Upper bound $ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4}...
$begingroup$
Is it possible to upper bound the probability bellow using Chernoff bound?
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace, $$
where, $t$ and $a_{n}$ are constants. Also, $mathbf{N}(x_{i})=Y_{i}+1$ is a random variable with $Y_{i} sim Bin(n-1, r_{n}).$ The terms $Y_{i}$ are mutually dependent.
The problem is that I do not know the distribution of $sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}$. I only know the distribution of $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$ but I could not have this term in the probability.
I could also do,
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace leq n max_{i} mathbb{P}leftlbrace dfrac{1}{mathbf{N(x_{i})}}> dfrac{t}{4} +dfrac{1}{a_{n}} rightrbrace$$
Then, I use the Chernoff bound but the obtained bound is not tight enough. So, I want to use the distribution of term $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$.
calculus probability sequences-and-series probability-theory probability-distributions
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
$endgroup$
|
show 4 more comments
$begingroup$
Is it possible to upper bound the probability bellow using Chernoff bound?
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace, $$
where, $t$ and $a_{n}$ are constants. Also, $mathbf{N}(x_{i})=Y_{i}+1$ is a random variable with $Y_{i} sim Bin(n-1, r_{n}).$ The terms $Y_{i}$ are mutually dependent.
The problem is that I do not know the distribution of $sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}$. I only know the distribution of $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$ but I could not have this term in the probability.
I could also do,
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace leq n max_{i} mathbb{P}leftlbrace dfrac{1}{mathbf{N(x_{i})}}> dfrac{t}{4} +dfrac{1}{a_{n}} rightrbrace$$
Then, I use the Chernoff bound but the obtained bound is not tight enough. So, I want to use the distribution of term $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$.
calculus probability sequences-and-series probability-theory probability-distributions
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
$endgroup$
$begingroup$
You never need to know the distribution of the random variables to apply Hoeffding, Chernoff, or similar bounds. You need, roughly speaking (there are strenghtenings) the summands to be independent and bounded say in $[0,1]$. This is clearly the case here (at least, based on what you write, I assume the $N(x_i)$ are independent?).
$endgroup$
– Clement C.
Dec 14 '18 at 0:57
$begingroup$
Now, Chernoff may not give you the best bound you could possibly obtain, so it's for you to see if what it gives is enough for what you have in mind.
$endgroup$
– Clement C.
Dec 14 '18 at 0:59
$begingroup$
I cannot apply Chernoff bound here because I do not have independence between $mathbf{N(x_{i})}$ for each $i$.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 1:00
$begingroup$
Why do you not have independence? In your question, this is very unclear: $N_i = N(x_i)$ is "1+ a binomial." Where are the dependences?
$endgroup$
– Clement C.
Dec 14 '18 at 1:02
1
$begingroup$
Why did you not mention these dependencies in the question? that's the very core of the question.
$endgroup$
– Clement C.
Dec 14 '18 at 1:07
|
show 4 more comments
$begingroup$
Is it possible to upper bound the probability bellow using Chernoff bound?
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace, $$
where, $t$ and $a_{n}$ are constants. Also, $mathbf{N}(x_{i})=Y_{i}+1$ is a random variable with $Y_{i} sim Bin(n-1, r_{n}).$ The terms $Y_{i}$ are mutually dependent.
The problem is that I do not know the distribution of $sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}$. I only know the distribution of $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$ but I could not have this term in the probability.
I could also do,
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace leq n max_{i} mathbb{P}leftlbrace dfrac{1}{mathbf{N(x_{i})}}> dfrac{t}{4} +dfrac{1}{a_{n}} rightrbrace$$
Then, I use the Chernoff bound but the obtained bound is not tight enough. So, I want to use the distribution of term $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$.
calculus probability sequences-and-series probability-theory probability-distributions
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
$endgroup$
Is it possible to upper bound the probability bellow using Chernoff bound?
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace, $$
where, $t$ and $a_{n}$ are constants. Also, $mathbf{N}(x_{i})=Y_{i}+1$ is a random variable with $Y_{i} sim Bin(n-1, r_{n}).$ The terms $Y_{i}$ are mutually dependent.
The problem is that I do not know the distribution of $sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}$. I only know the distribution of $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$ but I could not have this term in the probability.
I could also do,
$$ mathbb{P}leftlbrace sumlimits_{substack{i}}^{n} dfrac{1}{mathbf{N(x_{i})}}> dfrac{tn}{4} +dfrac{n}{a_{n}} rightrbrace leq n max_{i} mathbb{P}leftlbrace dfrac{1}{mathbf{N(x_{i})}}> dfrac{t}{4} +dfrac{1}{a_{n}} rightrbrace$$
Then, I use the Chernoff bound but the obtained bound is not tight enough. So, I want to use the distribution of term $ sumlimits_{substack{i}}^{n} mathbf{N(x_{i})}$.
calculus probability sequences-and-series probability-theory probability-distributions
calculus probability sequences-and-series probability-theory probability-distributions
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
edited Dec 14 '18 at 1:10
Mounia Hamidouche
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
asked Dec 14 '18 at 0:44
Mounia HamidoucheMounia Hamidouche
32419
32419
$begingroup$
You never need to know the distribution of the random variables to apply Hoeffding, Chernoff, or similar bounds. You need, roughly speaking (there are strenghtenings) the summands to be independent and bounded say in $[0,1]$. This is clearly the case here (at least, based on what you write, I assume the $N(x_i)$ are independent?).
