Mixing
Derivaton of Chernofff's bound on Bayes error for Multivariate Gaussian distribution
$begingroup$
I was following derivation of Chernoffs bound for Bayes error Given in the book Pattern recognition by Duda Hart and Stork. However there is a minor difference between the results in book and details i have worked out. Please guid me where I am wrong.
We know that Bayes probabilities for classifying $x$ in two classes $omega_1$ and $omega_2$ are given by $P(omega_1)p(x|omega_1)$ and $P(omega_2)p(x|omega_2)$. So Bayes error is $int_x min(P(omega_1)p(x|omega_1), P(omega_2)p(x|omega_2)) dx $.
We note that $min(a,b) le a^beta b^{(1-beta)}$ for $a, b>0$, so Bayes error is bounded by $P(omega_1)^beta P(omega_2)^ {1-beta}int_x p(x|omega_1)^beta p(x|omega_2)^{1-beta} dx$.
We consider a case where class conditional probabilities are Normal $p(x|omega_1)$ is given by $mathcal N(mu_1, Sigma_1)$ and $p(x|omega_2)$ is given by $mathcal N(mu_2, Sigma_2)$. then Bayes error is bounded by
$P(omega_1)^beta P(omega_2)^ {1-beta}int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$.
$int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$
$=frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx$ where $L =(x-mu_1)^tbeta Sigma_1^{-1}(x-mu_1)+ (x-mu_2)^t(1-beta) Sigma_2^{-1}(x-mu_12)$
Let $A= betaSigma_1^{-1}$ , $B= (1-beta)Sigma_2^{-1}$, $y=(x-mu_1)$ and $a=mu_2-mu_1$
Then $L=y^tAy+(y-a)^tB(y-a)$
$=y^t(A+B)y -a^tBy -y^tBa+a^tBa$
$=y^t(A+B)y-2a^tBy+a^tBa$ , as B is symmetric
As (A+B) is positive definite matrix it can be composed into a matrix $P$ such that $P^tP=(A+B)$
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy +a^tBa$
Note that $(a^tB(A+B)^{-1}P^t)(a^tB(A+B)^{-1}P^t)^t= a^tB(A+B)^{-1}P^tP(A+B)^{-1}Ba = a^tB(A+B)^{-1}Ba$
Also
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy + a^tB(A+B)^{-1}Ba - a^tB(A+B)^{-1}Ba +a^tBa$
Note that $-a^tB(A+B)^{-1}Ba +a^tBa= a^t[B-B(A+B)^{-1}B]a= a^t[B(A+B)^{-1}(A+B)-B(A+B)^{-1}B]a= a^tB(A+B)^{-1}Aa$
Let $z=y-(A+B)^{-1}Ba=x-mu_1-(A+B)^{-1}Ba$
$L= (Pz)^tPz+ a^tB(A+B)^{-1}Aa$
$= z^t(A+B)z+ a^tB(A+B)^{-1}Aa$
So,
$frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx=
exp^{-frac{1}{2}a^tB(A+B)^{-1}Aa}frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_zexp^{-frac{1}{2}z^t(A+B)z} dz $
Note that $(A+B)=A(A^{-1}+B^{-1})B$
So $(A+B)^{-1}=B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}$
Also it is known that $int_zexp^{-frac{1}{2}z^t(A+B)z} dz = sqrt{(2 pi)^d |(A+B)^{-1}|} $
So $exp^{-1/2L}$ is given by $exp^{-frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a}sqrt{frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} }$
Expressing $exp^{-1/2L}$ as $exp^{-k}$
$k= frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a - frac{1}{2} log frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} $
$A^{-1}+B^{-1}= frac{Sigma_1}{beta} + frac{Sigma_2}{1-beta}= frac{(1-beta) Sigma_1 + beta Sigma_2}{beta(1-beta)} $
$|(A+B)^{-1}|=|B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}|= frac{|Sigma_1||Sigma_2|}{|(1-beta) Sigma_1 + beta Sigma_2|}$
So $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_1 + beta Sigma_2]^{-1}a - frac{1}{2} log frac{|Sigma_1|^{1-beta}|Sigma_2|^beta}{|(1-beta) Sigma_1 + beta Sigma_2|} $
In the book $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_2 + beta Sigma_1]^{-1}a - frac{1}{2} log frac{|Sigma_2|^{1-beta}|Sigma_1|^beta}{|(1-beta) Sigma_2 + beta Sigma_1|} $. But from the diagrams function does not seem be be symmetric on $beta$. Please let me know where I have gone wrong
probability bayesian
asked Jan 11 at 7:46
CuriousCurious
889516
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add a comment |
$begingroup$
I was following derivation of Chernoffs bound for Bayes error Given in the book Pattern recognition by Duda Hart and Stork. However there is a minor difference between the results in book and details i have worked out. Please guid me where I am wrong.
