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I am concerned about the monotonicity of the following ratio

$ f(\eta)=\frac{\sum_{x=K}^{N}\left(\begin{array}{c} N\\ x \end{array}\right)\left(q_{G}\eta\right)^{x}\left(1-q_{G}\eta\right)^{N-x}}{\sum_{x=K}^{N}\left(\begin{array}{c} N\\ x \end{array}\right)\left(q_{B}\eta\right)^{x}\left(1-q_{B}\eta\right)^{N-x}} $

For $q_{G}>q_{B}$ and $0<K<N$, some numerical experiments imply that it is decreasing in $\eta$. I wonder how to prove this property? Thanks a lot.

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  • $\begingroup$ Perhaps it will help to know that the derivative of the denominator with respect to $\eta$ has a closed form, and similarly numerator. $\endgroup$ Apr 9, 2021 at 1:06

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Let $t:=\eta$, $n:=N\in\{1,2,\dots\}$, and $k:=K\in\{1,\dots,n\}$. We need to show that the ratio $$r(t):=\frac{G(t)}{B(t)}$$ decreases in $t$, where \begin{equation} B(t):=s(q_B\, t),\quad G(t):=s(q_G\, t), \end{equation} \begin{equation} s(p):=P(X_{n,p}\ge k), \end{equation} and $X_{n,p}$ is a random variable with the binomial distribution with parameters $n,p$; here we must assume that $q_B>0$ and $t\in(0,1/g_G)$, so that $q_G\,t$ and $q_B\,t$ be in the interval $(0,1)$.

Following the comment by Brendan McKay, note that \begin{equation} s'(p)=\frac{n!}{(k-1)! (n-k)!}\, p^{k-1} (1-p)^{n-k}. \end{equation} So, the "derivative ratio" \begin{equation} \rho(t):=\frac{G'(t)}{B'(t)}=C\Big(\frac{1-q_G\,t}{1-q_B\,t}\Big)^{n-k} \end{equation} is decreasing in $t\in(0,1/g_G)$; here, $C$ is a positive real number which does not depend on $t$. Also, $B(0+)=0=G(0+)$.

So, by the special-case l'Hospital-type rule for monotonicity (see e.g. Proposition 4.1), $r(t)$ is decreasing in $t\in(0,1/g_G)$, as desired.

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  • $\begingroup$ Thanks, Losif. The closed-form derivative is fantastic, and your paper is very helpful. $\endgroup$
    – Peter
    Apr 9, 2021 at 3:35

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