There are $N$ Bernoulli trials, where $m$ trails have probability of success $p$ and $N-m$ trails have probability of success $q=1-p$. Assume $p>q$. The number of successes is a random variable $S$ expressed as a sum of two binomial random variables:

\begin{align} S = Bin(p,m) + Bin(q,N-m) \end{align}

The distribution of $S$ is known to be a Poisson-Binomial Distribution. I am studying the behavior of the ratio between P(S=k) and P(S=k-1) with respect to $m$ and $k$. For a given $m$ and number of successes $k$, denote such ratio as $R(k,m)$:

\begin{align} R(k,m) = \frac{P(S=k \; | \;m)}{P(S=k-1 \; |\;m)} \end{align}

I am particularly interested in the behavior of this ratio for small quantiles, and ran numerical simulation for $R(k,m)$ when $k < mean$. It appears that for values of $k$ that are equally distant from the mean, the following holds:

Let $\mu_m$ be the mean of the corresponding distribution and $\alpha$ be the distance away from the mean. Then for $k=\mu_m - \alpha$:

\begin{align} R(\mu_0 - \alpha,0) > R(\mu_1 - \alpha,1) > \dots > R(\mu_{m-1} - \alpha,m-1) > R(\mu_{m} - \alpha,m) > \dots > R(\mu_{N} - \alpha,N) \end{align}

The ratio seems to be bounded by two binomial distribution for $m=0$ and $m=N$ respectively.

It is easy to see why $R(\mu_0 - \alpha,0) > R(\mu_{N} - \alpha,N)$.

\begin{align} R(k,0) = \frac{\binom{N}{k} q^kp^{N-k}}{\binom{N}{k-1} q^{k-1}p^{N-k+1}} = \frac{N-k+1}{k} \cdot \frac{q}{p} \\ R(k,N) = \frac{\binom{N}{k} q^kp^{N-k}}{\binom{N}{k-1} q^{k-1}p^{N-k+1}} = \frac{N-k+1}{k} \cdot \frac{p}{q} \end{align}

Setting $k=\mu - \alpha$ for each case, we have \begin{align} R(k,0) = \frac{N-qN + \alpha+1}{qN-\alpha} \cdot \frac{q}{p} \approx \frac{qp + \frac{\alpha q}{N}}{qp - \frac{\alpha p}{N}}\\ R(k,N) = \frac{N-pN + \alpha+1}{pN-\alpha} \cdot \frac{p}{q} \approx \frac{qp + \frac{\alpha p}{N}}{qp - \frac{\alpha q}{N}}\\ \frac{qp + \frac{\alpha q}{N}}{qp - \frac{\alpha p}{N}} > \frac{qp + \frac{\alpha p}{N}}{qp - \frac{\alpha q}{N}} \\ (qp + \frac{\alpha q}{N})(qp - \frac{\alpha q}{N}) > (qp + \frac{\alpha p}{N})(qp - \frac{\alpha p}{N}) \\ (qp)^2 - \left ( \frac{\alpha q}{N} \right )^2 > (qp)^2 - \left ( \frac{\alpha p}{N} \right )^2 \end{align}

Since $p>q$, the above inequality holds.

But I am unable to express $R(k,m)$ analytically, no prove that it is bounded by ratios corresponding to the limiting binomial distributions $m=0$ and $m=N$ (although I see it in simulations).

Also note that for a reasonable choice of $\alpha$ (for example, $3\cdot \sigma$), the difference between two limiting ratios shrinks proportionally to $\frac{1}{N}$, which makes me hope for some bounding expression that will analytically verify such behavior. My research will still be sound if I can show that $R(\mu_0 - \alpha,0)$ is within some reasonable bound for reasonable $\alpha$'s, compared to any other possible $R(\mu_m - \alpha,m)$, and the bound is approaching $0$ as $N$ grows.

I checked all-embracing paper by Wang (http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf), but could not extract a ready solution from it.

Any help with proving this and/or pointers to related papers will be much appreciated.


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