Lower bound for binomial tail probability conditional on composite parameter space

Question

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?

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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.

Answer 1

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:

$$P(X > c \, | \, \theta \leq \theta_0) \geq \frac{1}{\Theta(\theta_0) \sqrt{2n}} \left( \frac{n \, A_{\Theta}}{c}\right)^c $$