
Also, I would like to know why the statement of Hoeffding's inequality appears much weaker. Do Bernoulli distributions converge faster to their average than other distributions defined on the same space?
I found some pages like this one, Relation betweeen Hoeffding inequality and Chernoff bound?, but nothing really answered my question.
probability statistics algorithms
asked Jul 14 at 18:53
Michael Jarret
15312
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I am given two statistics $X,Y$ such that $E(X)+E(Y) = lambda$ for some unknown $lambda in [epsilon/2,frac32epsilon]$ that we ultimately want to determine. Furthermore, $X$ is Bernoulli with $E(X^n)=V^n-1 E(X)$ and $E(X) leq V/2$ for a constant $V$, and $Var(X)=Var(Y)$. I am interested in knowing the order of $V-lambda$.
Letting $widetildeX = frac1nsum_i=0^n X_i$ where $X_i$ are $i.i.d.$ copies of $X$, I know that if I normalize the statistic and apply a Chernoff bound, I can get
$$Pr(V-2widetildeX geq (1+delta)(V-2 widetildeX)) < e^fracdelta^2 n2(1-frac2E(X)V).$$
Thus, I need to simply take $n$ to scale inversely with $(1-frac2 E(x)V)$ to obtain a bound that scales independently of $E(X)/V$. I'd like to do something similar for the statistic $Y$, but a measurement of one statistic prevents me from measuring the other, so I can either measure only $X$ or only $Y$. $Y$ is not Bernoulli and I'm looking to determine $lambda$. Applying, say, Hoeffding's inequality yields poor scaling in $E(Y)$, however, since I seemingly need to take quadratically more samples. Now, I do know that $lambda in [epsilon/2,3/2epsilon]$ for some very small $epsilon$. It would be good enough if I could determine whether $lambda$ is bigger than or smaller than $epsilon$, which (maybe) allows me to apply the Chernoff bound to $Y$ as well, but I was wondering if there is a more efficient/useful method given that I know what I'm looking for.
Also, I would like to know why the statement of Hoeffding's inequality appears much weaker. Do Bernoulli distributions converge faster to their average than other distributions defined on the same space?
I found some pages like this one, Relation betweeen Hoeffding inequality and Chernoff bound?, but nothing really answered my question.
probability statistics algorithms
asked Jul 14 at 18:53
Michael Jarret
15312
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given two statistics $X,Y$ such that $E(X)+E(Y) = lambda$ for some unknown $lambda in [epsilon/2,frac32epsilon]$ that we ultimately want to determine. Furthermore, $X$ is Bernoulli with $E(X^n)=V^n-1 E(X)$ and $E(X) leq V/2$ for a constant $V$, and $Var(X)=Var(Y)$. I am interested in knowing the order of $V-lambda$.
Letting $widetildeX = frac1nsum_i=0^n X_i$ where $X_i$ are $i.i.d.$ copies of $X$, I know that if I normalize the statistic and apply a Chernoff bound, I can get
$$Pr(V-2widetildeX geq (1+delta)(V-2 widetildeX)) < e^fracdelta^2 n2(1-frac2E(X)V).$$
Thus, I need to simply take $n$ to scale inversely with $(1-frac2 E(x)V)$ to obtain a bound that scales independently of $E(X)/V$. I'd like to do something similar for the statistic $Y$, but a measurement of one statistic prevents me from measuring the other, so I can either measure only $X$ or only $Y$. $Y$ is not Bernoulli and I'm looking to determine $lambda$. Applying, say, Hoeffding's inequality yields poor scaling in $E(Y)$, however, since I seemingly need to take quadratically more samples. Now, I do know that $lambda in [epsilon/2,3/2epsilon]$ for some very small $epsilon$. It would be good enough if I could determine whether $lambda$ is bigger than or smaller than $epsilon$, which (maybe) allows me to apply the Chernoff bound to $Y$ as well, but I was wondering if there is a more efficient/useful method given that I know what I'm looking for.
Also, I would like to know why the statement of Hoeffding's inequality appears much weaker. Do Bernoulli distributions converge faster to their average than other distributions defined on the same space?
I found some pages like this one, Relation betweeen Hoeffding inequality and Chernoff bound?, but nothing really answered my question.
probability statistics algorithms
asked Jul 14 at 18:53
Michael Jarret
15312
I am given two statistics $X,Y$ such that $E(X)+E(Y) = lambda$ for some unknown $lambda in [epsilon/2,frac32epsilon]$ that we ultimately want to determine. Furthermore, $X$ is Bernoulli with $E(X^n)=V^n-1 E(X)$ and $E(X) leq V/2$ for a constant $V$, and $Var(X)=Var(Y)$. I am interested in knowing the order of $V-lambda$.
Letting $widetildeX = frac1nsum_i=0^n X_i$ where $X_i$ are $i.i.d.$ copies of $X$, I know that if I normalize the statistic and apply a Chernoff bound, I can get
$$Pr(V-2widetildeX geq (1+delta)(V-2 widetildeX)) < e^fracdelta^2 n2(1-frac2E(X)V).$$
Thus, I need to simply take $n$ to scale inversely with $(1-frac2 E(x)V)$ to obtain a bound that scales independently of $E(X)/V$. I'd like to do something similar for the statistic $Y$, but a measurement of one statistic prevents me from measuring the other, so I can either measure only $X$ or only $Y$. $Y$ is not Bernoulli and I'm looking to determine $lambda$. Applying, say, Hoeffding's inequality yields poor scaling in $E(Y)$, however, since I seemingly need to take quadratically more samples. Now, I do know that $lambda in [epsilon/2,3/2epsilon]$ for some very small $epsilon$. It would be good enough if I could determine whether $lambda$ is bigger than or smaller than $epsilon$, which (maybe) allows me to apply the Chernoff bound to $Y$ as well, but I was wondering if there is a more efficient/useful method given that I know what I'm looking for.
Also, I would like to know why the statement of Hoeffding's inequality appears much weaker. Do Bernoulli distributions converge faster to their average than other distributions defined on the same space?
I found some pages like this one, Relation betweeen Hoeffding inequality and Chernoff bound?, but nothing really answered my question.
probability statistics algorithms
asked Jul 14 at 18:53
Michael Jarret
15312
asked Jul 14 at 18:53
Michael Jarret
15312
asked Jul 14 at 18:53
Michael Jarret
15312
asked Jul 14 at 18:53
Michael Jarret
15312
15312
add a comment |Â
add a comment |Â
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