Hoeffding inequality for conditional probability (conditioned on event)
$begingroup$
Suppose I have independent $X_1simtext{Bin}(n,theta_1)$, $X_2simtext{Bin}(n,theta_2)$ with $X=X_1+X_2$. Suppose that $theta_1,theta_2in(0,1)$. Define the constant (but still depends on $n$) $a=ntheta_0+o(n^{2/3})$, for $theta_0in(0,1)$. I'd like to show the following:
$$
Pleft(X>cmid X_1leq a,X_2leq aright)to 0
$$
as $ntoinfty$ for some "large enough" $c$. Note that the parameters, $theta$'s, for $X_1$ and $X_2$ may not be the same, so that $X$ is not necessarily a binomial random variable.
Usually, in the marginal case, without the conditioning event, Hoeffding's inequality can be used to get the convergence. Is there a known inequality (like Chernoff or Hoeffding's) to show the exponential convergence for such a quantity conditioned on an event?
inequality probability-distributions reference-request asymptotics large-deviation-theory
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
$endgroup$
|
show 1 more comment
$begingroup$
Suppose I have independent $X_1simtext{Bin}(n,theta_1)$, $X_2simtext{Bin}(n,theta_2)$ with $X=X_1+X_2$. Suppose that $theta_1,theta_2in(0,1)$. Define the constant (but still depends on $n$) $a=ntheta_0+o(n^{2/3})$, for $theta_0in(0,1)$. I'd like to show the following:
$$
Pleft(X>cmid X_1leq a,X_2leq aright)to 0
$$
as $ntoinfty$ for some "large enough" $c$. Note that the parameters, $theta$'s, for $X_1$ and $X_2$ may not be the same, so that $X$ is not necessarily a binomial random variable.
Usually, in the marginal case, without the conditioning event, Hoeffding's inequality can be used to get the convergence. Is there a known inequality (like Chernoff or Hoeffding's) to show the exponential convergence for such a quantity conditioned on an event?
inequality probability-distributions reference-request asymptotics large-deviation-theory
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
$endgroup$
$begingroup$
Is $theta$ just restricted in $[0, 1]$, or do we know more (e.g. $theta < theta_1$).
$endgroup$
– Tom Chen
Jan 20 at 0:06
$begingroup$
$theta$ is bounded away from 0 and 1
$endgroup$
– stats134711
Jan 20 at 0:46
$begingroup$
What do you mean bounded "away from" zero and one? Do you mean $theta in (-infty, 0) cup (1, infty)$?
$endgroup$
– Lee David Chung Lin
Jan 20 at 16:06
$begingroup$
Sorry. @LeeDavidChungLin I meant $thetain(0,1)$. I am not interested in $theta=0$ or 1
$endgroup$
– stats134711
Jan 20 at 16:09
$begingroup$
Is $theta$ the same as $theta_0,theta_1,theta_2$? An easy case is that $P(X>c)$ is already exponentially small. Maybe that's all you want?
$endgroup$
– Dap
Jan 21 at 6:38
|
show 1 more comment
$begingroup$
Suppose I have independent $X_1simtext{Bin}(n,theta_1)$, $X_2simtext{Bin}(n,theta_2)$ with $X=X_1+X_2$. Suppose that $theta_1,theta_2in(0,1)$. Define the constant (but still depends on $n$) $a=ntheta_0+o(n^{2/3})$, for $theta_0in(0,1)$. I'd like to show the following:
$$
Pleft(X>cmid X_1leq a,X_2leq aright)to 0
$$
as $ntoinfty$ for some "large enough" $c$. Note that the parameters, $theta$'s, for $X_1$ and $X_2$ may not be the same, so that $X$ is not necessarily a binomial random variable.
Usually, in the marginal case, without the conditioning event, Hoeffding's inequality can be used to get the convergence. Is there a known inequality (like Chernoff or Hoeffding's) to show the exponential convergence for such a quantity conditioned on an event?
inequality probability-distributions reference-request asymptotics large-deviation-theory
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
$endgroup$
Suppose I have independent $X_1simtext{Bin}(n,theta_1)$, $X_2simtext{Bin}(n,theta_2)$ with $X=X_1+X_2$. Suppose that $theta_1,theta_2in(0,1)$. Define the constant (but still depends on $n$) $a=ntheta_0+o(n^{2/3})$, for $theta_0in(0,1)$. I'd like to show the following:
$$
Pleft(X>cmid X_1leq a,X_2leq aright)to 0
$$
as $ntoinfty$ for some "large enough" $c$. Note that the parameters, $theta$'s, for $X_1$ and $X_2$ may not be the same, so that $X$ is not necessarily a binomial random variable.
