Graduate Does Maximizing Likelihood Over Truncated Support Increase Probability?

[kullbach_liebler]

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 vector. 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 $S:\mathbf{x} \mapsto \mathbb{R}^n$. Moreover, suppose that,

$$\#(\mathcal{B} \cap \mathcal{A}) > \#(\neg \mathcal{B} \cap \mathcal{A})$$
and
$$!\exists \mathbf{x}^*>\tilde{\mathbf{x}}:=\arg \max_{\mathbf{x}}S(\mathbf{x}).$$

Then, it implies that,

\begin{align}
\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)
\end{align}

My questions:

1) Is the statement true?;
2) How could I improve the presentation of this proposition?;
3) What are the mildest possible conditions under which (1) will hold?

 

[Stephen Tashi]

2) How could I improve the presentation of this proposition?;

I don't know what a statistical journal would want, but for the purpose of getting an answer on an internet forum, you could give a more verbal statement of the proposition before presenting it using only notation.

\begin{align}
\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)
\end{align}

This appears to say the maximum value of a function ( which is a summation rather than an integration) when taken over a set is greater than the value of that function evaluated at the particular element $v_o$ in that set. Is that the general idea?

 

[pbuk]

1) Is the statement true?

Let $\mathbf{v_max}$ be a value of $\mathbf{v}$ which maximizes the sum. What happens when $\mathbf{v_0} = \mathbf{v_max}$?