Proof of a technical fact in the book of Schapire and Freund on boosting
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I am currently looking at Exercise 10.3, Chapter 10 of the book on Boosting by Schapire and Freund. More precisely, in the middle of the exercise they propose to use, without proof, the technical fact summarized below. Obviously, since it can be used without proof, I am now curious to know how to prove it!
To summarize the problem, let $mathcal{H}$ bet a set of functions $h : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. Define $text{co}(mathcal{H})$ as
begin{align*}
text{co}(mathcal{H}) = leftlbrace f : x,bar{y} mapsto sum_{t=1}^T a_t h_t(x,bar{y}) left| a_1,ldots,a_T geq 0; sum_{t=1}^Ta_t = 1; h_1,ldots h_T in mathcal{H}; Tgeq 1 right. rightrbracetext{.}
end{align*}
Notice that $f : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. For $f in text{co}left(mathcal{H}right)$, $eta > 0$, $bar{K} = |mathcal{bar{Y}}|$, and $(x,y) in mathcal{X} times mathcal{Y}$, let
begin{align*}
nu_{f,eta}(x,y) = - frac{1}{eta} lnleft(frac{1}{bar{K}} sum_{bar{y} in mathcal{bar{Y}}} expBig(-eta Omega(y,bar{y}) f(x,bar{y})Big)right)
end{align*}
where $Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$ and $-1$ otherwise. $Omega(y)$ associates each element from $mathcal{Y}$ to a subset of $mathcal{bar{Y}}$. Notice that $nu_{f,eta} : mathcal{X} times mathcal{Y} rightarrow [-1,1]$.
The technical fact is as follows. Let $1 geq theta > 0$ and define the grid:
begin{align*}
varepsilon_theta = leftlbrace frac{4lnbar{K}}{itheta} : i = 1, ldots, leftlceil frac{8lnbar{K}}{theta^2} rightrceil rightrbracetext{.}
end{align*}
For any $eta > 0$, let $hat{eta}$ be the closest value in $varepsilon_theta$ to $eta$. Then for all $f in text{co}(mathcal{H})$ and for all $(x,y) in mathcal{X} times mathcal{Y}$,
begin{align*}
left| nu_{f,eta}(x,y) - nu_{f,hat{eta}}(x,y) right| leq frac{theta}{4}text{.}
end{align*}
So far, I proved the statement when $eta > frac{4lnbar{K}}{theta}$ (using the properties of the LogSumExp function). Furthermore, using the grid, I showed that
begin{align*}
&& left| eta - hat{eta} right| leq frac{ln bar{K}}{theta} \
&Rightarrow& left| etanu_{f,eta}(x,y) - hat{eta}nu_{f,hat{eta}}(x,y) right| leq frac{ln bar{K}}{theta}text{.}
end{align*}
However, I did not manage to go further than that.
Am I going in the right direction? If yes, what would be the trick for the last step? If no, what method should I consider to prove this statement? Note that I am not asking for a full proof, but rather some hints on how to proceed to show the result.
real-analysis statistics inequality
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up vote
5
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I am currently looking at Exercise 10.3, Chapter 10 of the book on Boosting by Schapire and Freund. More precisely, in the middle of the exercise they propose to use, without proof, the technical fact summarized below. Obviously, since it can be used without proof, I am now curious to know how to prove it!
To summarize the problem, let $mathcal{H}$ bet a set of functions $h : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. Define $text{co}(mathcal{H})$ as
begin{align*}
text{co}(mathcal{H}) = leftlbrace f : x,bar{y} mapsto sum_{t=1}^T a_t h_t(x,bar{y}) left| a_1,ldots,a_T geq 0; sum_{t=1}^Ta_t = 1; h_1,ldots h_T in mathcal{H}; Tgeq 1 right. rightrbracetext{.}
end{align*}
Notice that $f : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. For $f in text{co}left(mathcal{H}right)$, $eta > 0$, $bar{K} = |mathcal{bar{Y}}|$, and $(x,y) in mathcal{X} times mathcal{Y}$, let
begin{align*}
nu_{f,eta}(x,y) = - frac{1}{eta} lnleft(frac{1}{bar{K}} sum_{bar{y} in mathcal{bar{Y}}} expBig(-eta Omega(y,bar{y}) f(x,bar{y})Big)right)
end{align*}
where $Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$ and $-1$ otherwise. $Omega(y)$ associates each element from $mathcal{Y}$ to a subset of $mathcal{bar{Y}}$. Notice that $nu_{f,eta} : mathcal{X} times mathcal{Y} rightarrow [-1,1]$.
