$X_1,..,X_n$ i.i.d random variable and discrete, local limit theorem
$begingroup$
Let $X_1,..., X_n$ be i.i.d discrete random variables which take their value in $mathbb{Z}$ with a non trivial and finite support. Let $S_n = X_1+...+ X_n$
Prove the existence of $ 0 < C_1 < C_2 < infty$ such that for all $ n geq 1$ :
$$ C_1/sqrt{n} leq sup_{k in mathbb{Z}} mathbb{P}(S_n = k) leq C_2/sqrt{n}$$
I know that this can be proved using the central limit theorem yet we didn’t see this theorem in class so this exercise can be solved without using CLT.
So far here are my thoughts :
The fact that it’s $sqrt{n}$ comes from the fact that the standard deviation of $S_n$ is $theta sqrt{n}$.
Hence one possible strategy is to study the random variable :
$frac{S_n- mu}{theta/sqrt{n}}$. Yet from now on I dind’t manage finding an upper bound on the probability.
For example using Markov inequality I get that :
$$mathbb{P}( mid frac{S_n}{n} -mu mid geq theta/sqrt{n}) leq frac{1}{n}$$
The problem is that it doesn’t help since I can’t say anything when $mid S_n/n -mu mid leq theta/sqrt{n}$.
Thank you !
probability probability-theory probability-limit-theorems
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
$endgroup$
|
show 6 more comments
$begingroup$
Let $X_1,..., X_n$ be i.i.d discrete random variables which take their value in $mathbb{Z}$ with a non trivial and finite support. Let $S_n = X_1+...+ X_n$
Prove the existence of $ 0 < C_1 < C_2 < infty$ such that for all $ n geq 1$ :
$$ C_1/sqrt{n} leq sup_{k in mathbb{Z}} mathbb{P}(S_n = k) leq C_2/sqrt{n}$$
I know that this can be proved using the central limit theorem yet we didn’t see this theorem in class so this exercise can be solved without using CLT.
So far here are my thoughts :
The fact that it’s $sqrt{n}$ comes from the fact that the standard deviation of $S_n$ is $theta sqrt{n}$.
Hence one possible strategy is to study the random variable :
$frac{S_n- mu}{theta/sqrt{n}}$. Yet from now on I dind’t manage finding an upper bound on the probability.
For example using Markov inequality I get that :
$$mathbb{P}( mid frac{S_n}{n} -mu mid geq theta/sqrt{n}) leq frac{1}{n}$$
The problem is that it doesn’t help since I can’t say anything when $mid S_n/n -mu mid leq theta/sqrt{n}$.
Thank you !
probability probability-theory probability-limit-theorems
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
$endgroup$
$begingroup$
You can apply your Markov/Chebyshev inequality if you know the random variables have a finite variance (else you might need some type of truncation argument). That inequality you give is useful since there is an order-$sqrt{n}$ region about the mean such that we are there with probability at least $1/2$. And of course we must always be an integer.
$endgroup$
– Michael
Jan 3 at 6:02
$begingroup$
@Michael Yes you are right about variance, so it’s even harder to prove iron the general case (when we don’t necessarily have variance). Yet I still don’t see how to prove it when we have variance.
$endgroup$
– Jebfiffkkfnfolzbd
Jan 3 at 17:01
$begingroup$
You can just try to get one of the bounds at first (either the lower or upper). For example, what happens when most of the probability is concentrated on only a few of the options? Say, if probability $p$ is spread over $k$ options, there must be some option with at least...
$endgroup$
– Michael
Jan 3 at 18:06
$begingroup$
I don’t think this is true without extra conditions such as finite variance. For intuition with continuous variables, if ${X_i}$ are iid Cauchy random variables with PDF $f(x)$ then the average $S_n/n$ is Cauchy with the same PDF. So begin{align}P[S_n in [a,a+1]] &= P[S_n/n in [a/n, a/n+1/n]]\ &= int_{a/n}^{a/n+1/n} f(x)dx \&leq frac{f_{max}}{n} = O(1/n)end{align} Likely a "discrete Cauchy-type" random variable would have similar $O(1/n)$ behavior.
