Chernoff bound from theory to pratice
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i have a question about Chernoff bound.
I think that i understood the sense of the formula, but i m not sure how to apply for my case.
I have $x=(x(1),...x(d))$ that i-th component $ x(i) $ chosen uniformly random in $[0,1]$
d is a distance from two point and $Theta(d)$ is the largest possibile distance d.
i want to show that my squared euclidean distance between $x$ and $y$ is at least
$(1-epsilon)cd $ with probability $1-e^{Theta(epsilon^2 d)}$ with $c<1$.
i think that i need to apply Markov in this way
assuming that $X=(x-y)^2$
$Pr(X geq (1-epsilon)cd leq frac{E(X)}{1-epsilon})$
i calculated expectation of $E(X)$
but i dont know where I need to use the probability $1-e^{Theta(epsilon^2 d)}$
as i read here https://stats.stackexchange.com/questions/20108/link-between-variance-and-pairwise-distances-within-a-variable for my problem i can use this formula
$E(X_1−X_2)^2=2Var(X_1)$ for calculate the variance, but i see that variance need for Chebyshev inequality
Sameone can help me to understand how to apply this bound?
i m following this book https://www.cs.cornell.edu/jeh/book.pdf page 12 (cap 2.2)
probability probability-theory inequality probability-distributions
asked Dec 4 at 15:38
theantomc
11
|
show 2 more comments
up vote
-2
down vote
favorite
i have a question about Chernoff bound.
I think that i understood the sense of the formula, but i m not sure how to apply for my case.
I have $x=(x(1),...x(d))$ that i-th component $ x(i) $ chosen uniformly random in $[0,1]$
d is a distance from two point and $Theta(d)$ is the largest possibile distance d.
i want to show that my squared euclidean distance between $x$ and $y$ is at least
$(1-epsilon)cd $ with probability $1-e^{Theta(epsilon^2 d)}$ with $c<1$.
i think that i need to apply Markov in this way
assuming that $X=(x-y)^2$
$Pr(X geq (1-epsilon)cd leq frac{E(X)}{1-epsilon})$
i calculated expectation of $E(X)$
but i dont know where I need to use the probability $1-e^{Theta(epsilon^2 d)}$
as i read here https://stats.stackexchange.com/questions/20108/link-between-variance-and-pairwise-distances-within-a-variable for my problem i can use this formula
$E(X_1−X_2)^2=2Var(X_1)$ for calculate the variance, but i see that variance need for Chebyshev inequality
Sameone can help me to understand how to apply this bound?
i m following this book https://www.cs.cornell.edu/jeh/book.pdf page 12 (cap 2.2)
probability probability-theory inequality probability-distributions
asked Dec 4 at 15:38
theantomc
11
I am (partially) working on probability theory, and have used Chernoff bounds myself. With that being said, your post is very confusing, to the point that I don't understand your problems at all. Could you rephrase your sentences, please. It would increase the probability that you get a meaningful answer. E.g.: "I have uniformly at random variable (that is a condition to apply the Chernoff B.)" What does this mean?
– A. Pongrácz
Dec 4 at 15:44
I reorganize and add my link resource where i m studying the argument. , thanks for your answer @A.Pongrácz
– theantomc
Dec 4 at 15:53
In order to get an exponential inequality, you would need to use the fact that $mathbb{P}(X geq t) = mathbb{P}(e^{sX} geq e^{st}), ; forall s > 0$.
– VHarisop
Dec 4 at 16:44
@VHarisop i dont caught the hint
– theantomc
Dec 4 at 16:58
The "hint" is that directly using the Markov inequality on the probability $ mathbb{P}(X geq t)$ is going to give you too weak of a result.
– VHarisop
Dec 4 at 17:14
|
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
i have a question about Chernoff bound.
I think that i understood the sense of the formula, but i m not sure how to apply for my case.
