it.information theory - Question regarding divergence

The inequality $D(P'|Q') ge D(P^star| Q^star)$ does not need to hold.

Here is an example.

Let $A$ be the set ${1,2,3,...,n}$. Let $E$ be the set of measures $P$ on $A$ such that $P({1}) = 0$. Projecting a measure $P$ on $E$ using $D$ is equivalent to conditioning $P$ on $A- {1}$. Choose $P'$ and $Q'$ such that they both put equal and nonzero mass on ${1}$. By direct computation one sees: $D(P^star| Q^star) = frac{1}{1-P'({1})} D(P'|Q') > D(P' | Q')$.

The details of the above computation are as follows.

For ease of notation set $n=3$. Let $E$ be the set of measures $P$ with $P({1}) =epsilon$; to obtain the example above, one sets $epsilon = 0$. Let us parametrize the measures on ${1,2,3}$ as follows: $P({1}) = p_1$, $P({2}) =p_2$ and $P({3}) = 1-p_1 -p_2$. Our problem is:

$$inf_{ Q in E}left[ p_1 log frac{p_1}{q_1} + p_2 log frac{p_2}{q_2} + (1-p_1 -p_2) logfrac{ 1- p_1 - p_2}{ 1- q_1 - q_2 } right].$$
Let $F$ denote the expression after the $inf$.
$F$ is strictly convex in $Q$ and therefore will have a unique optimizer. In the above coordinates, the normal to $E$ is the vector $(1,0)$. Then
$$frac{partial F} {partial q_1} = -frac{p_1}{q_1} + frac{1-p_1-p_2}{1-q_1-q_2} = lambda$$
and
$$frac{partial F} {partial q_2} = -frac{p_2}{q_2} + frac{1-p_1-p_2}{1-q_1-q_2} = 0.$$
We have the constraint that $Qin E$, i.e., $q_1 =epsilon$.
From the last two equalities one infers:
$$q_2 = frac{(1-epsilon) p_2}{ 1-p_1}.$$

Going back to the coordinates $(p_1,p_2,p_3)$ to denote a measure on ${1,2,3}$,
projecting a measure on $E$ using $D$ corresponds to the following map:

$$(p_1,p_2,p_3) rightarrow left(epsilon, (1-epsilon)frac{p_2}{p_2+p_3}, (1-epsilon)frac{p_3}{p_2 + p_3}right).$$
For $epsilon =0$, this is the same as conditioning $P$ on ${2,3}$.

One obtains the expression for the relative entropy given above by directly computing it using this formula for the projections.

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