The technical fact is as follows. Let $1 geq theta > 0$ and define the grid:
begin{align*}
varepsilon_theta = leftlbrace frac{4lnbar{K}}{itheta} : i = 1, ldots, leftlceil frac{8lnbar{K}}{theta^2} rightrceil rightrbracetext{.}
end{align*}
For any $eta > 0$, let $hat{eta}$ be the closest value in $varepsilon_theta$ to $eta$. Then for all $f in text{co}(mathcal{H})$ and for all $(x,y) in mathcal{X} times mathcal{Y}$,
begin{align*}
left| nu_{f,eta}(x,y) - nu_{f,hat{eta}}(x,y) right| leq frac{theta}{4}text{.}
end{align*}
So far, I proved the statement when $eta > frac{4lnbar{K}}{theta}$ (using the properties of the LogSumExp function). Furthermore, using the grid, I showed that
begin{align*}
&& left| eta - hat{eta} right| leq frac{ln bar{K}}{theta} \
&Rightarrow& left| etanu_{f,eta}(x,y) - hat{eta}nu_{f,hat{eta}}(x,y) right| leq frac{ln bar{K}}{theta}text{.}
end{align*}
However, I did not manage to go further than that.
Am I going in the right direction? If yes, what would be the trick for the last step? If no, what method should I consider to prove this statement? Note that I am not asking for a full proof, but rather some hints on how to proceed to show the result.
real-analysis statistics inequality
What do you mean by "$Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$"? $bar yin mathcal{bar Y}$ and $Omega(x)inmathcal{bar Y}$. How can $bar{y} in Omega(y)$?
– Hans
Nov 24 at 8:30
$Omega(y)$ is a function that associates each element in $mathcal{Y}$ to one or several elements in $bar{mathcal{Y}}$. In other words $Omega(y) subseteq bar{mathcal{Y}}$. I modified the question to clarify this point.
– M. P.
Nov 26 at 12:52
add a comment |
up vote
5
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up vote
5
down vote
favorite
I am currently looking at Exercise 10.3, Chapter 10 of the book on Boosting by Schapire and Freund. More precisely, in the middle of the exercise they propose to use, without proof, the technical fact summarized below. Obviously, since it can be used without proof, I am now curious to know how to prove it!
To summarize the problem, let $mathcal{H}$ bet a set of functions $h : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. Define $text{co}(mathcal{H})$ as
begin{align*}
text{co}(mathcal{H}) = leftlbrace f : x,bar{y} mapsto sum_{t=1}^T a_t h_t(x,bar{y}) left| a_1,ldots,a_T geq 0; sum_{t=1}^Ta_t = 1; h_1,ldots h_T in mathcal{H}; Tgeq 1 right. rightrbracetext{.}
end{align*}
Notice that $f : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. For $f in text{co}left(mathcal{H}right)$, $eta > 0$, $bar{K} = |mathcal{bar{Y}}|$, and $(x,y) in mathcal{X} times mathcal{Y}$, let
begin{align*}
nu_{f,eta}(x,y) = - frac{1}{eta} lnleft(frac{1}{bar{K}} sum_{bar{y} in mathcal{bar{Y}}} expBig(-eta Omega(y,bar{y}) f(x,bar{y})Big)right)
end{align*}
where $Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$ and $-1$ otherwise. $Omega(y)$ associates each element from $mathcal{Y}$ to a subset of $mathcal{bar{Y}}$. Notice that $nu_{f,eta} : mathcal{X} times mathcal{Y} rightarrow [-1,1]$.
The technical fact is as follows. Let $1 geq theta > 0$ and define the grid:
begin{align*}
varepsilon_theta = leftlbrace frac{4lnbar{K}}{itheta} : i = 1, ldots, leftlceil frac{8lnbar{K}}{theta^2} rightrceil rightrbracetext{.}
end{align*}
For any $eta > 0$, let $hat{eta}$ be the closest value in $varepsilon_theta$ to $eta$. Then for all $f in text{co}(mathcal{H})$ and for all $(x,y) in mathcal{X} times mathcal{Y}$,
begin{align*}
left| nu_{f,eta}(x,y) - nu_{f,hat{eta}}(x,y) right| leq frac{theta}{4}text{.}
end{align*}
So far, I proved the statement when $eta > frac{4lnbar{K}}{theta}$ (using the properties of the LogSumExp function). Furthermore, using the grid, I showed that
begin{align*}
&& left| eta - hat{eta} right| leq frac{ln bar{K}}{theta} \
&Rightarrow& left| etanu_{f,eta}(x,y) - hat{eta}nu_{f,hat{eta}}(x,y) right| leq frac{ln bar{K}}{theta}text{.}
end{align*}
However, I did not manage to go further than that.
Am I going in the right direction? If yes, what would be the trick for the last step? If no, what method should I consider to prove this statement? Note that I am not asking for a full proof, but rather some hints on how to proceed to show the result.
real-analysis statistics inequality
I am currently looking at Exercise 10.3, Chapter 10 of the book on Boosting by Schapire and Freund. More precisely, in the middle of the exercise they propose to use, without proof, the technical fact summarized below. Obviously, since it can be used without proof, I am now curious to know how to prove it!