Usually, in the marginal case, without the conditioning event, Hoeffding's inequality can be used to get the convergence. Is there a known inequality (like Chernoff or Hoeffding's) to show the exponential convergence for such a quantity conditioned on an event?
inequality probability-distributions reference-request asymptotics large-deviation-theory
inequality probability-distributions reference-request asymptotics large-deviation-theory
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
edited Jan 23 at 0:36
stats134711
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
asked Sep 5 '18 at 5:28
stats134711stats134711
187215
187215
$begingroup$
Is $theta$ just restricted in $[0, 1]$, or do we know more (e.g. $theta < theta_1$).
$endgroup$
– Tom Chen
Jan 20 at 0:06
$begingroup$
$theta$ is bounded away from 0 and 1
$endgroup$
– stats134711
Jan 20 at 0:46
$begingroup$
What do you mean bounded "away from" zero and one? Do you mean $theta in (-infty, 0) cup (1, infty)$?
$endgroup$
– Lee David Chung Lin
Jan 20 at 16:06
$begingroup$
Sorry. @LeeDavidChungLin I meant $thetain(0,1)$. I am not interested in $theta=0$ or 1
$endgroup$
– stats134711
Jan 20 at 16:09
$begingroup$
Is $theta$ the same as $theta_0,theta_1,theta_2$? An easy case is that $P(X>c)$ is already exponentially small. Maybe that's all you want?
$endgroup$
– Dap
Jan 21 at 6:38
|
show 1 more comment
$begingroup$
Is $theta$ just restricted in $[0, 1]$, or do we know more (e.g. $theta < theta_1$).
$endgroup$
– Tom Chen
Jan 20 at 0:06
$begingroup$
$theta$ is bounded away from 0 and 1
$endgroup$
– stats134711
Jan 20 at 0:46
$begingroup$
What do you mean bounded "away from" zero and one? Do you mean $theta in (-infty, 0) cup (1, infty)$?
$endgroup$
– Lee David Chung Lin
Jan 20 at 16:06
$begingroup$
Sorry. @LeeDavidChungLin I meant $thetain(0,1)$. I am not interested in $theta=0$ or 1
$endgroup$
– stats134711
Jan 20 at 16:09
$begingroup$
Is $theta$ the same as $theta_0,theta_1,theta_2$? An easy case is that $P(X>c)$ is already exponentially small. Maybe that's all you want?
$endgroup$
– Dap
Jan 21 at 6:38
$begingroup$
Is $theta$ just restricted in $[0, 1]$, or do we know more (e.g. $theta < theta_1$).
$endgroup$
– Tom Chen
Jan 20 at 0:06
$begingroup$
Is $theta$ just restricted in $[0, 1]$, or do we know more (e.g. $theta < theta_1$).
$endgroup$
– Tom Chen
Jan 20 at 0:06
$begingroup$
$theta$ is bounded away from 0 and 1
$endgroup$
– stats134711
Jan 20 at 0:46
$begingroup$
$theta$ is bounded away from 0 and 1
$endgroup$
– stats134711
Jan 20 at 0:46
$begingroup$
What do you mean bounded "away from" zero and one? Do you mean $theta in (-infty, 0) cup (1, infty)$?
$endgroup$
– Lee David Chung Lin
Jan 20 at 16:06
$begingroup$
What do you mean bounded "away from" zero and one? Do you mean $theta in (-infty, 0) cup (1, infty)$?
$endgroup$
– Lee David Chung Lin
Jan 20 at 16:06
$begingroup$
Sorry. @LeeDavidChungLin I meant $thetain(0,1)$. I am not interested in $theta=0$ or 1
$endgroup$
– stats134711
Jan 20 at 16:09
$begingroup$
Sorry. @LeeDavidChungLin I meant $thetain(0,1)$. I am not interested in $theta=0$ or 1
$endgroup$
– stats134711
Jan 20 at 16:09
$begingroup$
Is $theta$ the same as $theta_0,theta_1,theta_2$? An easy case is that $P(X>c)$ is already exponentially small. Maybe that's all you want?