We know that Bayes probabilities for classifying $x$ in two classes $omega_1$ and $omega_2$ are given by $P(omega_1)p(x|omega_1)$ and $P(omega_2)p(x|omega_2)$. So Bayes error is $int_x min(P(omega_1)p(x|omega_1), P(omega_2)p(x|omega_2)) dx $.
We note that $min(a,b) le a^beta b^{(1-beta)}$ for $a, b>0$, so Bayes error is bounded by $P(omega_1)^beta P(omega_2)^ {1-beta}int_x p(x|omega_1)^beta p(x|omega_2)^{1-beta} dx$.
We consider a case where class conditional probabilities are Normal $p(x|omega_1)$ is given by $mathcal N(mu_1, Sigma_1)$ and $p(x|omega_2)$ is given by $mathcal N(mu_2, Sigma_2)$. then Bayes error is bounded by
$P(omega_1)^beta P(omega_2)^ {1-beta}int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$.
$int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$
$=frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx$ where $L =(x-mu_1)^tbeta Sigma_1^{-1}(x-mu_1)+ (x-mu_2)^t(1-beta) Sigma_2^{-1}(x-mu_12)$
Let $A= betaSigma_1^{-1}$ , $B= (1-beta)Sigma_2^{-1}$, $y=(x-mu_1)$ and $a=mu_2-mu_1$
Then $L=y^tAy+(y-a)^tB(y-a)$
$=y^t(A+B)y -a^tBy -y^tBa+a^tBa$
$=y^t(A+B)y-2a^tBy+a^tBa$ , as B is symmetric
As (A+B) is positive definite matrix it can be composed into a matrix $P$ such that $P^tP=(A+B)$
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy +a^tBa$
Note that $(a^tB(A+B)^{-1}P^t)(a^tB(A+B)^{-1}P^t)^t= a^tB(A+B)^{-1}P^tP(A+B)^{-1}Ba = a^tB(A+B)^{-1}Ba$
Also
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy + a^tB(A+B)^{-1}Ba - a^tB(A+B)^{-1}Ba +a^tBa$
Note that $-a^tB(A+B)^{-1}Ba +a^tBa= a^t[B-B(A+B)^{-1}B]a= a^t[B(A+B)^{-1}(A+B)-B(A+B)^{-1}B]a= a^tB(A+B)^{-1}Aa$
Let $z=y-(A+B)^{-1}Ba=x-mu_1-(A+B)^{-1}Ba$
$L= (Pz)^tPz+ a^tB(A+B)^{-1}Aa$
$= z^t(A+B)z+ a^tB(A+B)^{-1}Aa$
So,
$frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx=
exp^{-frac{1}{2}a^tB(A+B)^{-1}Aa}frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_zexp^{-frac{1}{2}z^t(A+B)z} dz $
Note that $(A+B)=A(A^{-1}+B^{-1})B$
So $(A+B)^{-1}=B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}$
Also it is known that $int_zexp^{-frac{1}{2}z^t(A+B)z} dz = sqrt{(2 pi)^d |(A+B)^{-1}|} $
So $exp^{-1/2L}$ is given by $exp^{-frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a}sqrt{frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} }$
Expressing $exp^{-1/2L}$ as $exp^{-k}$
$k= frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a - frac{1}{2} log frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} $
$A^{-1}+B^{-1}= frac{Sigma_1}{beta} + frac{Sigma_2}{1-beta}= frac{(1-beta) Sigma_1 + beta Sigma_2}{beta(1-beta)} $
$|(A+B)^{-1}|=|B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}|= frac{|Sigma_1||Sigma_2|}{|(1-beta) Sigma_1 + beta Sigma_2|}$
So $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_1 + beta Sigma_2]^{-1}a - frac{1}{2} log frac{|Sigma_1|^{1-beta}|Sigma_2|^beta}{|(1-beta) Sigma_1 + beta Sigma_2|} $
In the book $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_2 + beta Sigma_1]^{-1}a - frac{1}{2} log frac{|Sigma_2|^{1-beta}|Sigma_1|^beta}{|(1-beta) Sigma_2 + beta Sigma_1|} $. But from the diagrams function does not seem be be symmetric on $beta$. Please let me know where I have gone wrong
probability bayesian
asked Jan 11 at 7:46
CuriousCurious
889516
$endgroup$
add a comment |
$begingroup$
I was following derivation of Chernoffs bound for Bayes error Given in the book Pattern recognition by Duda Hart and Stork. However there is a minor difference between the results in book and details i have worked out. Please guid me where I am wrong.
We know that Bayes probabilities for classifying $x$ in two classes $omega_1$ and $omega_2$ are given by $P(omega_1)p(x|omega_1)$ and $P(omega_2)p(x|omega_2)$. So Bayes error is $int_x min(P(omega_1)p(x|omega_1), P(omega_2)p(x|omega_2)) dx $.
