Coin flipping limit The Next CEO of Stack OverflowLaw of large numbers for non-identically distributed Bernoulli random variablesSeries limit involving Binomial coefficientsApplying Central Limit Theorem to show that $Eleft(fracsqrtnright) to sqrtfrac2pisigma$Convergence in distribution with exponential limit distributionProving the uniqueness of the weak limitLimit of an integral (Arrow theorem)Finding the Limit of Binomial CDFA question with a continuous limit to a series of discrete random variablesContinuity of a random variable and of its cumulative distribution functionFeller continuous Markov kernels on a compact metric space has an invariant distribution?Limit of density of $XsimtextGamma(alpha,alpha/mu)$ as $alphatoinfty$Series limit involving Binomial coefficients
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Traduction de « Life is a roller coaster »
Coin flipping limit
The Next CEO of Stack OverflowLaw of large numbers for non-identically distributed Bernoulli random variablesSeries limit involving Binomial coefficientsApplying Central Limit Theorem to show that $Eleft(fracsqrtnright) to sqrtfrac2pisigma$Convergence in distribution with exponential limit distributionProving the uniqueness of the weak limitLimit of an integral (Arrow theorem)Finding the Limit of Binomial CDFA question with a continuous limit to a series of discrete random variablesContinuity of a random variable and of its cumulative distribution functionFeller continuous Markov kernels on a compact metric space has an invariant distribution?Limit of density of $XsimtextGamma(alpha,alpha/mu)$ as $alphatoinfty$Series limit involving Binomial coefficients
$begingroup$
Let $S_nsimtextBinleft(n,pleft(nright)right)$ where $pleft(nright)$
is the unique solution to the equation $deltaleft(pleft(nright),nright)=0$
with $delta$ being continuous and bounded, strictly decreasing in
$p$ with $deltaleft(0,nright)>0$ and $deltaleft(1,nright)<0$
for all $ngeq2$. We also have that $pleft(nright)$ is strictly decreasing in $n$.
How to justify rigorously that there exists a unique
(fixed) $p$ such that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$,
where $p$ is the unique solution to the equation $lim_nrightarrowinftydeltaleft(pleft(nright),nright)=0$?
probability-theory random-variables binomial-distribution probability-limit-theorems
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
$endgroup$
add a comment |
$begingroup$
Let $S_nsimtextBinleft(n,pleft(nright)right)$ where $pleft(nright)$
is the unique solution to the equation $deltaleft(pleft(nright),nright)=0$
with $delta$ being continuous and bounded, strictly decreasing in
$p$ with $deltaleft(0,nright)>0$ and $deltaleft(1,nright)<0$
for all $ngeq2$. We also have that $pleft(nright)$ is strictly decreasing in $n$.
How to justify rigorously that there exists a unique
(fixed) $p$ such that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$,
where $p$ is the unique solution to the equation $lim_nrightarrowinftydeltaleft(pleft(nright),nright)=0$?
probability-theory random-variables binomial-distribution probability-limit-theorems
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
$endgroup$
$begingroup$
My guess is that we need a non-iid version of CLT and that the question is related to this answer. Any help is very much appreciated.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 1:47
add a comment |
$begingroup$
Let $S_nsimtextBinleft(n,pleft(nright)right)$ where $pleft(nright)$
is the unique solution to the equation $deltaleft(pleft(nright),nright)=0$
with $delta$ being continuous and bounded, strictly decreasing in
$p$ with $deltaleft(0,nright)>0$ and $deltaleft(1,nright)<0$
for all $ngeq2$. We also have that $pleft(nright)$ is strictly decreasing in $n$.
How to justify rigorously that there exists a unique
(fixed) $p$ such that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$,
where $p$ is the unique solution to the equation $lim_nrightarrowinftydeltaleft(pleft(nright),nright)=0$?
probability-theory random-variables binomial-distribution probability-limit-theorems
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
$endgroup$
Let $S_nsimtextBinleft(n,pleft(nright)right)$ where $pleft(nright)$
is the unique solution to the equation $deltaleft(pleft(nright),nright)=0$
with $delta$ being continuous and bounded, strictly decreasing in
$p$ with $deltaleft(0,nright)>0$ and $deltaleft(1,nright)<0$
for all $ngeq2$. We also have that $pleft(nright)$ is strictly decreasing in $n$.
