带分布与积分的不等式相关技术咨询

Hey everyone, let's break down this problem centered on probability distributions and integral equations. Here's the core setup:

Let $a \in (0,1)$ be the unique solution to the following integral equation:
$$\int_0^1 (\theta - a)e^{\dfrac{(\theta - a)^2}{\beta}}g(\theta)d\theta = 0 \tag{1}$$

The function $g(\theta)$ here is a continuously differentiable probability distribution with these defining properties:

• It's a valid probability density function: $\displaystyle \int_0^1 g(\theta)d\theta = 1$
• Its first derivative $g'(\theta)$ is continuous across the interval $[0,1]$
• The expected value of $\theta$ under $g$ satisfies: $\displaystyle \int_0^1 \theta g(\theta)d\theta \geq \dfrac{1}{2} \tag{2}$

We also note that $\beta$ is a positive real number: $\beta \in \mathbb{R}^{+} \tag{3}$

Condition (2) tells us a specific behavior as $\beta$ grows to infinity: the solution $a$ from equation (1) will fall within the interval $[\frac{1}{2},1)$. If, instead, the integral in (2) were less than $\frac{1}{2}$, inequality (a) would...

备注：内容来源于stack exchange，提问作者Dave299