Minimizing the sum of KL divergences
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Given a list of probability distributions $q_i$, what distribution $p$ minimizes the sum of KL divergences (if they exist) to and from each of them? That is, how do I determine
$$operatorname*{argmin}_p sum_i D_text{KL}(p mathbin{Vert} q_i)$$
and
$$operatorname*{argmin}_p sum_i D_text{KL}(q_i mathbin{Vert} p)$$
I recall reading somewhere that, in the Jensen-Shannon divergence,
begin{align*}
D_text{JS}(p, q) &= frac{D_text{KL}(p mathbin{Vert} r) + D_text{KL}(q mathbin{Vert} r)}{2} \
r = &frac{p + q}{2}
end{align*}
The midpoint distribution $r$ is precisely
$$r = operatorname*{argmin}_s frac{D_text{KL}(p mathbin{Vert} s) + D_text{KL}(q mathbin{Vert} s)}{2}$$
I can't find a reference for this, however.
probability probability-distributions reference-request
asked Dec 4 at 0:43
user76284
1,1651123
This question has an open bounty worth +50
reputation from user76284 ending in 3 days.
This question has not received enough attention.
add a comment |
up vote
6
down vote
favorite
Given a list of probability distributions $q_i$, what distribution $p$ minimizes the sum of KL divergences (if they exist) to and from each of them? That is, how do I determine
$$operatorname*{argmin}_p sum_i D_text{KL}(p mathbin{Vert} q_i)$$
and
$$operatorname*{argmin}_p sum_i D_text{KL}(q_i mathbin{Vert} p)$$
I recall reading somewhere that, in the Jensen-Shannon divergence,
begin{align*}
D_text{JS}(p, q) &= frac{D_text{KL}(p mathbin{Vert} r) + D_text{KL}(q mathbin{Vert} r)}{2} \
r = &frac{p + q}{2}
end{align*}
The midpoint distribution $r$ is precisely
$$r = operatorname*{argmin}_s frac{D_text{KL}(p mathbin{Vert} s) + D_text{KL}(q mathbin{Vert} s)}{2}$$
I can't find a reference for this, however.
probability probability-distributions reference-request
asked Dec 4 at 0:43
user76284
1,1651123
This question has an open bounty worth +50
reputation from user76284 ending in 3 days.
This question has not received enough attention.
Have you read "A new metric for probability distributions" by D.M. Endres and J.E. Schindelin? I think this may help you.
– Flying Dogfish
2 days ago
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Given a list of probability distributions $q_i$, what distribution $p$ minimizes the sum of KL divergences (if they exist) to and from each of them? That is, how do I determine
$$operatorname*{argmin}_p sum_i D_text{KL}(p mathbin{Vert} q_i)$$
and
$$operatorname*{argmin}_p sum_i D_text{KL}(q_i mathbin{Vert} p)$$
I recall reading somewhere that, in the Jensen-Shannon divergence,
begin{align*}
D_text{JS}(p, q) &= frac{D_text{KL}(p mathbin{Vert} r) + D_text{KL}(q mathbin{Vert} r)}{2} \
r = &frac{p + q}{2}
end{align*}
The midpoint distribution $r$ is precisely
$$r = operatorname*{argmin}_s frac{D_text{KL}(p mathbin{Vert} s) + D_text{KL}(q mathbin{Vert} s)}{2}$$
I can't find a reference for this, however.
probability probability-distributions reference-request
asked Dec 4 at 0:43
user76284
1,1651123
Given a list of probability distributions $q_i$, what distribution $p$ minimizes the sum of KL divergences (if they exist) to and from each of them? That is, how do I determine
$$operatorname*{argmin}_p sum_i D_text{KL}(p mathbin{Vert} q_i)$$
and
$$operatorname*{argmin}_p sum_i D_text{KL}(q_i mathbin{Vert} p)$$
I recall reading somewhere that, in the Jensen-Shannon divergence,
begin{align*}
D_text{JS}(p, q) &= frac{D_text{KL}(p mathbin{Vert} r) + D_text{KL}(q mathbin{Vert} r)}{2} \
r = &frac{p + q}{2}
end{align*}
The midpoint distribution $r$ is precisely
$$r = operatorname*{argmin}_s frac{D_text{KL}(p mathbin{Vert} s) + D_text{KL}(q mathbin{Vert} s)}{2}$$
I can't find a reference for this, however.