$endgroup$
– Michael
Jan 5 at 17:30
1
$begingroup$
@Jebfiffkkfnfolzbd : $X(Omega)$ is the set of all real numbers that $X$ can take, and it is always a subset of $mathbb{R}$ (for any random variable $X:Omega rightarrowmathbb{R}$). By "finite support" I think you mean that $X$ can only take a value from some finite interval $[a,b]$. This means that it indeed has a finite mean and variance. Since we also know that $X$ is integer valued, it means that $X$ can only take a finite number of (integer) values.
$endgroup$
– Michael
Jan 7 at 1:38
|
show 6 more comments
$begingroup$
Let $X_1,..., X_n$ be i.i.d discrete random variables which take their value in $mathbb{Z}$ with a non trivial and finite support. Let $S_n = X_1+...+ X_n$
Prove the existence of $ 0 < C_1 < C_2 < infty$ such that for all $ n geq 1$ :
$$ C_1/sqrt{n} leq sup_{k in mathbb{Z}} mathbb{P}(S_n = k) leq C_2/sqrt{n}$$
I know that this can be proved using the central limit theorem yet we didn’t see this theorem in class so this exercise can be solved without using CLT.
So far here are my thoughts :
The fact that it’s $sqrt{n}$ comes from the fact that the standard deviation of $S_n$ is $theta sqrt{n}$.
Hence one possible strategy is to study the random variable :
$frac{S_n- mu}{theta/sqrt{n}}$. Yet from now on I dind’t manage finding an upper bound on the probability.
For example using Markov inequality I get that :
$$mathbb{P}( mid frac{S_n}{n} -mu mid geq theta/sqrt{n}) leq frac{1}{n}$$
The problem is that it doesn’t help since I can’t say anything when $mid S_n/n -mu mid leq theta/sqrt{n}$.
Thank you !
probability probability-theory probability-limit-theorems
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
$endgroup$
Let $X_1,..., X_n$ be i.i.d discrete random variables which take their value in $mathbb{Z}$ with a non trivial and finite support. Let $S_n = X_1+...+ X_n$
Prove the existence of $ 0 < C_1 < C_2 < infty$ such that for all $ n geq 1$ :
$$ C_1/sqrt{n} leq sup_{k in mathbb{Z}} mathbb{P}(S_n = k) leq C_2/sqrt{n}$$
I know that this can be proved using the central limit theorem yet we didn’t see this theorem in class so this exercise can be solved without using CLT.
So far here are my thoughts :
The fact that it’s $sqrt{n}$ comes from the fact that the standard deviation of $S_n$ is $theta sqrt{n}$.
Hence one possible strategy is to study the random variable :
$frac{S_n- mu}{theta/sqrt{n}}$. Yet from now on I dind’t manage finding an upper bound on the probability.
For example using Markov inequality I get that :
$$mathbb{P}( mid frac{S_n}{n} -mu mid geq theta/sqrt{n}) leq frac{1}{n}$$
The problem is that it doesn’t help since I can’t say anything when $mid S_n/n -mu mid leq theta/sqrt{n}$.
Thank you !
probability probability-theory probability-limit-theorems
probability probability-theory probability-limit-theorems
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
edited Jan 6 at 10:49
Jebfiffkkfnfolzbd
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
asked Jan 2 at 14:45
JebfiffkkfnfolzbdJebfiffkkfnfolzbd
642
642
$begingroup$
You can apply your Markov/Chebyshev inequality if you know the random variables have a finite variance (else you might need some type of truncation argument). That inequality you give is useful since there is an order-$sqrt{n}$ region about the mean such that we are there with probability at least $1/2$. And of course we must always be an integer.
$endgroup$
– Michael
Jan 3 at 6:02
$begingroup$
@Michael Yes you are right about variance, so it’s even harder to prove iron the general case (when we don’t necessarily have variance). Yet I still don’t see how to prove it when we have variance.
$endgroup$
– Jebfiffkkfnfolzbd
Jan 3 at 17:01
$begingroup$
You can just try to get one of the bounds at first (either the lower or upper). For example, what happens when most of the probability is concentrated on only a few of the options? Say, if probability $p$ is spread over $k$ options, there must be some option with at least...
$endgroup$
– Michael
Jan 3 at 18:06
$begingroup$
I don’t think this is true without extra conditions such as finite variance. For intuition with continuous variables, if ${X_i}$ are iid Cauchy random variables with PDF $f(x)$ then the average $S_n/n$ is Cauchy with the same PDF. So begin{align}P[S_n in [a,a+1]] &= P[S_n/n in [a/n, a/n+1/n]]\ &= int_{a/n}^{a/n+1/n} f(x)dx \&leq frac{f_{max}}{n} = O(1/n)end{align} Likely a "discrete Cauchy-type" random variable would have similar $O(1/n)$ behavior.