$endgroup$
– Dap
Jan 21 at 6:38
$begingroup$
Is $theta$ the same as $theta_0,theta_1,theta_2$? An easy case is that $P(X>c)$ is already exponentially small. Maybe that's all you want?
$endgroup$
– Dap
Jan 21 at 6:38
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The FKG inequality, or more specifically Harris inequality, gives
$$P(X>c mid X_1 leq a, X_2 leq a) leq P ( X > c).$$
I will describe in more detail how FKG applies. Let $A$ be the event $X> c$ and let $B$ be the event that $X_1,X_2leq a.$ We want to prove $P(Amid B)leq P(A).$
The FKG lattice condition is always satisfied for a product measure $mu=mu_1timesmu_2$ defined on a Cartesian product of finite totally ordered sets, here $L={0,1,dots,n}^2.$ We are using $$(mu_1timesmu_1)({(x_1,x_1)})=mathbb P[X_1=x_1, X_2=x_2].$$
Let $1_A$ and $1_B$ denote the indicator functions of the events $A$ and $B$ respectively. Since $1_A$ is an increasing function on $L$ and $1_B$ is a decreasing function on $L,$ the FKG inequality says
$$sum_{xin L} 1_A(x)1_B(x)mu(x)sum_{xin L}mu(x)leq sum_{xin L} 1_A(x)mu(x)sum_{xin L} 1_B(x)mu(x),$$ which is just $P(Acap B)cdot 1leq P(A)P(B),$ so $P(Amid B)leq P(A).$
answered Jan 22 at 20:47
DapDap
20.2k842
$endgroup$
$begingroup$
I'm sorry. I accidentally rewarded the other post with the bounty. I meant to give it to you.
$endgroup$
– stats134711
Jan 23 at 1:36
$begingroup$
Ha, thanks anyway
$endgroup$
– Dap
Jan 23 at 4:58
add a comment |
$begingroup$
You might just try using the definition of conditional expectation. Since begin{align}P ( X > c | X _ { 1 } leq a , X _ { 2 } leq a ) &= frac{P ( X > c cap X _ { 1 } leq a , X _ { 2 } leq a )}{P(X_1 leq a, X_2 leq a)} \
&leq frac{P ( X > c)}{P(X_1 leq a, X_2 leq a)}, end{align}
you could just apply Hoeffding's inequality to the numerator assuming $a$ is fixed.
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
$endgroup$
$begingroup$
Unfortunately, $atoinfty$ as $ntoinfty$, and there is no guarantee that the denominator is $1-o(1)$ as $ntoinfty$.
$endgroup$
– stats134711
Jan 21 at 22:38
$begingroup$
The FKG inequality gives this without the denominator i.e. $Pleft(X>cmid X_1leq a,X_2leq aright)leq P(X>c).$ @stats134711 would this be enough for your situation?
$endgroup$
– Dap
Jan 22 at 20:23
$begingroup$
Yes, it seems that inequality would be enough. I'm not familiar with that inequality - could you please formalize it in the context of the scenario above?
$endgroup$
– stats134711
Jan 22 at 20:28
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
The FKG inequality, or more specifically Harris inequality, gives
$$P(X>c mid X_1 leq a, X_2 leq a) leq P ( X > c).$$
I will describe in more detail how FKG applies. Let $A$ be the event $X> c$ and let $B$ be the event that $X_1,X_2leq a.$ We want to prove $P(Amid B)leq P(A).$
The FKG lattice condition is always satisfied for a product measure $mu=mu_1timesmu_2$ defined on a Cartesian product of finite totally ordered sets, here $L={0,1,dots,n}^2.$ We are using $$(mu_1timesmu_1)({(x_1,x_1)})=mathbb P[X_1=x_1, X_2=x_2].$$
Let $1_A$ and $1_B$ denote the indicator functions of the events $A$ and $B$ respectively. Since $1_A$ is an increasing function on $L$ and $1_B$ is a decreasing function on $L,$ the FKG inequality says
$$sum_{xin L} 1_A(x)1_B(x)mu(x)sum_{xin L}mu(x)leq sum_{xin L} 1_A(x)mu(x)sum_{xin L} 1_B(x)mu(x),$$ which is just $P(Acap B)cdot 1leq P(A)P(B),$ so $P(Amid B)leq P(A).$
answered Jan 22 at 20:47
DapDap
20.2k842
$endgroup$
$begingroup$
I'm sorry. I accidentally rewarded the other post with the bounty. I meant to give it to you.