$endgroup$
– Clement C.
Dec 14 '18 at 0:57
$begingroup$
Now, Chernoff may not give you the best bound you could possibly obtain, so it's for you to see if what it gives is enough for what you have in mind.
$endgroup$
– Clement C.
Dec 14 '18 at 0:59
$begingroup$
I cannot apply Chernoff bound here because I do not have independence between $mathbf{N(x_{i})}$ for each $i$.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 1:00
$begingroup$
Why do you not have independence? In your question, this is very unclear: $N_i = N(x_i)$ is "1+ a binomial." Where are the dependences?
$endgroup$
– Clement C.
Dec 14 '18 at 1:02
1
$begingroup$
Why did you not mention these dependencies in the question? that's the very core of the question.
$endgroup$
– Clement C.
Dec 14 '18 at 1:07
|
show 4 more comments
$begingroup$
You never need to know the distribution of the random variables to apply Hoeffding, Chernoff, or similar bounds. You need, roughly speaking (there are strenghtenings) the summands to be independent and bounded say in $[0,1]$. This is clearly the case here (at least, based on what you write, I assume the $N(x_i)$ are independent?).
$endgroup$
– Clement C.
Dec 14 '18 at 0:57
$begingroup$
Now, Chernoff may not give you the best bound you could possibly obtain, so it's for you to see if what it gives is enough for what you have in mind.
$endgroup$
– Clement C.
Dec 14 '18 at 0:59
$begingroup$
I cannot apply Chernoff bound here because I do not have independence between $mathbf{N(x_{i})}$ for each $i$.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 1:00
$begingroup$
Why do you not have independence? In your question, this is very unclear: $N_i = N(x_i)$ is "1+ a binomial." Where are the dependences?
$endgroup$
– Clement C.
Dec 14 '18 at 1:02
1
$begingroup$
Why did you not mention these dependencies in the question? that's the very core of the question.
$endgroup$
– Clement C.
Dec 14 '18 at 1:07
$begingroup$
You never need to know the distribution of the random variables to apply Hoeffding, Chernoff, or similar bounds. You need, roughly speaking (there are strenghtenings) the summands to be independent and bounded say in $[0,1]$. This is clearly the case here (at least, based on what you write, I assume the $N(x_i)$ are independent?).
$endgroup$
– Clement C.
Dec 14 '18 at 0:57
$begingroup$
You never need to know the distribution of the random variables to apply Hoeffding, Chernoff, or similar bounds. You need, roughly speaking (there are strenghtenings) the summands to be independent and bounded say in $[0,1]$. This is clearly the case here (at least, based on what you write, I assume the $N(x_i)$ are independent?).
$endgroup$
– Clement C.
Dec 14 '18 at 0:57
$begingroup$
Now, Chernoff may not give you the best bound you could possibly obtain, so it's for you to see if what it gives is enough for what you have in mind.
$endgroup$
– Clement C.
Dec 14 '18 at 0:59
$begingroup$
Now, Chernoff may not give you the best bound you could possibly obtain, so it's for you to see if what it gives is enough for what you have in mind.
$endgroup$
– Clement C.
Dec 14 '18 at 0:59
$begingroup$
I cannot apply Chernoff bound here because I do not have independence between $mathbf{N(x_{i})}$ for each $i$.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 1:00
$begingroup$
I cannot apply Chernoff bound here because I do not have independence between $mathbf{N(x_{i})}$ for each $i$.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 1:00
$begingroup$
Why do you not have independence? In your question, this is very unclear: $N_i = N(x_i)$ is "1+ a binomial." Where are the dependences?
$endgroup$
– Clement C.
Dec 14 '18 at 1:02
$begingroup$
Why do you not have independence? In your question, this is very unclear: $N_i = N(x_i)$ is "1+ a binomial." Where are the dependences?
$endgroup$
– Clement C.
Dec 14 '18 at 1:02
1
1
$begingroup$
Why did you not mention these dependencies in the question? that's the very core of the question.
$endgroup$
– Clement C.
Dec 14 '18 at 1:07
$begingroup$
Why did you not mention these dependencies in the question? that's the very core of the question.
$endgroup$
– Clement C.
Dec 14 '18 at 1:07
|
show 4 more comments
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$begingroup$
You never need to know the distribution of the random variables to apply Hoeffding, Chernoff, or similar bounds. You need, roughly speaking (there are strenghtenings) the summands to be independent and bounded say in $[0,1]$. This is clearly the case here (at least, based on what you write, I assume the $N(x_i)$ are independent?).
$endgroup$
– Clement C.
Dec 14 '18 at 0:57
$begingroup$
Now, Chernoff may not give you the best bound you could possibly obtain, so it's for you to see if what it gives is enough for what you have in mind.
$endgroup$
– Clement C.
Dec 14 '18 at 0:59
$begingroup$
I cannot apply Chernoff bound here because I do not have independence between $mathbf{N(x_{i})}$ for each $i$.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 1:00
$begingroup$
Why do you not have independence? In your question, this is very unclear: $N_i = N(x_i)$ is "1+ a binomial." Where are the dependences?
$endgroup$
– Clement C.
Dec 14 '18 at 1:02
1
$begingroup$
Why did you not mention these dependencies in the question? that's the very core of the question.
$endgroup$
– Clement C.
Dec 14 '18 at 1:07