$endgroup$
– Michael
Jan 5 at 17:30
1
$begingroup$
@Jebfiffkkfnfolzbd : $X(Omega)$ is the set of all real numbers that $X$ can take, and it is always a subset of $mathbb{R}$ (for any random variable $X:Omega rightarrowmathbb{R}$). By "finite support" I think you mean that $X$ can only take a value from some finite interval $[a,b]$. This means that it indeed has a finite mean and variance. Since we also know that $X$ is integer valued, it means that $X$ can only take a finite number of (integer) values.
$endgroup$
– Michael
Jan 7 at 1:38
|
show 6 more comments
$begingroup$
You can apply your Markov/Chebyshev inequality if you know the random variables have a finite variance (else you might need some type of truncation argument). That inequality you give is useful since there is an order-$sqrt{n}$ region about the mean such that we are there with probability at least $1/2$. And of course we must always be an integer.
$endgroup$
– Michael
Jan 3 at 6:02
$begingroup$
@Michael Yes you are right about variance, so it’s even harder to prove iron the general case (when we don’t necessarily have variance). Yet I still don’t see how to prove it when we have variance.
$endgroup$
– Jebfiffkkfnfolzbd
Jan 3 at 17:01
$begingroup$
You can just try to get one of the bounds at first (either the lower or upper). For example, what happens when most of the probability is concentrated on only a few of the options? Say, if probability $p$ is spread over $k$ options, there must be some option with at least...
$endgroup$
– Michael
Jan 3 at 18:06
$begingroup$
I don’t think this is true without extra conditions such as finite variance. For intuition with continuous variables, if ${X_i}$ are iid Cauchy random variables with PDF $f(x)$ then the average $S_n/n$ is Cauchy with the same PDF. So begin{align}P[S_n in [a,a+1]] &= P[S_n/n in [a/n, a/n+1/n]]\ &= int_{a/n}^{a/n+1/n} f(x)dx \&leq frac{f_{max}}{n} = O(1/n)end{align} Likely a "discrete Cauchy-type" random variable would have similar $O(1/n)$ behavior.
$endgroup$
– Michael
Jan 5 at 17:30
1
$begingroup$
@Jebfiffkkfnfolzbd : $X(Omega)$ is the set of all real numbers that $X$ can take, and it is always a subset of $mathbb{R}$ (for any random variable $X:Omega rightarrowmathbb{R}$). By "finite support" I think you mean that $X$ can only take a value from some finite interval $[a,b]$. This means that it indeed has a finite mean and variance. Since we also know that $X$ is integer valued, it means that $X$ can only take a finite number of (integer) values.
$endgroup$
– Michael
Jan 7 at 1:38
$begingroup$
You can apply your Markov/Chebyshev inequality if you know the random variables have a finite variance (else you might need some type of truncation argument). That inequality you give is useful since there is an order-$sqrt{n}$ region about the mean such that we are there with probability at least $1/2$. And of course we must always be an integer.
$endgroup$
– Michael
Jan 3 at 6:02
$begingroup$
You can apply your Markov/Chebyshev inequality if you know the random variables have a finite variance (else you might need some type of truncation argument). That inequality you give is useful since there is an order-$sqrt{n}$ region about the mean such that we are there with probability at least $1/2$. And of course we must always be an integer.
$endgroup$
– Michael
Jan 3 at 6:02
$begingroup$
@Michael Yes you are right about variance, so it’s even harder to prove iron the general case (when we don’t necessarily have variance). Yet I still don’t see how to prove it when we have variance.
$endgroup$
– Jebfiffkkfnfolzbd
Jan 3 at 17:01
$begingroup$
@Michael Yes you are right about variance, so it’s even harder to prove iron the general case (when we don’t necessarily have variance). Yet I still don’t see how to prove it when we have variance.
$endgroup$
– Jebfiffkkfnfolzbd
Jan 3 at 17:01
$begingroup$
You can just try to get one of the bounds at first (either the lower or upper). For example, what happens when most of the probability is concentrated on only a few of the options? Say, if probability $p$ is spread over $k$ options, there must be some option with at least...