I have $x=(x(1),...x(d))$ that i-th component $ x(i) $ chosen uniformly random in $[0,1]$
d is a distance from two point and $Theta(d)$ is the largest possibile distance d.
i want to show that my squared euclidean distance between $x$ and $y$ is at least
$(1-epsilon)cd $ with probability $1-e^{Theta(epsilon^2 d)}$ with $c<1$.
i think that i need to apply Markov in this way
assuming that $X=(x-y)^2$
$Pr(X geq (1-epsilon)cd leq frac{E(X)}{1-epsilon})$
i calculated expectation of $E(X)$
but i dont know where I need to use the probability $1-e^{Theta(epsilon^2 d)}$
as i read here https://stats.stackexchange.com/questions/20108/link-between-variance-and-pairwise-distances-within-a-variable for my problem i can use this formula
$E(X_1−X_2)^2=2Var(X_1)$ for calculate the variance, but i see that variance need for Chebyshev inequality
Sameone can help me to understand how to apply this bound?
i m following this book https://www.cs.cornell.edu/jeh/book.pdf page 12 (cap 2.2)
probability probability-theory inequality probability-distributions
asked Dec 4 at 15:38
theantomc
11
i have a question about Chernoff bound.
I think that i understood the sense of the formula, but i m not sure how to apply for my case.
I have $x=(x(1),...x(d))$ that i-th component $ x(i) $ chosen uniformly random in $[0,1]$
d is a distance from two point and $Theta(d)$ is the largest possibile distance d.
i want to show that my squared euclidean distance between $x$ and $y$ is at least
$(1-epsilon)cd $ with probability $1-e^{Theta(epsilon^2 d)}$ with $c<1$.
i think that i need to apply Markov in this way
assuming that $X=(x-y)^2$
$Pr(X geq (1-epsilon)cd leq frac{E(X)}{1-epsilon})$
i calculated expectation of $E(X)$
but i dont know where I need to use the probability $1-e^{Theta(epsilon^2 d)}$
as i read here https://stats.stackexchange.com/questions/20108/link-between-variance-and-pairwise-distances-within-a-variable for my problem i can use this formula
$E(X_1−X_2)^2=2Var(X_1)$ for calculate the variance, but i see that variance need for Chebyshev inequality
Sameone can help me to understand how to apply this bound?
i m following this book https://www.cs.cornell.edu/jeh/book.pdf page 12 (cap 2.2)
probability probability-theory inequality probability-distributions
probability probability-theory inequality probability-distributions
asked Dec 4 at 15:38
theantomc
11
asked Dec 4 at 15:38
theantomc
11
edited Dec 4 at 16:56
asked Dec 4 at 15:38
theantomc
11
asked Dec 4 at 15:38
theantomc
11
asked Dec 4 at 15:38
theantomc
11
11
I am (partially) working on probability theory, and have used Chernoff bounds myself. With that being said, your post is very confusing, to the point that I don't understand your problems at all. Could you rephrase your sentences, please. It would increase the probability that you get a meaningful answer. E.g.: "I have uniformly at random variable (that is a condition to apply the Chernoff B.)" What does this mean?
– A. Pongrácz
Dec 4 at 15:44
I reorganize and add my link resource where i m studying the argument. , thanks for your answer @A.Pongrácz
– theantomc
Dec 4 at 15:53
In order to get an exponential inequality, you would need to use the fact that $mathbb{P}(X geq t) = mathbb{P}(e^{sX} geq e^{st}), ; forall s > 0$.
– VHarisop
Dec 4 at 16:44
@VHarisop i dont caught the hint
– theantomc
Dec 4 at 16:58
The "hint" is that directly using the Markov inequality on the probability $ mathbb{P}(X geq t)$ is going to give you too weak of a result.
– VHarisop
Dec 4 at 17:14
|
show 2 more comments
I am (partially) working on probability theory, and have used Chernoff bounds myself. With that being said, your post is very confusing, to the point that I don't understand your problems at all. Could you rephrase your sentences, please. It would increase the probability that you get a meaningful answer. E.g.: "I have uniformly at random variable (that is a condition to apply the Chernoff B.)" What does this mean?