$endgroup$
– stats134711
Jan 23 at 1:36
$begingroup$
Ha, thanks anyway
$endgroup$
– Dap
Jan 23 at 4:58
add a comment |
$begingroup$
The FKG inequality, or more specifically Harris inequality, gives
$$P(X>c mid X_1 leq a, X_2 leq a) leq P ( X > c).$$
I will describe in more detail how FKG applies. Let $A$ be the event $X> c$ and let $B$ be the event that $X_1,X_2leq a.$ We want to prove $P(Amid B)leq P(A).$
The FKG lattice condition is always satisfied for a product measure $mu=mu_1timesmu_2$ defined on a Cartesian product of finite totally ordered sets, here $L={0,1,dots,n}^2.$ We are using $$(mu_1timesmu_1)({(x_1,x_1)})=mathbb P[X_1=x_1, X_2=x_2].$$
Let $1_A$ and $1_B$ denote the indicator functions of the events $A$ and $B$ respectively. Since $1_A$ is an increasing function on $L$ and $1_B$ is a decreasing function on $L,$ the FKG inequality says
$$sum_{xin L} 1_A(x)1_B(x)mu(x)sum_{xin L}mu(x)leq sum_{xin L} 1_A(x)mu(x)sum_{xin L} 1_B(x)mu(x),$$ which is just $P(Acap B)cdot 1leq P(A)P(B),$ so $P(Amid B)leq P(A).$
answered Jan 22 at 20:47
DapDap
20.2k842
$endgroup$
$begingroup$
I'm sorry. I accidentally rewarded the other post with the bounty. I meant to give it to you.
$endgroup$
– stats134711
Jan 23 at 1:36
$begingroup$
Ha, thanks anyway
$endgroup$
– Dap
Jan 23 at 4:58
add a comment |
$begingroup$
The FKG inequality, or more specifically Harris inequality, gives
$$P(X>c mid X_1 leq a, X_2 leq a) leq P ( X > c).$$
I will describe in more detail how FKG applies. Let $A$ be the event $X> c$ and let $B$ be the event that $X_1,X_2leq a.$ We want to prove $P(Amid B)leq P(A).$
The FKG lattice condition is always satisfied for a product measure $mu=mu_1timesmu_2$ defined on a Cartesian product of finite totally ordered sets, here $L={0,1,dots,n}^2.$ We are using $$(mu_1timesmu_1)({(x_1,x_1)})=mathbb P[X_1=x_1, X_2=x_2].$$
Let $1_A$ and $1_B$ denote the indicator functions of the events $A$ and $B$ respectively. Since $1_A$ is an increasing function on $L$ and $1_B$ is a decreasing function on $L,$ the FKG inequality says
$$sum_{xin L} 1_A(x)1_B(x)mu(x)sum_{xin L}mu(x)leq sum_{xin L} 1_A(x)mu(x)sum_{xin L} 1_B(x)mu(x),$$ which is just $P(Acap B)cdot 1leq P(A)P(B),$ so $P(Amid B)leq P(A).$
answered Jan 22 at 20:47
DapDap
20.2k842
$endgroup$
The FKG inequality, or more specifically Harris inequality, gives
$$P(X>c mid X_1 leq a, X_2 leq a) leq P ( X > c).$$
I will describe in more detail how FKG applies. Let $A$ be the event $X> c$ and let $B$ be the event that $X_1,X_2leq a.$ We want to prove $P(Amid B)leq P(A).$
The FKG lattice condition is always satisfied for a product measure $mu=mu_1timesmu_2$ defined on a Cartesian product of finite totally ordered sets, here $L={0,1,dots,n}^2.$ We are using $$(mu_1timesmu_1)({(x_1,x_1)})=mathbb P[X_1=x_1, X_2=x_2].$$
Let $1_A$ and $1_B$ denote the indicator functions of the events $A$ and $B$ respectively. Since $1_A$ is an increasing function on $L$ and $1_B$ is a decreasing function on $L,$ the FKG inequality says
$$sum_{xin L} 1_A(x)1_B(x)mu(x)sum_{xin L}mu(x)leq sum_{xin L} 1_A(x)mu(x)sum_{xin L} 1_B(x)mu(x),$$ which is just $P(Acap B)cdot 1leq P(A)P(B),$ so $P(Amid B)leq P(A).$
answered Jan 22 at 20:47
DapDap
20.2k842
answered Jan 22 at 20:47
DapDap
20.2k842
answered Jan 22 at 20:47
DapDap
20.2k842
answered Jan 22 at 20:47
DapDap
20.2k842
20.2k842
$begingroup$
I'm sorry. I accidentally rewarded the other post with the bounty. I meant to give it to you.