We note that $min(a,b) le a^beta b^{(1-beta)}$ for $a, b>0$, so Bayes error is bounded by $P(omega_1)^beta P(omega_2)^ {1-beta}int_x p(x|omega_1)^beta p(x|omega_2)^{1-beta} dx$.
We consider a case where class conditional probabilities are Normal $p(x|omega_1)$ is given by $mathcal N(mu_1, Sigma_1)$ and $p(x|omega_2)$ is given by $mathcal N(mu_2, Sigma_2)$. then Bayes error is bounded by
$P(omega_1)^beta P(omega_2)^ {1-beta}int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$.
$int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$
$=frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx$ where $L =(x-mu_1)^tbeta Sigma_1^{-1}(x-mu_1)+ (x-mu_2)^t(1-beta) Sigma_2^{-1}(x-mu_12)$
Let $A= betaSigma_1^{-1}$ , $B= (1-beta)Sigma_2^{-1}$, $y=(x-mu_1)$ and $a=mu_2-mu_1$
Then $L=y^tAy+(y-a)^tB(y-a)$
$=y^t(A+B)y -a^tBy -y^tBa+a^tBa$
$=y^t(A+B)y-2a^tBy+a^tBa$ , as B is symmetric
As (A+B) is positive definite matrix it can be composed into a matrix $P$ such that $P^tP=(A+B)$
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy +a^tBa$
Note that $(a^tB(A+B)^{-1}P^t)(a^tB(A+B)^{-1}P^t)^t= a^tB(A+B)^{-1}P^tP(A+B)^{-1}Ba = a^tB(A+B)^{-1}Ba$
Also
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy + a^tB(A+B)^{-1}Ba - a^tB(A+B)^{-1}Ba +a^tBa$
Note that $-a^tB(A+B)^{-1}Ba +a^tBa= a^t[B-B(A+B)^{-1}B]a= a^t[B(A+B)^{-1}(A+B)-B(A+B)^{-1}B]a= a^tB(A+B)^{-1}Aa$
Let $z=y-(A+B)^{-1}Ba=x-mu_1-(A+B)^{-1}Ba$
$L= (Pz)^tPz+ a^tB(A+B)^{-1}Aa$
$= z^t(A+B)z+ a^tB(A+B)^{-1}Aa$
So,
$frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx=
exp^{-frac{1}{2}a^tB(A+B)^{-1}Aa}frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_zexp^{-frac{1}{2}z^t(A+B)z} dz $
Note that $(A+B)=A(A^{-1}+B^{-1})B$
So $(A+B)^{-1}=B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}$
Also it is known that $int_zexp^{-frac{1}{2}z^t(A+B)z} dz = sqrt{(2 pi)^d |(A+B)^{-1}|} $
So $exp^{-1/2L}$ is given by $exp^{-frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a}sqrt{frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} }$
Expressing $exp^{-1/2L}$ as $exp^{-k}$
$k= frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a - frac{1}{2} log frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} $
$A^{-1}+B^{-1}= frac{Sigma_1}{beta} + frac{Sigma_2}{1-beta}= frac{(1-beta) Sigma_1 + beta Sigma_2}{beta(1-beta)} $
$|(A+B)^{-1}|=|B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}|= frac{|Sigma_1||Sigma_2|}{|(1-beta) Sigma_1 + beta Sigma_2|}$
So $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_1 + beta Sigma_2]^{-1}a - frac{1}{2} log frac{|Sigma_1|^{1-beta}|Sigma_2|^beta}{|(1-beta) Sigma_1 + beta Sigma_2|} $
In the book $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_2 + beta Sigma_1]^{-1}a - frac{1}{2} log frac{|Sigma_2|^{1-beta}|Sigma_1|^beta}{|(1-beta) Sigma_2 + beta Sigma_1|} $. But from the diagrams function does not seem be be symmetric on $beta$. Please let me know where I have gone wrong
probability bayesian
asked Jan 11 at 7:46
CuriousCurious
889516
$endgroup$
I was following derivation of Chernoffs bound for Bayes error Given in the book Pattern recognition by Duda Hart and Stork. However there is a minor difference between the results in book and details i have worked out. Please guid me where I am wrong.
We know that Bayes probabilities for classifying $x$ in two classes $omega_1$ and $omega_2$ are given by $P(omega_1)p(x|omega_1)$ and $P(omega_2)p(x|omega_2)$. So Bayes error is $int_x min(P(omega_1)p(x|omega_1), P(omega_2)p(x|omega_2)) dx $.