How to justify rigorously that there exists a unique
(fixed) $p$ such that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$,
where $p$ is the unique solution to the equation $lim_nrightarrowinftydeltaleft(pleft(nright),nright)=0$?
probability-theory random-variables binomial-distribution probability-limit-theorems
probability-theory random-variables binomial-distribution probability-limit-theorems
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
edited Mar 28 at 1:44
Konstantinos I. Stouras
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
asked Mar 28 at 0:38
Konstantinos I. StourasKonstantinos I. Stouras
436
436
$begingroup$
My guess is that we need a non-iid version of CLT and that the question is related to this answer. Any help is very much appreciated.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 1:47
add a comment |
$begingroup$
My guess is that we need a non-iid version of CLT and that the question is related to this answer. Any help is very much appreciated.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 1:47
$begingroup$
My guess is that we need a non-iid version of CLT and that the question is related to this answer. Any help is very much appreciated.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 1:47
$begingroup$
My guess is that we need a non-iid version of CLT and that the question is related to this answer. Any help is very much appreciated.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 1:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not true without additional assumptions. Here is a counterexample.
For fixed $n$, Let $delta(x,n)$ be the linear interpotation between $1$ and $0$ between $x=0$ and $x=frac 12 + frac 1n$, and the linear interpolation between $0$ and $-1$ between $x=frac 12 + frac 1n$ and $x=1$.
Then $delta$ satisfies all the requirements and $p(n) = frac 12 + frac 1n$ is decreasing to $p=frac 12$.
However,
$$E[S_n] - np = n (frac 12 + frac 1n) - n frac 12 =1.$$
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
$endgroup$
$begingroup$
+1, thank you so much. But looking at your answer and Example 14.5 in these notes made me more confused now. The only assumption that this Example requires is that the limit $p_n (1-p_n)$ to be non-zero. Why can't I use it?
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:27
$begingroup$
If it helps, the $delta$ function I am referring to is stated in this related question.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:31
$begingroup$
I see now what I misunderstood. The non-iid version of CLT applies in my setting and implies specifically that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pleft(nright)right)=0$. To show that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$ indeed needs more information.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 19:33
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not true without additional assumptions. Here is a counterexample.
For fixed $n$, Let $delta(x,n)$ be the linear interpotation between $1$ and $0$ between $x=0$ and $x=frac 12 + frac 1n$, and the linear interpolation between $0$ and $-1$ between $x=frac 12 + frac 1n$ and $x=1$.
Then $delta$ satisfies all the requirements and $p(n) = frac 12 + frac 1n$ is decreasing to $p=frac 12$.
However,
$$E[S_n] - np = n (frac 12 + frac 1n) - n frac 12 =1.$$
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
$endgroup$
$begingroup$
+1, thank you so much. But looking at your answer and Example 14.5 in these notes made me more confused now. The only assumption that this Example requires is that the limit $p_n (1-p_n)$ to be non-zero. Why can't I use it?
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:27
$begingroup$
If it helps, the $delta$ function I am referring to is stated in this related question.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:31
$begingroup$
I see now what I misunderstood. The non-iid version of CLT applies in my setting and implies specifically that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pleft(nright)right)=0$. To show that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$ indeed needs more information.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 19:33
add a comment |
$begingroup$
It is not true without additional assumptions. Here is a counterexample.
For fixed $n$, Let $delta(x,n)$ be the linear interpotation between $1$ and $0$ between $x=0$ and $x=frac 12 + frac 1n$, and the linear interpolation between $0$ and $-1$ between $x=frac 12 + frac 1n$ and $x=1$.
Then $delta$ satisfies all the requirements and $p(n) = frac 12 + frac 1n$ is decreasing to $p=frac 12$.
