Given $$\int_{\mathbb{R}}\frac{1}{p(\theta)e^{(v-\theta)}+1-p(\theta)}dG(v) = 1$$ and function $p(\cdot)$, is it possible to find out $G(v)$?
$p:\mathbb{R}\to [0,1]$ decreasing, $G$ is a CDF.
Given $$\int_{\mathbb{R}}\frac{1}{p(\theta)e^{(v-\theta)}+1-p(\theta)}dG(v) = 1$$ and function $p(\cdot)$, is it possible to find out $G(v)$?
$p:\mathbb{R}\to [0,1]$ decreasing, $G$ is a CDF.
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