$endgroup$
– Michael
Jan 3 at 18:06
$begingroup$
You can just try to get one of the bounds at first (either the lower or upper). For example, what happens when most of the probability is concentrated on only a few of the options? Say, if probability $p$ is spread over $k$ options, there must be some option with at least...
$endgroup$
– Michael
Jan 3 at 18:06
$begingroup$
I don’t think this is true without extra conditions such as finite variance. For intuition with continuous variables, if ${X_i}$ are iid Cauchy random variables with PDF $f(x)$ then the average $S_n/n$ is Cauchy with the same PDF. So begin{align}P[S_n in [a,a+1]] &= P[S_n/n in [a/n, a/n+1/n]]\ &= int_{a/n}^{a/n+1/n} f(x)dx \&leq frac{f_{max}}{n} = O(1/n)end{align} Likely a "discrete Cauchy-type" random variable would have similar $O(1/n)$ behavior.
$endgroup$
– Michael
Jan 5 at 17:30
$begingroup$
I don’t think this is true without extra conditions such as finite variance. For intuition with continuous variables, if ${X_i}$ are iid Cauchy random variables with PDF $f(x)$ then the average $S_n/n$ is Cauchy with the same PDF. So begin{align}P[S_n in [a,a+1]] &= P[S_n/n in [a/n, a/n+1/n]]\ &= int_{a/n}^{a/n+1/n} f(x)dx \&leq frac{f_{max}}{n} = O(1/n)end{align} Likely a "discrete Cauchy-type" random variable would have similar $O(1/n)$ behavior.
$endgroup$
– Michael
Jan 5 at 17:30
1
1
$begingroup$
@Jebfiffkkfnfolzbd : $X(Omega)$ is the set of all real numbers that $X$ can take, and it is always a subset of $mathbb{R}$ (for any random variable $X:Omega rightarrowmathbb{R}$). By "finite support" I think you mean that $X$ can only take a value from some finite interval $[a,b]$. This means that it indeed has a finite mean and variance. Since we also know that $X$ is integer valued, it means that $X$ can only take a finite number of (integer) values.
$endgroup$
– Michael
Jan 7 at 1:38
$begingroup$
@Jebfiffkkfnfolzbd : $X(Omega)$ is the set of all real numbers that $X$ can take, and it is always a subset of $mathbb{R}$ (for any random variable $X:Omega rightarrowmathbb{R}$). By "finite support" I think you mean that $X$ can only take a value from some finite interval $[a,b]$. This means that it indeed has a finite mean and variance. Since we also know that $X$ is integer valued, it means that $X$ can only take a finite number of (integer) values.
$endgroup$
– Michael
Jan 7 at 1:38
|
show 6 more comments
1 Answer
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I can prove the lower bound as follows:
$P(|S_n-nmu| geq cn^atheta) leq frac{1}{c^2n^{2a-1}}$.
Thus the largest $P(S_n=k)$ is roughly greater than $(1-c^{-2}n^{1-2a})(2cn^atheta+1)^{-1}$.
(Because this quantity is not greater than $P(|S_n-nmu| leq cn^atheta)|[nmu-cn^atheta,nmu+cn^atheta] cap mathbb{Z}|^{-1}$).
Take $a=1/2, c=2$, this quantity is greater than $frac{3/4}{4thetasqrt{n}+1}$, which proves the lower bound.
Edit: this was done without looking at Michael’s comment.
answered Jan 10 at 0:34
MindlackMindlack
4,910211
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add a comment |
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$begingroup$
I can prove the lower bound as follows:
$P(|S_n-nmu| geq cn^atheta) leq frac{1}{c^2n^{2a-1}}$.
Thus the largest $P(S_n=k)$ is roughly greater than $(1-c^{-2}n^{1-2a})(2cn^atheta+1)^{-1}$.
(Because this quantity is not greater than $P(|S_n-nmu| leq cn^atheta)|[nmu-cn^atheta,nmu+cn^atheta] cap mathbb{Z}|^{-1}$).
Take $a=1/2, c=2$, this quantity is greater than $frac{3/4}{4thetasqrt{n}+1}$, which proves the lower bound.
Edit: this was done without looking at Michael’s comment.
answered Jan 10 at 0:34
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
I can prove the lower bound as follows:
$P(|S_n-nmu| geq cn^atheta) leq frac{1}{c^2n^{2a-1}}$.
