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Properties of Conditional Entropies of Degree s

We have the following properties:

Property 6.17. For all $s\geq 1$, we have

(i) $^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s((X,Y)\wedge Z) = \, ^4{\ensurema......wedge Y) + \, ^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s(Y \wedge Z\vert X);$

(ii) $^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s(X \wedge Y) = {\ensuremath{\bold......r{H}}}}_s^s(X,Y) = \,^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(Y \wedge X);$

(iii) ${\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X) + {\ensuremath{\boldsymbol{\mat......((X,Y) \wedge Z) + \,^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s(X \wedge Y);$

(iv) $^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s((X,Y) \wedge Z\vert V) = \,^4{\e......ert V) + \, ^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s(Y \wedge Z\vert X,V).$

Property 6.18. For all $s\geq 1$, we have $$^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s((X,Y) \wedge Z) \geq \, ^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s(X \wedgeZ),$$ with equality iff $X$ and $Y$ are independent given $Z$.

Property 6.19. For all $s\geq 1$, we have

(i) $^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X\vert Z) \leq \, ^4{\ensuremath......}}}_s^s(X\vert Y) + \, ^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(Y\vert Z);$

(ii) $^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X,Y\vert Z) \leq \, ^4{\ensurema......}}}_s^s(X\vert Z) + \, ^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(Y\vert Z);$

(iii) If $^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(\cdot,\cdot) \neq 0$, then

$$\frac{^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X\vert Y)}{^4......thscr{H}}}}_s^s(X\vert Z)}{^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X.Z)}.$$

Property 6.20. Let  $$d_s^1(X,Y) = \,^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X\vert Y) + \,^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(Y\vert X),$$ $$d^s_2(X,Y) = \left\{\begin{array}{ll}1- \frac{^4{\ensuremath{\b...... & ^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X,Y) = 0\end{array}\right.$$ and $$d^s_3(X,Y) = \left\{\begin{array}{ll}\frac{^4{\ensuremath{\bold......) \geq {\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X) > 0\end{array}\right.$$ then, we have

(i) $d_s^t(X,Y) + d_s^t(Y,Z) \geq d_s^t(X,Z), \, t=1,2\,\,$   and$\,\, 3;$

(ii) $\vert{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X) - {\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(Y)\vert \leq d_s^1(X,Y);$

(iii) $\vert\,^4{\ensuremath{\boldsymbol{\mathscr{H}}}}_s^s(X_1\vert Y_1) -\, ^4{\ens......H}}}}_s^s(X_2\vert Y_2)\vert \leq d_s^1(X_1,X_2) + d_s^1(Y_1,Y_2), \, s \geq 1;$

(iv) $\vert\,^4{\ensuremath{\boldsymbol{\mathscr{F}}}}_s^s(X_1 \wedge Y_1) -\, ^4{\e......}}}_s^s(X_2 \wedge Y_2)\vert \leq d_s^1(X_1,X_2) + d_s^1(Y_1,Y_2), \, s \geq 1.$