We note that $min(a,b) le a^beta b^{(1-beta)}$ for $a, b>0$, so Bayes error is bounded by $P(omega_1)^beta P(omega_2)^ {1-beta}int_x p(x|omega_1)^beta p(x|omega_2)^{1-beta} dx$.
We consider a case where class conditional probabilities are Normal $p(x|omega_1)$ is given by $mathcal N(mu_1, Sigma_1)$ and $p(x|omega_2)$ is given by $mathcal N(mu_2, Sigma_2)$. then Bayes error is bounded by
$P(omega_1)^beta P(omega_2)^ {1-beta}int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$.
$int_x mathcal N(mu_1, Sigma_1)^beta mathcal N(mu_2, Sigma_2)^{1-beta} dx$
$=frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx$ where $L =(x-mu_1)^tbeta Sigma_1^{-1}(x-mu_1)+ (x-mu_2)^t(1-beta) Sigma_2^{-1}(x-mu_12)$
Let $A= betaSigma_1^{-1}$ , $B= (1-beta)Sigma_2^{-1}$, $y=(x-mu_1)$ and $a=mu_2-mu_1$
Then $L=y^tAy+(y-a)^tB(y-a)$
$=y^t(A+B)y -a^tBy -y^tBa+a^tBa$
$=y^t(A+B)y-2a^tBy+a^tBa$ , as B is symmetric
As (A+B) is positive definite matrix it can be composed into a matrix $P$ such that $P^tP=(A+B)$
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy +a^tBa$
Note that $(a^tB(A+B)^{-1}P^t)(a^tB(A+B)^{-1}P^t)^t= a^tB(A+B)^{-1}P^tP(A+B)^{-1}Ba = a^tB(A+B)^{-1}Ba$
Also
$L= (Py)^t(Py) -2a^tB(A+B)^{-1}P^tPy + a^tB(A+B)^{-1}Ba - a^tB(A+B)^{-1}Ba +a^tBa$
Note that $-a^tB(A+B)^{-1}Ba +a^tBa= a^t[B-B(A+B)^{-1}B]a= a^t[B(A+B)^{-1}(A+B)-B(A+B)^{-1}B]a= a^tB(A+B)^{-1}Aa$
Let $z=y-(A+B)^{-1}Ba=x-mu_1-(A+B)^{-1}Ba$
$L= (Pz)^tPz+ a^tB(A+B)^{-1}Aa$
$= z^t(A+B)z+ a^tB(A+B)^{-1}Aa$
So,
$frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_xexp^{-1/2L} dx=
exp^{-frac{1}{2}a^tB(A+B)^{-1}Aa}frac {1}{sqrt{(2 pi)^d |Sigma_1|^beta|Sigma_2|^{(1-beta)}}}int_zexp^{-frac{1}{2}z^t(A+B)z} dz $
Note that $(A+B)=A(A^{-1}+B^{-1})B$
So $(A+B)^{-1}=B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}$
Also it is known that $int_zexp^{-frac{1}{2}z^t(A+B)z} dz = sqrt{(2 pi)^d |(A+B)^{-1}|} $
So $exp^{-1/2L}$ is given by $exp^{-frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a}sqrt{frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} }$
Expressing $exp^{-1/2L}$ as $exp^{-k}$
$k= frac{1}{2}a^t(A^{-1}+B^{-1})^{-1}a - frac{1}{2} log frac {|(A+B)^{-1}|}{ |Sigma_1|^beta|Sigma_2|^{(1-beta)}} $
$A^{-1}+B^{-1}= frac{Sigma_1}{beta} + frac{Sigma_2}{1-beta}= frac{(1-beta) Sigma_1 + beta Sigma_2}{beta(1-beta)} $
$|(A+B)^{-1}|=|B^{-1}(A^{-1}+B^{-1})^{-1}A^{-1}|= frac{|Sigma_1||Sigma_2|}{|(1-beta) Sigma_1 + beta Sigma_2|}$
So $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_1 + beta Sigma_2]^{-1}a - frac{1}{2} log frac{|Sigma_1|^{1-beta}|Sigma_2|^beta}{|(1-beta) Sigma_1 + beta Sigma_2|} $
In the book $k= frac{beta(1-beta)}{2}a^t[(1-beta)Sigma_2 + beta Sigma_1]^{-1}a - frac{1}{2} log frac{|Sigma_2|^{1-beta}|Sigma_1|^beta}{|(1-beta) Sigma_2 + beta Sigma_1|} $. But from the diagrams function does not seem be be symmetric on $beta$. Please let me know where I have gone wrong
probability bayesian
probability bayesian
asked Jan 11 at 7:46
CuriousCurious
889516
asked Jan 11 at 7:46
CuriousCurious
889516
asked Jan 11 at 7:46
CuriousCurious
889516
asked Jan 11 at 7:46
CuriousCurious
889516
asked Jan 11 at 7:46
CuriousCurious
889516
889516
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