Is Kullback-Leibler Divergennce not equal to Relative Entropy?












0












$begingroup$


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?










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  • 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
















0












$begingroup$


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?










share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






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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














  • 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










1 Answer
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$begingroup$

They're equivalent because $sum_i u_i=sum_i v_i=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

They're equivalent because $sum_i u_i=sum_i v_i=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

They're equivalent because $sum_i u_i=sum_i v_i=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

They're equivalent because $sum_i u_i=sum_i v_i=1$.






share|cite|improve this answer









$endgroup$



They're equivalent because $sum_i u_i=sum_i v_i=1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 23:00









J.G.J.G.

32.6k23250




32.6k23250












  • $begingroup$
    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










  • $begingroup$
    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


















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