Binomial mass ratios and geometric tail domination

The factorial formula in [F3] makes every $\left(\genfrac{}{}{0px}{}{N}{j}\right)$ positive for $0\le j\le N$. Since $p$ and $1-p$ are positive, [F2] gives $f_j>0$. For $1\le j\le N$, cancellation in [F2] and [F3] gives $f_{j-1}\mathrm{/}{f}_j=j(1-p)\mathrm{/}((N-j+1)p)$. All factors canceled are positive. Taking the reciprocal and replacing $j$ by $j+1$ gives the displayed forward ratio for $0\le j<N$; its denominator is positive. When $N=0$ both ratio ranges are empty.

Let $E_0,\dots ,E_m$ be pairwise disjoint measurable events, and set $E_r=\mathrm{\varnothing }$ for $r>m$. Applying countable additivity from [F4] to this sequence gives $\mathbb{P}({\bigcup }_{r=0}^m{E}_r)={\sum }_{r=0}^{\mathrm{\infty }}\mathbb{P}(E_r)={\sum }_{r=0}^m\mathbb{P}(E_r)$, because each appended term has probability zero. This establishes finite additivity on the underlying probability space without requiring its sample space to be finite. If $A\subseteq C$ are measurable, applying this to and using nonnegativity gives $\mathbb{P}(A)\le \mathbb{P}(C)$. Applying it to gives $\mathbb{P}(\mathrm{\Omega }\setminus A)=1-\mathbb{P}(A)$.

Write $B={\sum }_{i=1}^N{X}_i$ as in [F1]. For each $1\le i\le N$, the measurable events $\{X_i=0\}$ and $\{X_i=1\}$ are disjoint and have probabilities $1-p$ and $p$, so step 1.2 gives $\mathbb{P}(C_i)=0$ for $C_i=\{X_i\mathrm{\notin }\{0,1\}\}$. Put $D_i=C_i\setminus {\bigcup }_{h<i}{C}_h$. These measurable sets are pairwise disjoint, satisfy $D_i\subseteq C_i$, and have union $E={\bigcup }_{i=1}^N{C}_i$. By monotonicity and finite additivity in step 1.2, $\mathbb{P}(E)={\sum }_{i=1}^N\mathbb{P}(D_i)=0$. Outside $E$, every summand is $0$ or $1$, so $B\in \{0,\dots ,N\}$. The measurable event $S=\{B\mathrm{\notin }\{0,\dots ,N\}\}$ is therefore contained in $E$ and has probability zero by monotonicity. When $N=0$, the unions are empty and $B=0$, so the same conclusion holds.

Under the lower-tail hypothesis, induction on $r$ gives $f_{k-r}\le q^r{f}_k$ for each $0\le r\le k$: the base case is equality, and the successor step uses $f_{k-r-1}\le q{f}_{k-r}$ and $q\ge 0$. The measurable residual event $R_{-}=\{B\le k\}\setminus {\bigcup }_{j=0}^k\{B=j\}$ is contained in $S$ from step 2.1 and hence has probability zero by step 1.2. Thus the tail event is the disjoint union of $R_{-}$ and the level events $\{B=j\}$ for $0\le j\le k$. Finite additivity gives $\mathbb{P}(B\le k)={\sum }_{j=0}^k{f}_j\le f_k{\sum }_{r=0}^k{q}^r$. The omitted terms of the convergent series in [F6] are nonnegative, so this is at most $f_k\mathrm{/}(1-q)$. This reasoning includes $q=0$ and the empty adjacent-mass range when $k=0$.

Under the upper-tail hypothesis, induction on $r$ gives $f_{k+r}\le q^r{f}_k$ for $0\le r\le N-k$: the base case is equality, and for $r<N-k$ the successor step uses $f_{k+r+1}\le q{f}_{k+r}$ and $q\ge 0$. The measurable residual event $R_{+}=\{B\ge k\}\setminus {\bigcup }_{j=k}^N\{B=j\}$ is contained in $S$ from step 2.1 and hence has probability zero by step 1.2. Thus the tail event is the disjoint union of $R_{+}$ and the level events $\{B=j\}$ for $k\le j\le N$. Finite additivity gives $\mathbb{P}(B\ge k)={\sum }_{j=k}^N{f}_j\le f_k{\sum }_{r=0}^{N-k}{q}^r$. Since the omitted terms are nonnegative, [F6] gives ${\sum }_{r=0}^{N-k}{q}^r\le 1\mathrm{/}(1-q)$. This includes the empty adjacent-mass range when $k=N$. Both tail denominators are positive because $q<1$, completing the proof.