Suppose $X\sim \operatorname{Bin}(n,\theta)$ and define the function $$ \lambda(x)=\frac{P(X=x,\theta>\theta_0)}{P(X=x,\theta\leq\theta_0)} $$ for some constant $0<\theta_0<1$. Suppose also a prior distribution on the parameters, $\Theta\sim G$. I'm trying to find a condition on $\theta_0$ such that $\lambda(x)$ is monotonically increasing in $x$.
\begin{align} \nabla_x\lambda(x)&=\nabla_x\frac{\int_{\theta_0}^1 s^x(1-s)^{n-x}dG(s)}{\int_{0}^{\theta_0} t^x(1-t)^{n-x}dG(t)}\\ &=\nabla_x\frac{\int_{\theta_0}^1 s^x(1-s)^{n-x}dG(s)}{\int_{0}^{\theta_0} t^x(1-t)^{n-x}dG(t)}\\ &=\frac{\int_{0}^{\theta_0} t^x(1-t)^{n-x}dG(t)\int_{\theta_0}^1 s^x(1-s)^{n-x}\log\left(\frac{s}{1-s}\right)dG(s)-\int_{\theta_0}^1 s^x(1-s)^{n-x}dG(s)\int_{0}^{\theta_0} t^x(1-t)^{n-x}\log\left(\frac{t}{1-t}\right)dG(t)}{[\int_{0}^{\theta_0} t^x(1-t)^{n-x}dG(t)]^2}\\ \end{align}
I can bring the differential operator inside the integral by dominated convergence, since the integrands are bounded by 1. Furthermore, the denominator is always greater than zero, so we can consider only the numerator.
\begin{align} \text{numerator}(\nabla_x\lambda(x))=&\int_{0}^{\theta_0} \int_{\theta_0}^1 s^x(1-s)^{n-x}\log\left(\frac{s}{1-s}\right)t^x(1-t)^{n-x}dG(s)dG(t)-\int_{0}^{\theta_0}\int_{\theta_0}^1 s^x(1-s)^{n-x} t^x(1-t)^{n-x}\log\left(\frac{t}{1-t}\right)dG(s)dG(t)\\ =&\int_{0}^{\theta_0}\int_{\theta_0}^1 s^x(1-s)^{n-x} t^x(1-t)^{n-x}\left[\log\left(\frac{s}{1-s}\right)-\log\left(\frac{t}{1-t}\right)\right]dG(s)dG(t)\\ =&\int_{0}^{\theta_0}\int_{\theta_0}^1 s^x(1-s)^{n-x} t^x(1-t)^{n-x}\log\frac{s/(1-s)}{t/(1-t)}dG(s)dG(t) \end{align}
This is where I got stuck. Even when I imposed a uniform distribution on $G$ (i.e. $dG(s)=ds$, $dG(t)=dt$), I had issues deriving a condition on $\theta_0$ such that $\text{numerator}(\nabla_x\lambda(x))>0$. I tried to use integration by parts for this specific case, but to no avail.
Am I missing something fundamental in my derivation?
EDIT Per @MANMAID's suggestion, I went through the algebra for $\lambda(x+1)-\lambda(x)$. For clear presentation, I set $G$ to be the uniform distribution.
\begin{align} \lambda(x+1)-\lambda(x)&=\frac{P(X=x+1,\theta>\theta_0)}{P(X=x+1,\theta\leq\theta_0)}-\frac{P(X=x,\theta>\theta_0)}{P(X=x,\theta\leq\theta_0)}\\ &=\frac{\int_{\theta_0}^1 s^{x+1}(1-s)^{n-x-1}ds}{\int_{0}^{\theta_0} t^{x+1}(1-t)^{n-x-1}dt}-\frac{\int_{\theta_0}^1 s^{x}(1-s)^{n-x}ds}{\int_{0}^{\theta_0} t^{x}(1-t)^{n-x}dt}\\ &=\frac{\int_{\theta_0}^1 s^{x+1}(1-s)^{n-x-1}ds\int_{0}^{\theta_0} t^{x}(1-t)^{n-x}dt-\int_{\theta_0}^1 s^{x}(1-s)^{n-x}ds\int_{0}^{\theta_0} t^{x+1}(1-t)^{n-x-1}dt}{\int_{0}^{\theta_0} t^{x+1}(1-t)^{n-x-1}dt\int_{0}^{\theta_0} t^{x}(1-t)^{n-x}dt} \end{align}
Again, considering only the numerator, as the denominator is positive:
\begin{align} \text{numerator}(\lambda(x+1)-\lambda(x))=&\left(\frac{\Gamma(x+2)\Gamma(n-x)}{\Gamma(n+2)}-B_{\theta_0}(x+2,n-x) \right)B_{\theta_0}(x+1,n-x+1) - \left(\frac{\Gamma(x+1)\Gamma(n-x+1)}{\Gamma(n+2)}-B_{\theta_0}(x+1,n-x+1) \right)B_{\theta_0}(x+2,n-x) \\ =&\frac{\Gamma(x+2)\Gamma(n-x)}{\Gamma(n+2)}B_{\theta_0}(x+1,n-x+1)-\frac{\Gamma(x+1)\Gamma(n-x+1)}{\Gamma(n+2)}B_{\theta_0}(x+2,n-x)\\ =&\frac{\Gamma(x+1)\Gamma(n-x)}{\Gamma(n+2)}\left\{(x+1)B_{\theta_0}(x+1,n-x+1)-(n-x)B_{\theta_0}(x+2,n-x) \right\} \end{align}
where $\Gamma(y)=(y-1)!$ for positive integer $y$ and $B_{\theta_0}(q,r)=\int_0^{\theta_0}u^{q-1}(1-u)^{r-1}du$.
The last expression above is positive $iff$ the expression in $\{\cdot\}$ is positive. However, is there a way to simplify that condition? I've tried plugging in various values for $x$ and $n$, and it seems that the condition I need takes the form of $0<\theta_0<\frac{n-x-1}{n+2}$. However, this condition on $\theta_0$ depends on $x$, which is not really something that is desired.