- #1

noowutah

- 57

- 3

[tex]d(C,\lambda{}A+(1-\lambda)B)<d(C,A)=d(C,B)[/tex]

[itex]d[/itex] is the Euclidean distance measure (taking [itex]a_{i}[/itex] to be the coordinates of A in [itex]\mathbb{R}^{n}[/itex])

[tex]d(A,B)=\sum_{i=1}^{n}\sqrt{(a_{i}-b_{i})^{2}}[/tex]

I want to show that this is also true if our notion of distance is the Kullback-Leibler divergence from information theory. So, let A, B, C be points in n-dimensional space with

[tex]D_{KL}(C,A)=D_{KL}(C,B)[/tex]

where

[tex]D_{KL}(X,Y)=\sum_{i=1}^{n}x_{i}\ln\frac{x_{i}}{y_{i}}[/tex]

Let F be a point between A and B in the sense that

[tex]F=\lambda{}A+(1-\lambda)B,0<\lambda<1[/tex]

Then I want to prove that

[tex]D_{KL}(C,F)<D_{KL}(C,A)=D_{KL}(C,B)[/tex]

Two points that may be helpful are (1) the Gibbs inequality ([itex]p\ln{}p<p\ln{}q[/itex]); and (2) the convexity of the logarithm ([itex]\ln(\lambda{}x+(1-\lambda)y)<\lambda\ln{}x+(1-\lambda)\ln{}y[/itex]), but I haven't been able to get anywhere. I'd love some help.