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M-Dimensional Unified (r,s)-Information Radii

The $ M-$dimensional unified$ (r,s)-$information radii or $ M-$dimensional unified$ (r,s)-I-$divergence measures are given by
$\displaystyle ^\alpha{\ensuremath{\boldsymbol{\mathscr{I}}}}^s_r(P_1,P_2,...,P_......P_M), & r\neq 1,\ \ s=1 \\ I(P_1,P_2,...,P_M), &r=1,\ \ s=1\end{array}\right.$
    (5.12)

$ \alpha=1,2$ and 3, where for $ \alpha =1$, we have 

$\displaystyle ^1I^s_r(P_1,P_2,...,P_M)= (1-2^{1-s})^{-1} \Big\{ \Big[\sum_{......lambda_k p_{ki}}\Big)^{1-r}}\Big]^{s-1\over r-1} -1\Big\},\ r\neq 1,\ s\neq 1$
$\displaystyle ^1 I^s_1(P_1,P_2,...,P_M)=(1-2^{1-s})^{-1} \Big\{ 2^{(s-1)I(P_1,P_2,...,P_M)}-1 \Big\},\s\neq 1$
and 
$\displaystyle ^1 I^1_r(P_1,P_2,...,P_M)= (r-1)^{-1}\log_2 \Big\{\sum_{i=1}^n{......^r_{ji}}\Big) \Big(\sum_{k=1}^M{\lambda_k p_{ki}}\Big)^{1-r}}\Big\},\ r\neq 1$
For$ \alpha=2$ we have
$\displaystyle ^2{\ensuremath{\boldsymbol{\mathscr{I}}}}^s_r(P_1,P_2,...,P_M)=\s......symbol{\mathscr{D}}}}^s_r\Big( P_j\vert\vert\sum_{k=1}^M{\lambda_k P_k}\Big) },$
    (5.13)

For $ \alpha =3$ we have

$\displaystyle ^3{\ensuremath{\boldsymbol{\mathscr{I}}}}^s_r(P_1,P_2,...,P_M)={\......ig) - \sum_{j=1}^M{\lambda_j {\ensuremath{\boldsymbol{\mathscr{H}}}}^s_r(P_j)},$
    (5.14)

where $ {\ensuremath{\boldsymbol{\mathscr{H}}}}^s_r$ and $ {\ensuremath{\boldsymbol{\mathscr{D}}}}^s_r$ are as given in chapter 3 and chapter 4 by expressions (3.8) and (4.1) respectively.

In particular we have 

$\displaystyle ^1{\ensuremath{\boldsymbol{\mathscr{I}}}}^s_s(P_1,P_2,...,P_M)= \......P_1,P_2,...,P_M) = {\ensuremath{\boldsymbol{\mathscr{I}}}}^s_s(P_1,P_2,...,P_M)$
where
$\displaystyle {\ensuremath{\boldsymbol{\mathscr{I}}}}^s_s(P_1,P_2,...,P_M) = \l......P_1,P_2,...,P_M), & s\neq 1 \\  I(P_1,P_2,...,P_M), & s=1\end{array}\right.$
with
$\displaystyle I^s_s(P_1,P_2,...,P_M)=(1-2^{1-s})^{-1}\Big\{\sum_{i=1}^n{\Big(\s......{ji}}\Big) \Big( \sum_{k=1}^M{\lambda_kp_{ki}}\Big)^{1-s}}-1\Big\}, \ s\neq 1$

For two-dimensional case, i.e., when M=2, $ \lambda_1=\lambda_2={1\over 2}$$ P_1=P$ and$ P_2=Q$, we have
$ ^1I^s_r(P_1,P_2)=\ ^1I^s_r(P\vert\vert Q)$

$\displaystyle = (1-2^{1-s})^{-1}\Big\{ \Big[ \sum_{i=1}^n{ \Big({p^r_i+q^r_i\over 2}\Big)\Big({p_i +q_i\over2}\Big)^{1-r}}\Big]^{s-1\over r-1}-1\Big\},$
    (5.15)

$ ^2I^s_r(P_1,P_2)=\ ^2I^s_r(P\vert\vert Q)$

$\displaystyle = (1-2^{1-s})^{-1}\, {1\over 2} \Big\{ \Big[ \sum_{i=1}^n{p^r_i\......um_{i=1}^n{q^r_i\Big({p_i +q_i\over2}\Big)^{1-r}}\Big]^{s-1\over r-1}-2\Big\},$
    (5.16)

$ ^3I^s_r(P_1,P_2)=\ ^3I^s_r(P\vert\vert Q)$

$\displaystyle = (1-2^{1-s})^{-1}\Big\{ {1\over 2} \Big[ \Big( \sum_{i=1}^n{p^r......g] -\Big[\sum_{i=1}^n{\Big({p_i+q_i\over 2}\Big)^r}\Big]^{s-1\over r-1}\Big\},$
    (5.17)

for all $ r\neq 1, \ s\neq 1$.

Note 5.2.The measures (5.15) and (5.16) are the same as given in (4.10) and (4.6) respectively. While the measure (5.17).
 


21-06-2001
Inder Jeet Taneja
Departamento de Matemática - UFSC
88.040-900 Florianópolis, SC - Brazil