In the beginning of the book Train (2009, p.4) on "Discrete choice methods with simulation" we read:
Define an indicator function $I[h(x,ε) = y]$ that takes the value of
$1$ when the statement in brackets is true and $0$ when the statement is
false. That is, $I[·] = 1$ if the value of $ε$, combined with $x$,
induces the agent to choose outcome $y$, and $I[·] = 0$ if the value
of $ε$, combined with $x$, induces the agent to choose some other
outcome.
Then the following statement is presented.
$$
P(y | x) = Prob(I[h(x,ε) = y] = 1)\\
= \int I[h(x,ε) = y] f (ε) dε
$$
My question is about ($=$) inside the indicator function under integration.
(In the rest of the book all examples are with inequality, because disctrete choice models are built on comparisions of utilitites of different alternatives, like $u_j>u_k$, for all $k\neq j$).
Perhaps, I'm missing something, but I thought that indicator function just trancated the intergral function (change its bounds), so equal sign in expression must lead such integral to $0$ value. What am I missing? A counterexample is welcome.
Best Answer
The expected value of the indicator fuction $I\{A\}$ (where $A\in\Omega$) is $P(A)$. In this case, the integral of the joint probability indicator function over the whole $\varepsilon$ domain gives us the conditional probability $P(y|x)$ (see more here).
For your case, assume $h(x,\epsilon)=x^T\beta+\epsilon$ (which is the first of three possible proofs in your book). Note that it is given that $x$ is observed - meaning that given $x$ and $\epsilon$, we can determine $y$. Moreover, it is clearly stated that $P(y|x)=P(\epsilon~~\text{s.t.}~~h(x,\epsilon)=y)$.
$$\int_{\mathcal{E}}I\{y=h(x,\epsilon)\}f(\epsilon)d\epsilon=\int_{\mathcal{E}}I\{y=x^T\beta+\epsilon\}f(\epsilon)d\epsilon=\int_{\epsilon=y-x^T\beta}{f(\epsilon)}=P(\epsilon=y-x\beta)=\\P(\epsilon~~\text{s.t.}~~x^T\beta+\epsilon=y)=P(y|x)$$
Note that we didn't actually use the realization $h(x,\epsilon)=x^T\beta+\epsilon$.