Entropy dominance of certain restricted sequenes Say you have positive $\{a_i\}_{i=1}^n$ and you have $p_i=\frac{a_i}{\sum_{i=1}^na_i}$, then assume you have a $C$ such that $C<2a_n\ll\sum_{i=1}^na_i$ (that is $C$ is not very large), then define $q_i=\frac{a_i}{C+\sum_{i=1}^na_i}$ and $q_{n+1}=\frac{C}{C+\sum_{i=1}^na_i}$. Does Shannon entropy of $q$ dominate entropy of $p$?
Take $C=a_n+\log_2^ka_n$ for any fixed $k$. Now does Shannon entropy of $q$ dominate entropy of $p$ after certain $n$?
There are two cases to consider. Case $(1)$ $a_{i+1}=a_i+O(\log^ka_i)$ Case $(2)$ $a_{i+1}=a_i+O(a_i)$.
When can one expect Shannon entropy to be dominant in sequences of these types? Clearly in this post Entropy dominance, one criteria  for negative result is given.
 A: I'll show that Yes, if you rule out an event of sufficiently small probability then the entropy decreases.
I'll change the notation around a bit so that my $p_1,\dots,p_n$ correspond to your $q_{n+1},\dots,q_1$.
Suppose $\sum_{i=1}^n p_i=1$ and we are given that the event corresponding to $p_1$ did not occur.
Then the new probability of the event corresponding to $p_k$, $k\ne 1$, is
$$
\hat p_k = \frac{p_k}{1-p_1}.
$$
So the new entropy $H(\hat p)$ is (where $S(p)=-\log_2(p)$ and $q$ is the experiment that only determines whether $p_1$ occurs)
$$
\sum_{k=2}^n S\left(\frac{p_k}{1-p_1}\right)\frac{p_k}{1-p_1}
= \frac1{1-p_1}\sum_{k=2}^n [S(p_k)-S(1-p_1)]{p_k}
$$
$$
= \frac1{1-p_1}\left(\sum_{k=2}^n S(p_k)p_k - (1-p_1)S(1-p_1)\right)
= \frac1{1-p_1}\left(H(p) - S(p_1)p_1 - (1-p_1)S(1-p_1)\right)
$$
$$
= \frac1{1-p_1}\left(H(p) - H(q)\right) \le H(p)\quad\text{(as we want) iff }
$$
$$
\frac{H(q)}{p_1}\ge H(p).
$$
The function $p_1\mapsto H(q)/p_1$ is
$$
p_1\mapsto -\log_2(p_1) - \frac{1-p_1}{p_1}\log_2(1-p_1)
$$
which goes to infinity as $p_1\rightarrow 0$, whereas $H(p)$ is bounded by $\log_2 n $,
so it is enough to take $p_1\le 1/n$.