– A. Pongrácz
Dec 4 at 15:44
I reorganize and add my link resource where i m studying the argument. , thanks for your answer @A.Pongrácz
– theantomc
Dec 4 at 15:53
In order to get an exponential inequality, you would need to use the fact that $mathbb{P}(X geq t) = mathbb{P}(e^{sX} geq e^{st}), ; forall s > 0$.
– VHarisop
Dec 4 at 16:44
@VHarisop i dont caught the hint
– theantomc
Dec 4 at 16:58
The "hint" is that directly using the Markov inequality on the probability $ mathbb{P}(X geq t)$ is going to give you too weak of a result.
– VHarisop
Dec 4 at 17:14
I am (partially) working on probability theory, and have used Chernoff bounds myself. With that being said, your post is very confusing, to the point that I don't understand your problems at all. Could you rephrase your sentences, please. It would increase the probability that you get a meaningful answer. E.g.: "I have uniformly at random variable (that is a condition to apply the Chernoff B.)" What does this mean?
– A. Pongrácz
Dec 4 at 15:44
I am (partially) working on probability theory, and have used Chernoff bounds myself. With that being said, your post is very confusing, to the point that I don't understand your problems at all. Could you rephrase your sentences, please. It would increase the probability that you get a meaningful answer. E.g.: "I have uniformly at random variable (that is a condition to apply the Chernoff B.)" What does this mean?
– A. Pongrácz
Dec 4 at 15:44
I reorganize and add my link resource where i m studying the argument. , thanks for your answer @A.Pongrácz
– theantomc
Dec 4 at 15:53
I reorganize and add my link resource where i m studying the argument. , thanks for your answer @A.Pongrácz
– theantomc
Dec 4 at 15:53
In order to get an exponential inequality, you would need to use the fact that $mathbb{P}(X geq t) = mathbb{P}(e^{sX} geq e^{st}), ; forall s > 0$.
– VHarisop
Dec 4 at 16:44
In order to get an exponential inequality, you would need to use the fact that $mathbb{P}(X geq t) = mathbb{P}(e^{sX} geq e^{st}), ; forall s > 0$.
– VHarisop
Dec 4 at 16:44
@VHarisop i dont caught the hint
– theantomc
Dec 4 at 16:58
@VHarisop i dont caught the hint
– theantomc
Dec 4 at 16:58
The "hint" is that directly using the Markov inequality on the probability $ mathbb{P}(X geq t)$ is going to give you too weak of a result.
– VHarisop
Dec 4 at 17:14
The "hint" is that directly using the Markov inequality on the probability $ mathbb{P}(X geq t)$ is going to give you too weak of a result.
– VHarisop
Dec 4 at 17:14
|
show 2 more comments
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I am (partially) working on probability theory, and have used Chernoff bounds myself. With that being said, your post is very confusing, to the point that I don't understand your problems at all. Could you rephrase your sentences, please. It would increase the probability that you get a meaningful answer. E.g.: "I have uniformly at random variable (that is a condition to apply the Chernoff B.)" What does this mean?
– A. Pongrácz
Dec 4 at 15:44
I reorganize and add my link resource where i m studying the argument. , thanks for your answer @A.Pongrácz
– theantomc
Dec 4 at 15:53
In order to get an exponential inequality, you would need to use the fact that $mathbb{P}(X geq t) = mathbb{P}(e^{sX} geq e^{st}), ; forall s > 0$.
– VHarisop
Dec 4 at 16:44
@VHarisop i dont caught the hint
– theantomc
Dec 4 at 16:58
The "hint" is that directly using the Markov inequality on the probability $ mathbb{P}(X geq t)$ is going to give you too weak of a result.
– VHarisop
Dec 4 at 17:14