probability probability-distributions reference-request
probability probability-distributions reference-request
asked Dec 4 at 0:43
user76284
1,1651123
asked Dec 4 at 0:43
user76284
1,1651123
edited Dec 5 at 0:40
asked Dec 4 at 0:43
user76284
1,1651123
asked Dec 4 at 0:43
user76284
1,1651123
asked Dec 4 at 0:43
user76284
1,1651123
1,1651123
This question has an open bounty worth +50
reputation from user76284 ending in 3 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from user76284 ending in 3 days.
This question has not received enough attention.
Have you read "A new metric for probability distributions" by D.M. Endres and J.E. Schindelin? I think this may help you.
– Flying Dogfish
2 days ago
add a comment |
Have you read "A new metric for probability distributions" by D.M. Endres and J.E. Schindelin? I think this may help you.
– Flying Dogfish
2 days ago
Have you read "A new metric for probability distributions" by D.M. Endres and J.E. Schindelin? I think this may help you.
– Flying Dogfish
2 days ago
Have you read "A new metric for probability distributions" by D.M. Endres and J.E. Schindelin? I think this may help you.
– Flying Dogfish
2 days ago
add a comment |
1 Answer
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A rather brute-force approach, using Lagrange multipliers. Defining the objetive functional
$$g(p_x)=sum_{i=1}^k D(p||q^{(i)})=sum_{i=1}^k sum_{x} p_x logleft( frac{p_x}{q_x^{(i)}} right) tag{1}$$
and the restriction $sum_x p_x = 1$ we get the critical point at
$$ n+sum_{i=1}^k logleft( frac{p_x}{q_x^{(i)}} right) +lambda =0 tag{2}$$
Rearraging we get
$$ p_x = gamma left(prod_{i=1}^k q_x^{(i)}right)^{1/k} tag{3}$$
where $gamma$ is the normalizing constant. Hence the critical distribution is the normalized geometric mean of the given $q_i$ distributions.
Because the KL divergence is convex in both arguments, this critical point must be a global minimum.
Or simpler and better: changing the sum order in $(1)$ :
$$g(p_x)= sum_{x} p_x log prod_{i=1}^k left( frac{p_x}{q_x^{(i)}}right)
=k sum_{x} p_x log left( frac{p_x}{overline{q_x}}right) tag{4}$$
where $overline{q_x}$ is the geometric mean, as in $(3)$. Notice, however that
$overline{q_x}$ is not in general a probability function. Defining the normalization constant $gamma=1/sum overline{q_x}$ we get
$$g(p_x)=k sum_{x} p_x log left( frac{gamma p_x}{ gamma overline{q_x}}right)=k sum_{x} p_x log left( frac{ p_x}{ gamma overline{q_x}}right) + k log(gamma) =\=k D(p||gamma overline{q}) + k log(gamma) tag{5}$$
The variable term is the first, which is a true KL-divergence, and is minimized (at zero) by $p=gamma overline{q}$, in agreement with $(3)$. The residual term gives the value of this minimum.
BTW: that $gammage 1$ (with equality only for all $q_i$ identical) is easily proved by GM-AM inequality.
answered 7 hours ago
leonbloy
40k645107
add a comment |
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1 Answer
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A rather brute-force approach, using Lagrange multipliers. Defining the objetive functional
$$g(p_x)=sum_{i=1}^k D(p||q^{(i)})=sum_{i=1}^k sum_{x} p_x logleft( frac{p_x}{q_x^{(i)}} right) tag{1}$$
and the restriction $sum_x p_x = 1$ we get the critical point at
$$ n+sum_{i=1}^k logleft( frac{p_x}{q_x^{(i)}} right) +lambda =0 tag{2}$$
Rearraging we get
$$ p_x = gamma left(prod_{i=1}^k q_x^{(i)}right)^{1/k} tag{3}$$
where $gamma$ is the normalizing constant. Hence the critical distribution is the normalized geometric mean of the given $q_i$ distributions.