To summarize the problem, let $mathcal{H}$ bet a set of functions $h : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. Define $text{co}(mathcal{H})$ as
begin{align*}
text{co}(mathcal{H}) = leftlbrace f : x,bar{y} mapsto sum_{t=1}^T a_t h_t(x,bar{y}) left| a_1,ldots,a_T geq 0; sum_{t=1}^Ta_t = 1; h_1,ldots h_T in mathcal{H}; Tgeq 1 right. rightrbracetext{.}
end{align*}
Notice that $f : mathcal{X} times mathcal{bar{Y}} rightarrow [-1,1]$. For $f in text{co}left(mathcal{H}right)$, $eta > 0$, $bar{K} = |mathcal{bar{Y}}|$, and $(x,y) in mathcal{X} times mathcal{Y}$, let
begin{align*}
nu_{f,eta}(x,y) = - frac{1}{eta} lnleft(frac{1}{bar{K}} sum_{bar{y} in mathcal{bar{Y}}} expBig(-eta Omega(y,bar{y}) f(x,bar{y})Big)right)
end{align*}
where $Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$ and $-1$ otherwise. $Omega(y)$ associates each element from $mathcal{Y}$ to a subset of $mathcal{bar{Y}}$. Notice that $nu_{f,eta} : mathcal{X} times mathcal{Y} rightarrow [-1,1]$.
The technical fact is as follows. Let $1 geq theta > 0$ and define the grid:
begin{align*}
varepsilon_theta = leftlbrace frac{4lnbar{K}}{itheta} : i = 1, ldots, leftlceil frac{8lnbar{K}}{theta^2} rightrceil rightrbracetext{.}
end{align*}
For any $eta > 0$, let $hat{eta}$ be the closest value in $varepsilon_theta$ to $eta$. Then for all $f in text{co}(mathcal{H})$ and for all $(x,y) in mathcal{X} times mathcal{Y}$,
begin{align*}
left| nu_{f,eta}(x,y) - nu_{f,hat{eta}}(x,y) right| leq frac{theta}{4}text{.}
end{align*}
So far, I proved the statement when $eta > frac{4lnbar{K}}{theta}$ (using the properties of the LogSumExp function). Furthermore, using the grid, I showed that
begin{align*}
&& left| eta - hat{eta} right| leq frac{ln bar{K}}{theta} \
&Rightarrow& left| etanu_{f,eta}(x,y) - hat{eta}nu_{f,hat{eta}}(x,y) right| leq frac{ln bar{K}}{theta}text{.}
end{align*}
However, I did not manage to go further than that.
Am I going in the right direction? If yes, what would be the trick for the last step? If no, what method should I consider to prove this statement? Note that I am not asking for a full proof, but rather some hints on how to proceed to show the result.
real-analysis statistics inequality
real-analysis statistics inequality
edited yesterday
asked Nov 16 at 11:04
M. P.
37117
37117
What do you mean by "$Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$"? $bar yin mathcal{bar Y}$ and $Omega(x)inmathcal{bar Y}$. How can $bar{y} in Omega(y)$?
– Hans
Nov 24 at 8:30
$Omega(y)$ is a function that associates each element in $mathcal{Y}$ to one or several elements in $bar{mathcal{Y}}$. In other words $Omega(y) subseteq bar{mathcal{Y}}$. I modified the question to clarify this point.
– M. P.
Nov 26 at 12:52
add a comment |
What do you mean by "$Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$"? $bar yin mathcal{bar Y}$ and $Omega(x)inmathcal{bar Y}$. How can $bar{y} in Omega(y)$?
– Hans
Nov 24 at 8:30
$Omega(y)$ is a function that associates each element in $mathcal{Y}$ to one or several elements in $bar{mathcal{Y}}$. In other words $Omega(y) subseteq bar{mathcal{Y}}$. I modified the question to clarify this point.
– M. P.
Nov 26 at 12:52
What do you mean by "$Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$"? $bar yin mathcal{bar Y}$ and $Omega(x)inmathcal{bar Y}$. How can $bar{y} in Omega(y)$?
– Hans
Nov 24 at 8:30
What do you mean by "$Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$"? $bar yin mathcal{bar Y}$ and $Omega(x)inmathcal{bar Y}$. How can $bar{y} in Omega(y)$?
– Hans
Nov 24 at 8:30
$Omega(y)$ is a function that associates each element in $mathcal{Y}$ to one or several elements in $bar{mathcal{Y}}$. In other words $Omega(y) subseteq bar{mathcal{Y}}$. I modified the question to clarify this point.
– M. P.
Nov 26 at 12:52
$Omega(y)$ is a function that associates each element in $mathcal{Y}$ to one or several elements in $bar{mathcal{Y}}$. In other words $Omega(y) subseteq bar{mathcal{Y}}$. I modified the question to clarify this point.
– M. P.
Nov 26 at 12:52
add a comment |
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What do you mean by "$Omega(y,bar{y}) = 1$ if $bar{y} in Omega(y)$"? $bar yin mathcal{bar Y}$ and $Omega(x)inmathcal{bar Y}$. How can $bar{y} in Omega(y)$?
– Hans
Nov 24 at 8:30
$Omega(y)$ is a function that associates each element in $mathcal{Y}$ to one or several elements in $bar{mathcal{Y}}$. In other words $Omega(y) subseteq bar{mathcal{Y}}$. I modified the question to clarify this point.
– M. P.
Nov 26 at 12:52