Is Kullback-Leibler Divergennce not equal to Relative Entropy?
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In many books, Kullback-Leibler Divergence is equal to Relative Entropy.
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i).
$$
However, I find in the book, Convex Optimization (Stephen Boyd) page 90, the KL Divergence is defined as,
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i)-u_i+v_i).
$$
Why KL Divergence has these two different definition? Which one is correct?
convex-optimization machine-learning information-theory
asked Dec 19 '18 at 22:36
AlanAlan
32
$endgroup$
add a comment |
$begingroup$
In many books, Kullback-Leibler Divergence is equal to Relative Entropy.
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i).
$$
However, I find in the book, Convex Optimization (Stephen Boyd) page 90, the KL Divergence is defined as,
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i)-u_i+v_i).
$$
Why KL Divergence has these two different definition? Which one is correct?
convex-optimization machine-learning information-theory
asked Dec 19 '18 at 22:36
AlanAlan
32
$endgroup$
1
$begingroup$
The book (also written by Vandenberghe) clearly mentions "Note that the relative entropy and the Kullback-Leibler divergence are the same when u and v are probability vectors, i.e., satisfy 1Tu = 1Tv = 1.)"
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
$begingroup$
In many books, Kullback-Leibler Divergence is equal to Relative Entropy.
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i).
$$
However, I find in the book, Convex Optimization (Stephen Boyd) page 90, the KL Divergence is defined as,
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i)-u_i+v_i).
$$
Why KL Divergence has these two different definition? Which one is correct?
convex-optimization machine-learning information-theory
asked Dec 19 '18 at 22:36
AlanAlan
32
$endgroup$
In many books, Kullback-Leibler Divergence is equal to Relative Entropy.
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i).
$$
However, I find in the book, Convex Optimization (Stephen Boyd) page 90, the KL Divergence is defined as,
$$
D_{kl}(u,v) = sum_{i=1}^n(u_ilog(u_i/v_i)-u_i+v_i).
$$
Why KL Divergence has these two different definition? Which one is correct?
convex-optimization machine-learning information-theory
convex-optimization machine-learning information-theory
asked Dec 19 '18 at 22:36
AlanAlan
32
asked Dec 19 '18 at 22:36
AlanAlan
32
asked Dec 19 '18 at 22:36
AlanAlan
32
asked Dec 19 '18 at 22:36
AlanAlan
32
asked Dec 19 '18 at 22:36
AlanAlan
32
32
1
$begingroup$
The book (also written by Vandenberghe) clearly mentions "Note that the relative entropy and the Kullback-Leibler divergence are the same when u and v are probability vectors, i.e., satisfy 1Tu = 1Tv = 1.)"
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
1
$begingroup$
The book (also written by Vandenberghe) clearly mentions "Note that the relative entropy and the Kullback-Leibler divergence are the same when u and v are probability vectors, i.e., satisfy 1Tu = 1Tv = 1.)"
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
1
1
$begingroup$
The book (also written by Vandenberghe) clearly mentions "Note that the relative entropy and the Kullback-Leibler divergence are the same when u and v are probability vectors, i.e., satisfy 1Tu = 1Tv = 1.)"
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
$begingroup$
The book (also written by Vandenberghe) clearly mentions "Note that the relative entropy and the Kullback-Leibler divergence are the same when u and v are probability vectors, i.e., satisfy 1Tu = 1Tv = 1.)"
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
1 Answer
1
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oldest
votes
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They're equivalent because $sum_i u_i=sum_i v_i=1$.
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– Alan
Dec 19 '18 at 23:15
$begingroup$
equivalent only for probability vectors, as the book mentions
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
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$begingroup$
They're equivalent because $sum_i u_i=sum_i v_i=1$.
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– Alan
Dec 19 '18 at 23:15
$begingroup$
equivalent only for probability vectors, as the book mentions
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
$begingroup$
They're equivalent because $sum_i u_i=sum_i v_i=1$.
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– Alan
Dec 19 '18 at 23:15
$begingroup$
equivalent only for probability vectors, as the book mentions
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
$begingroup$
They're equivalent because $sum_i u_i=sum_i v_i=1$.
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
$endgroup$
They're equivalent because $sum_i u_i=sum_i v_i=1$.
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
answered Dec 19 '18 at 23:00
J.G.J.G.
27.9k22843
27.9k22843
$begingroup$
thank you so much.
$endgroup$
– Alan
Dec 19 '18 at 23:15
$begingroup$
equivalent only for probability vectors, as the book mentions
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
$begingroup$
thank you so much.
$endgroup$
– Alan
Dec 19 '18 at 23:15
$begingroup$
equivalent only for probability vectors, as the book mentions
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
$begingroup$
thank you so much.
$endgroup$
– Alan
Dec 19 '18 at 23:15
$begingroup$
thank you so much.
$endgroup$
– Alan
Dec 19 '18 at 23:15
$begingroup$
equivalent only for probability vectors, as the book mentions
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
$begingroup$
equivalent only for probability vectors, as the book mentions
$endgroup$
– LinAlg
Dec 20 '18 at 2:45
add a comment |
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$begingroup$
The book (also written by Vandenberghe) clearly mentions "Note that the relative entropy and the Kullback-Leibler divergence are the same when u and v are probability vectors, i.e., satisfy 1Tu = 1Tv = 1.)"
$endgroup$
– LinAlg
Dec 20 '18 at 2:45