<p>This has nothing to do with any specific model. For any event <span class="math-container">$A$</span>, let <span class="math-container">$I_A$</span> be the indicator function such that <span class="math-container">$I_A(\omega)=1$</span> if <span class="math-container">$\omega\in A$</span> and <span class="math-container">$I_A(\omega)=0$</span> if <span class="math-container">$\omega\notin A.$</span></p>
<p>Then <span class="math-container">$\mathbb{P}(A)=\mathbb{E}[I_A],$</span>
and here the expectation is given in terms of a density function.</p>

CC BY-SA 4.0
2021-04-14T17:12:28.393Z
2021-04-14T17:12:28.393Z
Michael Greinecker
<p>Given a utility function, <span class="math-container">$U_{nj} = V_{nj} + \varepsilon_{nj}$</span>, it makes sense that we can find the probability the decision maker <span class="math-container">$n$</span> chooses alternative <span class="math-container">$i$</span> as:</p>
<p><span class="math-container">$$Pr(U_{ni} &gt; U_{nj} \forall j \neq i)$$</span>
<span class="math-container">$$=Pr(V_{ni} + \varepsilon_{ni} &gt; V_{nj} + \varepsilon_{nj} \forall j \neq i)$$</span>
<span class="math-container">$$=Pr(\varepsilon_{nj} - \varepsilon_{ni} &lt; V_{ni} - V_{nj} \forall j \neq i)$$</span></p>
<p>I am having an issue understanding the next step though, where:</p>
<p><span class="math-container">$$Pr(\varepsilon_{nj} - \varepsilon_{ni} &lt; V_{ni} - V_{nj} \forall j \neq i)$$</span>
<span class="math-container">$$= \int_{\varepsilon} \text{I}(\varepsilon_{nj} - \varepsilon_{ni} &lt; V_{ni} - V_{nj} \forall j \neq i)f(\varepsilon_n)d\varepsilon_n$$</span></p>
<p>Specifically, I am having trouble wrapping my head around why the joint density function, <span class="math-container">$f(\varepsilon_n)$</span>, comes into play. I understand that this is one of the crucial aspects to understanding discrete choice models, since the choice of this joint density function underlies what model is to be used. Hence, I would like to gain a better understanding of why it is in here. If anyone would maybe have an example or a dumbed-down explanation of why this integral is what it is, I would appreciate it.</p>

CC BY-SA 4.0
2021-04-14T17:09:08.580Z
Did you mean $\forall j \ne i$?
Herr K.
<microeconomics><econometrics><utility><discrete-preferences><logit>
Trouble Understanding the Integral Underlying Random Utility Models
jl_1995_95
# Trouble Understanding the Integral Underlying Random Utility Models

(Asked by: username_0 on 2021-04-14)

Given a utility function, $U\_{nj} = V\_{nj} + \varepsilon\_{nj}$, it makes sense that we can find the probability the decision maker $n$ chooses alternative $i$ as:

$$Pr(U\_{ni} > U\_{nj} \forall j \neq i)$$
$$=Pr(V\_{ni} + \varepsilon\_{ni} > V\_{nj} + \varepsilon\_{nj} \forall j \neq i)$$
$$=Pr(\varepsilon\_{nj} - \varepsilon\_{ni} < V\_{ni} - V\_{nj} \forall j \neq i)$$

I am having an issue understanding the next step though, where:

$$Pr(\varepsilon\_{nj} - \varepsilon\_{ni} < V\_{ni} - V\_{nj} \forall j \neq i)$$
$$= \int\_{\varepsilon} \text{I}(\varepsilon\_{nj} - \varepsilon\_{ni} < V\_{ni} - V\_{nj} \forall j \neq i)f(\varepsilon\_n)d\varepsilon\_n$$

Specifically, I am having trouble wrapping my head around why the joint density function, $f(\varepsilon\_n)$, comes into play. I understand that this is one of the crucial aspects to understanding discrete choice models, since the choice of this joint density function underlies what model is to be used. Hence, I would like to gain a better understanding of why it is in here. If anyone would maybe have an example or a dumbed-down explanation of why this integral is what it is, I would appreciate it.

**Tags:** microeconomics, econometrics, utility, discrete-preferences, logit

> **Comments:**
> **username_1**: Did you mean $\forall j \ne i$?  

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## 1 answers

### Answer by username_2 on 2021-04-14. Score: 2

This has nothing to do with any specific model. For any event $A$, let $I\_A$ be the indicator function such that $I\_A(\omega)=1$ if $\omega\in A$ and $I\_A(\omega)=0$ if $\omega\notin A.$

Then $\mathbb{P}(A)=\mathbb{E}[I\_A],$
and here the expectation is given in terms of a density function.


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