Because the KL divergence is convex in both arguments, this critical point must be a global minimum.
Or simpler and better: changing the sum order in $(1)$ :
$$g(p_x)= sum_{x} p_x log prod_{i=1}^k left( frac{p_x}{q_x^{(i)}}right)
=k sum_{x} p_x log left( frac{p_x}{overline{q_x}}right) tag{4}$$
where $overline{q_x}$ is the geometric mean, as in $(3)$. Notice, however that
$overline{q_x}$ is not in general a probability function. Defining the normalization constant $gamma=1/sum overline{q_x}$ we get
$$g(p_x)=k sum_{x} p_x log left( frac{gamma p_x}{ gamma overline{q_x}}right)=k sum_{x} p_x log left( frac{ p_x}{ gamma overline{q_x}}right) + k log(gamma) =\=k D(p||gamma overline{q}) + k log(gamma) tag{5}$$
The variable term is the first, which is a true KL-divergence, and is minimized (at zero) by $p=gamma overline{q}$, in agreement with $(3)$. The residual term gives the value of this minimum.
BTW: that $gammage 1$ (with equality only for all $q_i$ identical) is easily proved by GM-AM inequality.
answered 7 hours ago
leonbloy
40k645107
add a comment |
up vote
1
down vote
A rather brute-force approach, using Lagrange multipliers. Defining the objetive functional
$$g(p_x)=sum_{i=1}^k D(p||q^{(i)})=sum_{i=1}^k sum_{x} p_x logleft( frac{p_x}{q_x^{(i)}} right) tag{1}$$
and the restriction $sum_x p_x = 1$ we get the critical point at
$$ n+sum_{i=1}^k logleft( frac{p_x}{q_x^{(i)}} right) +lambda =0 tag{2}$$
Rearraging we get
$$ p_x = gamma left(prod_{i=1}^k q_x^{(i)}right)^{1/k} tag{3}$$
where $gamma$ is the normalizing constant. Hence the critical distribution is the normalized geometric mean of the given $q_i$ distributions.
Because the KL divergence is convex in both arguments, this critical point must be a global minimum.
Or simpler and better: changing the sum order in $(1)$ :
$$g(p_x)= sum_{x} p_x log prod_{i=1}^k left( frac{p_x}{q_x^{(i)}}right)
=k sum_{x} p_x log left( frac{p_x}{overline{q_x}}right) tag{4}$$
where $overline{q_x}$ is the geometric mean, as in $(3)$. Notice, however that
$overline{q_x}$ is not in general a probability function. Defining the normalization constant $gamma=1/sum overline{q_x}$ we get
$$g(p_x)=k sum_{x} p_x log left( frac{gamma p_x}{ gamma overline{q_x}}right)=k sum_{x} p_x log left( frac{ p_x}{ gamma overline{q_x}}right) + k log(gamma) =\=k D(p||gamma overline{q}) + k log(gamma) tag{5}$$
The variable term is the first, which is a true KL-divergence, and is minimized (at zero) by $p=gamma overline{q}$, in agreement with $(3)$. The residual term gives the value of this minimum.
BTW: that $gammage 1$ (with equality only for all $q_i$ identical) is easily proved by GM-AM inequality.
answered 7 hours ago
leonbloy
40k645107
add a comment |
up vote
1
down vote
up vote
1
down vote
A rather brute-force approach, using Lagrange multipliers. Defining the objetive functional
$$g(p_x)=sum_{i=1}^k D(p||q^{(i)})=sum_{i=1}^k sum_{x} p_x logleft( frac{p_x}{q_x^{(i)}} right) tag{1}$$
and the restriction $sum_x p_x = 1$ we get the critical point at
$$ n+sum_{i=1}^k logleft( frac{p_x}{q_x^{(i)}} right) +lambda =0 tag{2}$$
Rearraging we get
$$ p_x = gamma left(prod_{i=1}^k q_x^{(i)}right)^{1/k} tag{3}$$
where $gamma$ is the normalizing constant. Hence the critical distribution is the normalized geometric mean of the given $q_i$ distributions.
