Suppose $\mathbf{X}$ is a random variable with a finite support $\Omega$ and with some pdf $f(\cdot; \mathbf{v}_0)$ where $\mathbf{v}_0$ is the parameter. Define, $\mathcal{A}:= \{\mathbf{x}:S(\mathbf{x}) \geq \gamma\} \subset \Omega$ and $\tilde{\mathbf{x}}:=S(\tilde{\mathbf{x}}) = \gamma$, $\mathcal{B}:=\{\mathbf{x} \geq \tilde{\mathbf{x}}\}$ where $S:\mathbf{x} \mapsto \mathbb{R}^1$. Moreover, suppose that,
$$\#(\mathcal{B} \cap \mathcal{A}) > \#(\neg \mathcal{B} \cap \mathcal{A})\tag{1}\label{eq1}$$ and
$$!\exists \mathbf{x}^*>\tilde{\mathbf{x}}:=\arg \max_{\mathbf{x}}S(\mathbf{x})\tag{2}\label{eq2}.$$
Then, it implies that,
$$\max_{\mathbf{v}}\sum_{\mathbf{x} :S(\mathbf{x}) \geq \gamma} f(\mathbf{x};\mathbf{v}) > \sum_{\mathbf{x} :S(\mathbf{x}) \geq \gamma} f(\mathbf{x};\mathbf{v}_0) \tag{3}\label{eq3}$$
My questions:
- Is the statement true?;
- How could I improve the presentation of this proposition?;
- What are the mildest possible conditions under which \eqref{eq3} will hold?
Addendum:
My potential solution is:
The set $\{\mathbf{x} :S(\mathbf{x}) \geq \gamma\}$ = $(\mathcal{A}\cap\mathcal{B})\cup(\mathcal{A}\cap(\neg\mathcal{B}) \subset \Omega$. Thus, \eqref{eq3} can be re-written as:
$$\max_{\mathbf{v}}\big[\sum_{\mathbf{x}\in\mathcal{A}\cap\mathcal{B}} f(\mathbf{x};\mathbf{v})+\sum_{\mathbf{x}\in\mathcal{A}\cap\neg\mathcal{B}} f(\mathbf{x};\mathbf{v})\big] > \big[\sum_{\mathbf{x}\in\mathcal{A}\cap\mathcal{B}} f(\mathbf{x};\mathbf{v}_0)+\sum_{\mathbf{x}\in\mathcal{A}\cap\neg\mathcal{B}} f(\mathbf{x};\mathbf{v}_0)\big] \tag{4}\label{eq4}$$
By \eqref{eq1}, the cardinality of the first set over which we sum is greater than that of the second set ($\mathcal{A}\cap\neg\mathcal{B}$). We can always place all the probability mass on the first one and \eqref{eq3} will be satisfied.