However,
$$E[S_n] - np = n (frac 12 + frac 1n) - n frac 12 =1.$$
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
$endgroup$
$begingroup$
+1, thank you so much. But looking at your answer and Example 14.5 in these notes made me more confused now. The only assumption that this Example requires is that the limit $p_n (1-p_n)$ to be non-zero. Why can't I use it?
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:27
$begingroup$
If it helps, the $delta$ function I am referring to is stated in this related question.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:31
$begingroup$
I see now what I misunderstood. The non-iid version of CLT applies in my setting and implies specifically that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pleft(nright)right)=0$. To show that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$ indeed needs more information.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 19:33
add a comment |
$begingroup$
It is not true without additional assumptions. Here is a counterexample.
For fixed $n$, Let $delta(x,n)$ be the linear interpotation between $1$ and $0$ between $x=0$ and $x=frac 12 + frac 1n$, and the linear interpolation between $0$ and $-1$ between $x=frac 12 + frac 1n$ and $x=1$.
Then $delta$ satisfies all the requirements and $p(n) = frac 12 + frac 1n$ is decreasing to $p=frac 12$.
However,
$$E[S_n] - np = n (frac 12 + frac 1n) - n frac 12 =1.$$
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
$endgroup$
It is not true without additional assumptions. Here is a counterexample.
For fixed $n$, Let $delta(x,n)$ be the linear interpotation between $1$ and $0$ between $x=0$ and $x=frac 12 + frac 1n$, and the linear interpolation between $0$ and $-1$ between $x=frac 12 + frac 1n$ and $x=1$.
Then $delta$ satisfies all the requirements and $p(n) = frac 12 + frac 1n$ is decreasing to $p=frac 12$.
However,
$$E[S_n] - np = n (frac 12 + frac 1n) - n frac 12 =1.$$
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
answered Mar 28 at 2:07
FnacoolFnacool
5,281611
5,281611
$begingroup$
+1, thank you so much. But looking at your answer and Example 14.5 in these notes made me more confused now. The only assumption that this Example requires is that the limit $p_n (1-p_n)$ to be non-zero. Why can't I use it?
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:27
$begingroup$
If it helps, the $delta$ function I am referring to is stated in this related question.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:31
$begingroup$
I see now what I misunderstood. The non-iid version of CLT applies in my setting and implies specifically that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pleft(nright)right)=0$. To show that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$ indeed needs more information.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 19:33
add a comment |
$begingroup$
+1, thank you so much. But looking at your answer and Example 14.5 in these notes made me more confused now. The only assumption that this Example requires is that the limit $p_n (1-p_n)$ to be non-zero. Why can't I use it?
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:27
$begingroup$
If it helps, the $delta$ function I am referring to is stated in this related question.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:31
$begingroup$
I see now what I misunderstood. The non-iid version of CLT applies in my setting and implies specifically that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pleft(nright)right)=0$. To show that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$ indeed needs more information.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 19:33
$begingroup$
+1, thank you so much. But looking at your answer and Example 14.5 in these notes made me more confused now. The only assumption that this Example requires is that the limit $p_n (1-p_n)$ to be non-zero. Why can't I use it?
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:27
$begingroup$
+1, thank you so much. But looking at your answer and Example 14.5 in these notes made me more confused now. The only assumption that this Example requires is that the limit $p_n (1-p_n)$ to be non-zero. Why can't I use it?
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:27
$begingroup$
If it helps, the $delta$ function I am referring to is stated in this related question.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:31
$begingroup$
If it helps, the $delta$ function I am referring to is stated in this related question.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 2:31
$begingroup$
I see now what I misunderstood. The non-iid version of CLT applies in my setting and implies specifically that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pleft(nright)right)=0$. To show that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$ indeed needs more information.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 19:33
$begingroup$
I see now what I misunderstood. The non-iid version of CLT applies in my setting and implies specifically that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pleft(nright)right)=0$. To show that $lim_nrightarrowinftyleft(Eleft[S_nright]-ncdot pright)=0$ indeed needs more information.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 19:33
add a comment |
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$begingroup$
My guess is that we need a non-iid version of CLT and that the question is related to this answer. Any help is very much appreciated.
$endgroup$
– Konstantinos I. Stouras
Mar 28 at 1:47