$endgroup$
– stats134711
Jan 23 at 1:36
$begingroup$
Ha, thanks anyway
$endgroup$
– Dap
Jan 23 at 4:58
add a comment |
$begingroup$
I'm sorry. I accidentally rewarded the other post with the bounty. I meant to give it to you.
$endgroup$
– stats134711
Jan 23 at 1:36
$begingroup$
Ha, thanks anyway
$endgroup$
– Dap
Jan 23 at 4:58
$begingroup$
I'm sorry. I accidentally rewarded the other post with the bounty. I meant to give it to you.
$endgroup$
– stats134711
Jan 23 at 1:36
$begingroup$
I'm sorry. I accidentally rewarded the other post with the bounty. I meant to give it to you.
$endgroup$
– stats134711
Jan 23 at 1:36
$begingroup$
Ha, thanks anyway
$endgroup$
– Dap
Jan 23 at 4:58
$begingroup$
Ha, thanks anyway
$endgroup$
– Dap
Jan 23 at 4:58
add a comment |
$begingroup$
You might just try using the definition of conditional expectation. Since begin{align}P ( X > c | X _ { 1 } leq a , X _ { 2 } leq a ) &= frac{P ( X > c cap X _ { 1 } leq a , X _ { 2 } leq a )}{P(X_1 leq a, X_2 leq a)} \
&leq frac{P ( X > c)}{P(X_1 leq a, X_2 leq a)}, end{align}
you could just apply Hoeffding's inequality to the numerator assuming $a$ is fixed.
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
$endgroup$
$begingroup$
Unfortunately, $atoinfty$ as $ntoinfty$, and there is no guarantee that the denominator is $1-o(1)$ as $ntoinfty$.
$endgroup$
– stats134711
Jan 21 at 22:38
$begingroup$
The FKG inequality gives this without the denominator i.e. $Pleft(X>cmid X_1leq a,X_2leq aright)leq P(X>c).$ @stats134711 would this be enough for your situation?
$endgroup$
– Dap
Jan 22 at 20:23
$begingroup$
Yes, it seems that inequality would be enough. I'm not familiar with that inequality - could you please formalize it in the context of the scenario above?
$endgroup$
– stats134711
Jan 22 at 20:28
add a comment |
$begingroup$
You might just try using the definition of conditional expectation. Since begin{align}P ( X > c | X _ { 1 } leq a , X _ { 2 } leq a ) &= frac{P ( X > c cap X _ { 1 } leq a , X _ { 2 } leq a )}{P(X_1 leq a, X_2 leq a)} \
&leq frac{P ( X > c)}{P(X_1 leq a, X_2 leq a)}, end{align}
you could just apply Hoeffding's inequality to the numerator assuming $a$ is fixed.
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
$endgroup$
$begingroup$
Unfortunately, $atoinfty$ as $ntoinfty$, and there is no guarantee that the denominator is $1-o(1)$ as $ntoinfty$.
$endgroup$
– stats134711
Jan 21 at 22:38
$begingroup$
The FKG inequality gives this without the denominator i.e. $Pleft(X>cmid X_1leq a,X_2leq aright)leq P(X>c).$ @stats134711 would this be enough for your situation?
$endgroup$
– Dap
Jan 22 at 20:23
$begingroup$
Yes, it seems that inequality would be enough. I'm not familiar with that inequality - could you please formalize it in the context of the scenario above?
$endgroup$
– stats134711
Jan 22 at 20:28
add a comment |
$begingroup$
You might just try using the definition of conditional expectation. Since begin{align}P ( X > c | X _ { 1 } leq a , X _ { 2 } leq a ) &= frac{P ( X > c cap X _ { 1 } leq a , X _ { 2 } leq a )}{P(X_1 leq a, X_2 leq a)} \
&leq frac{P ( X > c)}{P(X_1 leq a, X_2 leq a)}, end{align}
you could just apply Hoeffding's inequality to the numerator assuming $a$ is fixed.
