I am wondering if the function $L[n]$ defined on $n=0,1,2,\ldots,N$ below is "monotonically" non-decreasing in $n$. I put monotonically in quotes because the function is not continuous and I am not sure if I can safely use the usual definition of monotone function here.
$$\begin{array}{rcl}L[n]&=&\frac{1}{\binom{N}{n}}\sum_{t=0}^{\min(n,M)}\binom{M}{t}p^{t-n}(1-p)^{M-t-N+n}\binom{N-M}{n-t}q^{n-t}(1-q)^{N-M-n+t}\\ &=&\frac{(f_A\ast f_B)[n]}{f_X[n]}\end{array}$$ where $M$, $N$ are integers satisfying $0\leq M\leq N$, and $p$ and $q$ satisfy $0<p<q<1$. The second line gives the definition of $L[n]$ as the likelihood ratio between likelihoods of an observation of a random variable $X\sim\text{Binomial}(N,p)$ and $Y=A+B$, where $A\sim\text{Binomial}(M,p)$ and $B\sim\text{Binomial}(N-M,q)$ (hence the convolution). This is related to a question I posted on stats.SE. Any ideas?
WHAT I'VE DONE
I took the difference $L[n+1]-L[n]$ and came up with the following form:
$$L[n+1]-L[n]=\left[\frac{1}{\binom{N}{n}}\sum_{t=0}^{\min(n,M)}l[t]\left(\frac{(N+1)(N-n-M+t)}{(N-n)(n+t+1)}\frac{q(1-p)}{p(1-q)}-1\right)\right]+\frac{\binom{M}{n+1}}{\binom{N}{n+1}}\left(\frac{1-q}{1-p}\right)^{N-M}$$
where $l[t]=\binom{M}{t}p^{t-n}(1-p)^{M-t-N+n}\binom{N-M}{n-t}q^{n-t}(1-q)^{N-M-n+t}$ is the summand in $L[n]$. It's easy to show that $\frac{q(1-p)}{p(1-q)}>1$ when $q>p$, and if I can show that $\frac{(N+1)(N-n-M+t)}{(N-n)(n+t+1)}>1$ then I am done. However, I am not sure if that is always true. Perhaps there is a different way, or I am overlooking something?