Because the KL divergence is convex in both arguments, this critical point must be a global minimum.
Or simpler and better: changing the sum order in $(1)$ :
$$g(p_x)= sum_{x} p_x log prod_{i=1}^k left( frac{p_x}{q_x^{(i)}}right)
=k sum_{x} p_x log left( frac{p_x}{overline{q_x}}right) tag{4}$$
where $overline{q_x}$ is the geometric mean, as in $(3)$. Notice, however that
$overline{q_x}$ is not in general a probability function. Defining the normalization constant $gamma=1/sum overline{q_x}$ we get
$$g(p_x)=k sum_{x} p_x log left( frac{gamma p_x}{ gamma overline{q_x}}right)=k sum_{x} p_x log left( frac{ p_x}{ gamma overline{q_x}}right) + k log(gamma) =\=k D(p||gamma overline{q}) + k log(gamma) tag{5}$$
The variable term is the first, which is a true KL-divergence, and is minimized (at zero) by $p=gamma overline{q}$, in agreement with $(3)$. The residual term gives the value of this minimum.
BTW: that $gammage 1$ (with equality only for all $q_i$ identical) is easily proved by GM-AM inequality.
answered 7 hours ago
leonbloy
40k645107
A rather brute-force approach, using Lagrange multipliers. Defining the objetive functional
$$g(p_x)=sum_{i=1}^k D(p||q^{(i)})=sum_{i=1}^k sum_{x} p_x logleft( frac{p_x}{q_x^{(i)}} right) tag{1}$$
and the restriction $sum_x p_x = 1$ we get the critical point at
$$ n+sum_{i=1}^k logleft( frac{p_x}{q_x^{(i)}} right) +lambda =0 tag{2}$$
Rearraging we get
$$ p_x = gamma left(prod_{i=1}^k q_x^{(i)}right)^{1/k} tag{3}$$
where $gamma$ is the normalizing constant. Hence the critical distribution is the normalized geometric mean of the given $q_i$ distributions.
Because the KL divergence is convex in both arguments, this critical point must be a global minimum.
Or simpler and better: changing the sum order in $(1)$ :
$$g(p_x)= sum_{x} p_x log prod_{i=1}^k left( frac{p_x}{q_x^{(i)}}right)
=k sum_{x} p_x log left( frac{p_x}{overline{q_x}}right) tag{4}$$
where $overline{q_x}$ is the geometric mean, as in $(3)$. Notice, however that
$overline{q_x}$ is not in general a probability function. Defining the normalization constant $gamma=1/sum overline{q_x}$ we get
$$g(p_x)=k sum_{x} p_x log left( frac{gamma p_x}{ gamma overline{q_x}}right)=k sum_{x} p_x log left( frac{ p_x}{ gamma overline{q_x}}right) + k log(gamma) =\=k D(p||gamma overline{q}) + k log(gamma) tag{5}$$
The variable term is the first, which is a true KL-divergence, and is minimized (at zero) by $p=gamma overline{q}$, in agreement with $(3)$. The residual term gives the value of this minimum.
BTW: that $gammage 1$ (with equality only for all $q_i$ identical) is easily proved by GM-AM inequality.
answered 7 hours ago
leonbloy
40k645107
edited 6 hours ago
answered 7 hours ago
leonbloy
40k645107
answered 7 hours ago
leonbloy
40k645107
answered 7 hours ago
leonbloy
40k645107
40k645107
add a comment |
add a comment |
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Have you read "A new metric for probability distributions" by D.M. Endres and J.E. Schindelin? I think this may help you.
– Flying Dogfish
2 days ago