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
$endgroup$
You might just try using the definition of conditional expectation. Since begin{align}P ( X > c | X _ { 1 } leq a , X _ { 2 } leq a ) &= frac{P ( X > c cap X _ { 1 } leq a , X _ { 2 } leq a )}{P(X_1 leq a, X_2 leq a)} \
&leq frac{P ( X > c)}{P(X_1 leq a, X_2 leq a)}, end{align}
you could just apply Hoeffding's inequality to the numerator assuming $a$ is fixed.
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
answered Jan 21 at 22:17
OldGodzillaOldGodzilla
57427
57427
$begingroup$
Unfortunately, $atoinfty$ as $ntoinfty$, and there is no guarantee that the denominator is $1-o(1)$ as $ntoinfty$.
$endgroup$
– stats134711
Jan 21 at 22:38
$begingroup$
The FKG inequality gives this without the denominator i.e. $Pleft(X>cmid X_1leq a,X_2leq aright)leq P(X>c).$ @stats134711 would this be enough for your situation?
$endgroup$
– Dap
Jan 22 at 20:23
$begingroup$
Yes, it seems that inequality would be enough. I'm not familiar with that inequality - could you please formalize it in the context of the scenario above?
$endgroup$
– stats134711
Jan 22 at 20:28
add a comment |
$begingroup$
Unfortunately, $atoinfty$ as $ntoinfty$, and there is no guarantee that the denominator is $1-o(1)$ as $ntoinfty$.
$endgroup$
– stats134711
Jan 21 at 22:38
$begingroup$
The FKG inequality gives this without the denominator i.e. $Pleft(X>cmid X_1leq a,X_2leq aright)leq P(X>c).$ @stats134711 would this be enough for your situation?
$endgroup$
– Dap
Jan 22 at 20:23
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Yes, it seems that inequality would be enough. I'm not familiar with that inequality - could you please formalize it in the context of the scenario above?
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– stats134711
Jan 22 at 20:28
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Unfortunately, $atoinfty$ as $ntoinfty$, and there is no guarantee that the denominator is $1-o(1)$ as $ntoinfty$.
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– stats134711
Jan 21 at 22:38
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Unfortunately, $atoinfty$ as $ntoinfty$, and there is no guarantee that the denominator is $1-o(1)$ as $ntoinfty$.
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– stats134711
Jan 21 at 22:38
$begingroup$
The FKG inequality gives this without the denominator i.e. $Pleft(X>cmid X_1leq a,X_2leq aright)leq P(X>c).$ @stats134711 would this be enough for your situation?
$endgroup$
– Dap
Jan 22 at 20:23
$begingroup$
The FKG inequality gives this without the denominator i.e. $Pleft(X>cmid X_1leq a,X_2leq aright)leq P(X>c).$ @stats134711 would this be enough for your situation?
$endgroup$
– Dap
Jan 22 at 20:23
$begingroup$
Yes, it seems that inequality would be enough. I'm not familiar with that inequality - could you please formalize it in the context of the scenario above?
$endgroup$
– stats134711
Jan 22 at 20:28
$begingroup$
Yes, it seems that inequality would be enough. I'm not familiar with that inequality - could you please formalize it in the context of the scenario above?
$endgroup$
– stats134711
Jan 22 at 20:28
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Is $theta$ just restricted in $[0, 1]$, or do we know more (e.g. $theta < theta_1$).
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– Tom Chen
Jan 20 at 0:06
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$theta$ is bounded away from 0 and 1
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– stats134711
Jan 20 at 0:46
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What do you mean bounded "away from" zero and one? Do you mean $theta in (-infty, 0) cup (1, infty)$?
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– Lee David Chung Lin
Jan 20 at 16:06
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Sorry. @LeeDavidChungLin I meant $thetain(0,1)$. I am not interested in $theta=0$ or 1
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– stats134711
Jan 20 at 16:09
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Is $theta$ the same as $theta_0,theta_1,theta_2$? An easy case is that $P(X>c)$ is already exponentially small. Maybe that's all you want?
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– Dap
Jan 21 at 6:38