Lower bound for binomial tail probability conditional on composite parameter space Suppose that $X\sim \text{Bin}(n,\theta)$ is a binomial random variable and we are interested in the quantity
$$
P(X>c\mid\theta\leq\theta_0)
$$
where $\theta_0\in(0,1)$ is a fixed, known quantity. Is it possible to find a lower bound for this probability? So far, I can express the following:
\begin{eqnarray*}
P(X>c\mid\theta\leq\theta_0)&=&\frac{P(X>c,\theta\leq\theta_0)}{P(\theta\leq\theta_0)}\\
&=&\frac{1}{\Theta(\theta_0)}\int_0^{\theta_0}P(X>c\mid t)\, d\Theta(t)
\end{eqnarray*}
where $\Theta(t)=P(\theta\leq t)$ is some unspecified weight function (distribution) on $\theta$. 
From this post I can bound the integrand further from below, assuming that $c/n>t$:
$$
P(X>c\mid t)\geq \frac{1}{\sqrt{2n}} \exp\left(-c\log\frac{c}{nt}\right)
$$
This is where I get stuck. I can't seem to bound this further, preferably with something that depends on $\theta_0$. Are there any references to bound this type of quantity?

EDIT
One possible idea is to note that $\theta_0$ is the supremum of $\theta$. So, for some $d_n\downarrow 0$, we can consider the set $\mathcal{B}=\{\theta\in(0,1):\theta>\theta_0-d_n\}$. Then
\begin{eqnarray*}
P(X>c\mid\theta\leq\theta_0)&=&\frac{1}{\Theta(\theta_0)}\int_0^{\theta_0}P(X>c\mid t)\, d\Theta(t)\\
&\geq&\frac{1}{\Theta(\theta_0)}\int_{t\in(0,\theta_0)\cap\mathcal{B}}P(X>c\mid t)\, d\Theta(t)\\
&>&\frac{\Theta(\{\theta\in (0,\theta_0)\cap\mathcal{B} \})}{\Theta(\theta_0)}P(X>c\mid \theta_0-d_n)\\
&\geq& \frac{1}{\sqrt{2n}} \exp\left(-c\log\frac{c}{n(\theta_0-d_n)}\right)\frac{\Theta(\{\theta\in (0,\theta_0)\cap\mathcal{B} \})}{\Theta(\theta_0)}
\end{eqnarray*}
where the second inequality follows from the fact that $P(X>c\mid \theta)$ increases with $\theta$, so that $P(X>c\mid \theta)>P(X>c\mid\theta_0-d_n)$ on $\mathcal{B}$. Since $\Theta(\theta_0)\leq 1$, I can also remove the quantity from the denominator above and get a smaller bound.
 A: Update
In general I do not think you can find a lower bound that is dependent on $\theta_0$.
If we assume that $\Theta \equiv 1$, describing an atomic distribution such that $\theta = 0$ almost surely. Regardless of the value of $\theta_0$, in this case we know for $c \geq 0$
$$ P(X > c \, | \, \theta \leq \theta_0) = 0$$
So what this shows is that unless you know more specific information about $\Theta$, and in particular can rule out the case $\Theta \equiv 1$, you cannot find a lower bound other than $0$.
In my original proof below, I show that if you know $E[\theta] = A_{\Theta}$ then this can be used to derive a lower bound.
For $\Theta \equiv 1$, $A_\Theta = 0$, and we obtain the result above, but if $A_{\Theta} > 0$ then we derive a non-trivial bound.
In heuristic terms the above is saying we cannot turn an upper bound on an increasing function (i.e. $\text{Bin}(n,\theta)$ is `increasing' in $\theta$, in some heuristic sense) into a lower bound.
Original
Assuming the inequality you provide for $P( X > c\, | \, t)$, then for $c/n > \theta_0$
\begin{eqnarray*}
P(X>c\mid\theta\leq\theta_0)&=&
\frac{1}{\Theta(\theta_0)}\int_0^{\theta_0}P(X>c\mid t)\, d\Theta(t) \\
&\geq&
\frac{1}{\Theta(\theta_0)\sqrt{2n} } \int_0^{\theta_0}\exp\left(-c\log\frac{c}{nt}\right) d \Theta(t) \\
&=&
\frac{1}{\Theta(\theta_0)\sqrt{2n} }\left(\frac{n}{c}\right)^c \int_0^{\theta_0}t^c d \Theta(t)
\end{eqnarray*}
The integral in the last line is equivalent to the expectation $E_{\Theta}[\theta^c]$; if we make the additional constraint that $c > 1$, then Jensen's inequality implies:
$$
E_{\Theta}[\theta^c] \geq E_{\Theta}[\theta]^c$$
If $\Theta$ is known, but unspecified then you at least know that there is a constant $A_\Theta$ such that $E_{\Theta}[\theta] = A_\Theta$, from which you can bound the probability:
$$P(X > c \, | \, \theta \leq \theta_0) \geq \frac{1}{\Theta(\theta_0) \sqrt{2n}} \left( \frac{n \, A_{\Theta}}{c}\right)^c
$$