Thus the largest $P(S_n=k)$ is roughly greater than $(1-c^{-2}n^{1-2a})(2cn^atheta+1)^{-1}$.
(Because this quantity is not greater than $P(|S_n-nmu| leq cn^atheta)|[nmu-cn^atheta,nmu+cn^atheta] cap mathbb{Z}|^{-1}$).
Take $a=1/2, c=2$, this quantity is greater than $frac{3/4}{4thetasqrt{n}+1}$, which proves the lower bound.
Edit: this was done without looking at Michael’s comment.
answered Jan 10 at 0:34
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
I can prove the lower bound as follows:
$P(|S_n-nmu| geq cn^atheta) leq frac{1}{c^2n^{2a-1}}$.
Thus the largest $P(S_n=k)$ is roughly greater than $(1-c^{-2}n^{1-2a})(2cn^atheta+1)^{-1}$.
(Because this quantity is not greater than $P(|S_n-nmu| leq cn^atheta)|[nmu-cn^atheta,nmu+cn^atheta] cap mathbb{Z}|^{-1}$).
Take $a=1/2, c=2$, this quantity is greater than $frac{3/4}{4thetasqrt{n}+1}$, which proves the lower bound.
Edit: this was done without looking at Michael’s comment.
answered Jan 10 at 0:34
MindlackMindlack
4,910211
$endgroup$
I can prove the lower bound as follows:
$P(|S_n-nmu| geq cn^atheta) leq frac{1}{c^2n^{2a-1}}$.
Thus the largest $P(S_n=k)$ is roughly greater than $(1-c^{-2}n^{1-2a})(2cn^atheta+1)^{-1}$.
(Because this quantity is not greater than $P(|S_n-nmu| leq cn^atheta)|[nmu-cn^atheta,nmu+cn^atheta] cap mathbb{Z}|^{-1}$).
Take $a=1/2, c=2$, this quantity is greater than $frac{3/4}{4thetasqrt{n}+1}$, which proves the lower bound.
Edit: this was done without looking at Michael’s comment.
answered Jan 10 at 0:34
MindlackMindlack
4,910211
answered Jan 10 at 0:34
MindlackMindlack
4,910211
answered Jan 10 at 0:34
MindlackMindlack
4,910211
answered Jan 10 at 0:34
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
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$begingroup$
You can apply your Markov/Chebyshev inequality if you know the random variables have a finite variance (else you might need some type of truncation argument). That inequality you give is useful since there is an order-$sqrt{n}$ region about the mean such that we are there with probability at least $1/2$. And of course we must always be an integer.
$endgroup$
– Michael
Jan 3 at 6:02
$begingroup$
@Michael Yes you are right about variance, so it’s even harder to prove iron the general case (when we don’t necessarily have variance). Yet I still don’t see how to prove it when we have variance.
$endgroup$
– Jebfiffkkfnfolzbd
Jan 3 at 17:01
$begingroup$
You can just try to get one of the bounds at first (either the lower or upper). For example, what happens when most of the probability is concentrated on only a few of the options? Say, if probability $p$ is spread over $k$ options, there must be some option with at least...
$endgroup$
– Michael
Jan 3 at 18:06
$begingroup$
I don’t think this is true without extra conditions such as finite variance. For intuition with continuous variables, if ${X_i}$ are iid Cauchy random variables with PDF $f(x)$ then the average $S_n/n$ is Cauchy with the same PDF. So begin{align}P[S_n in [a,a+1]] &= P[S_n/n in [a/n, a/n+1/n]]\ &= int_{a/n}^{a/n+1/n} f(x)dx \&leq frac{f_{max}}{n} = O(1/n)end{align} Likely a "discrete Cauchy-type" random variable would have similar $O(1/n)$ behavior.
$endgroup$
– Michael
Jan 5 at 17:30
1
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@Jebfiffkkfnfolzbd : $X(Omega)$ is the set of all real numbers that $X$ can take, and it is always a subset of $mathbb{R}$ (for any random variable $X:Omega rightarrowmathbb{R}$). By "finite support" I think you mean that $X$ can only take a value from some finite interval $[a,b]$. This means that it indeed has a finite mean and variance. Since we also know that $X$ is integer valued, it means that $X$ can only take a finite number of (integer) values.
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– Michael
Jan 7 at 1:38