Problem Statement: Does this probability statement have a closed form solution?
$\mathbb{P}\left(\min\left\{ w,p\right\} >c\right)=\mathbb{P}\left(\min\left\{ \left(\frac{a+\epsilon_{1}-\epsilon_{0}}{b}\right),p\right\} >c\right)$,
where:
- $a,b,c$ and $p$ are known (constants).
- $\epsilon_{1}$ and $\epsilon_{0}$ are both extreme value random variables (assumed to be independent of each other).
My Attempt: ($\wedge$ denotes “and” below)
$\mathbb{P}\left(\min\left\{ w,p\right\} >c\right)=\mathbb{P}(w<p\wedge w>c)+\mathbb{P}(w>p\wedge p>c)$
by Baye's formula I can decompose the joint probability into:
$=\mathbb{P}(w<p)\cdot\mathbb{P}(w>c\mid w<p)+\mathbb{P}(w>p)\cdot \mathbb{P}(p>c\mid w>p)$
where (I think) that
$\mathbb{P}(w<p)=\mathbb{P}(\epsilon_{1}-\epsilon_{0}<p\cdot b-a)=\displaystyle\frac{1}{exp(-(p \cdot b-a))},$ $\mathbb{P}(w>p)=\mathbb{P}(\epsilon_{0}-\epsilon_{1}<a-b \cdot p)=\displaystyle\frac{1}{exp(-(a-b \cdot p))}$
(where the final step is due to the Logit formula).
However, I am not sure that the conditional probabilities $\mathbb{P}(w>c\mid w<p)$ and $\mathbb{P}(p>c\mid w>p)$ have a closed form? Here is my attempt for the first one:
$\mathbb{P}(w>c\mid w<p)=\mathbb{P}(\epsilon_{0}-\epsilon_{1}<a-b\cdot c\mid\epsilon_{0}-\epsilon_{1}>a-b \cdot p)$
Is this a truncated Logit probaility? What is its closed form (if it has one)?
I am trying to learn this, so please point out any flaws/mistakes in my reasoning. Any help is greatly appreciated. Thanks!