<p>This derivation follows the same procedure as the one used in discrete logit model.</p>
<p>Let's denote the index of the maximum given <span class="math-container">$y$</span> is selected is <span class="math-container">$(y_i^m,\varepsilon_i^m)$</span>,  where <span class="math-container">$y_i^m = y$</span>. We have <span class="math-container">$$\operatorname{Pr}_i = \operatorname{Pr}\left(U\left(x, y_k^m\right)+\varepsilon_k^m \leq U\left(x, y_i^m\right)+\varepsilon_i^m \ \forall k \neq i \right) \\ = \exp \left( -\exp \left[-U\left(x, y_i^m\right)-\varepsilon_i^m+\log \int_{\mathcal{Y}} \exp U(x, y) d y\right] \right)$$</span>​.</p>
<p>Then we can write the choice probabilities as <span class="math-container">$$\pi(y \mid x) =  \int \operatorname{Pr}_i  e^{-\varepsilon_i} d \varepsilon_i =\int \exp \left( -\exp \left[-U\left(x, y \right)-\varepsilon_i +\log \int_{\mathcal{Y}} \exp U(x, y') d y'\right] \right) e^{-\varepsilon_i} d \varepsilon_i \\ = \int \exp \left( -\exp \left[-\varepsilon_i\right]  \exp \left[-U\left(x, y \right)\right]  \int_{\mathcal{Y}} \exp U(x, y') d y'  \right) e^{-\varepsilon_i} d \varepsilon_i \\ = \int \exp \left( -\exp   \left[-\varepsilon_i\right]   \int_{\mathcal{Y}} \exp \left[ U(x, y')-U\left(x, y \right) \right] d y  \right) e^{-\varepsilon_i} d \varepsilon_i$$</span></p>
<p>Then define  <span class="math-container">$t=\exp(-\varepsilon_i)$</span> such that <span class="math-container">$-\exp (-\varepsilon_i) d \varepsilon_i=d t$</span> , we have <span class="math-container">$$\pi(y \mid x) = \int_{\infty}^0 \exp \left( - t   \int_{\mathcal{Y}} \exp \left[ U(x, y')-U\left(x, y \right) \right] d y'  \right) (-dt) \\ = \int_0^{\infty} \exp \left( - t   \int_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right]d y'  \right) dt \\ = \frac{\exp \left( - t   \int_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right] d y'  \right)}{  -   \int_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right] d y'  } \left.\right|_0 ^{\infty} \\ = \frac{1}{ \int_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right] d y'  } =\frac{\exp U\left(x, y \right)}{ \int_{\mathcal{Y}} \exp U(x, y')  d y'  }$$</span></p>
<p>Any comments are welcome.</p>

CC BY-SA 4.0
2024-02-14T17:04:41.733Z
2024-02-14T17:16:23.233Z
2024-02-14T17:16:23.233Z
Alalalalaki
<p>As they state, <span class="math-container">$\pi(y|x)$</span> is the density of <span class="math-container">$\ln\left(\int_Y \exp(U(x,y) dy\right)$</span> (the density being the derivative of the cdf).</p>
<p>If we take the derivative of <span class="math-container">$\ln\left(\int_Y \exp(U(x,y')) dy' \right)$</span> with respect to <span class="math-container">$y$</span>, we obtain:</p>
<p><span class="math-container">$$
\pi(y|x) = \frac{\exp(U(x,y))}{\int_Y \exp(U(x,y') dy'}
$$</span>
The denominator follows from the <span class="math-container">$\ln$</span> operator. The numerator follows from the chain rule given that <span class="math-container">$\dfrac{d \left(\int_Y exp(U(x,y')) dy'\right)}{dy} = \exp(U(x,y))$</span>.</p>

CC BY-SA 4.0
2024-02-14T12:50:34.293Z
Thanks very much for your answer as always. The derivation is clear but can you explain why $\log \int_Y \exp U(x, y) d y$ is the cdf? And their statement is "choice probabilities are given by their density with respect to the Lebesgue measure"  but not the choice probabilities are the density function itself?
Alalalalaki
CC BY-SA 4.0
2024-02-14T07:06:02.063Z
2024-02-14T07:06:02.063Z
tdm
<p>I am reading <a href="https://www.journals.uchicago.edu/doi/full/10.1086/677191" rel="nofollow noreferrer">Dupuy &amp; Galichon (2014)</a>, which extend the estimation of matching model in Choo &amp; Siow (2006) to continuous types. The way they build the continuous logit model is based on the insights of Cosslett (1988) and Dagsvik (1994) which &quot;have independently suggested using max-stable processes to model continuous choice.&quot;</p>
<p>The detail of the continuous logit model model is described in the Appendix A:</p>
<p>&quot;In this paragraph, we expound the main ideas of Cosslett (1988) and Dagsvik (1994), who show how to obtain a continuous version of the multinomial logit model. Assume that <span class="math-container">$\lbrace\left(y_k^m, \varepsilon_k^m\right), k \in \mathbb{N} \rbrace$</span> are the points of a Poisson point process on <span class="math-container">$\mathcal{Y} \times \mathbb{R}$</span> of intensity <span class="math-container">$d y \times e^{-\varepsilon} d \varepsilon$</span>. We recall that this implies that for <span class="math-container">$S$</span> a subset of <span class="math-container">$\mathcal{Y} \times \mathbb{R}$</span>, the probability that man <span class="math-container">$m$</span> has no acquaintance in set <span class="math-container">$S$</span> is <span class="math-container">$\exp \left(-\int_S e^{-\varepsilon} d y d \varepsilon\right)$</span>. From (2), man <span class="math-container">$m$</span> chooses woman <span class="math-container">$k$</span> among his acquaintances such that his utility is maximized; that is, man <span class="math-container">$m$</span> solves
<span class="math-container">$$\max_k \lbrace U\left(x,y_k^m\right)+\varepsilon_k^m \rbrace$$</span>.</p>
<p>Letting <span class="math-container">$Z$</span> be the value of this maximum, one has for any <span class="math-container">$c \in \mathbb{R}$</span>
<span class="math-container">$$
\operatorname{Pr}(Z \leq c)=\operatorname{Pr}\left(U\left(x, y_k^m\right)+\varepsilon_k^m \leq c \forall k\right),
$$</span>
which is exactly the probability that the Poisson point process <span class="math-container">$\left(y_k, \varepsilon_k^m\right)$</span> has no point in <span class="math-container">$\{(y, \varepsilon): U(x, y)+\varepsilon&gt;c\}$</span>; thus
<span class="math-container">$$
\begin{aligned}
\log \operatorname{Pr}(Z \leq c) &amp; =-\iint_{\mathcal{Y} \times \mathbb{R}} 1(U(x, y)+\varepsilon&gt;c) d y e^{-\varepsilon} d \varepsilon \\
&amp; =-\int_{\mathcal{Y}} \int_{c-U(x, y)} e^{-\varepsilon} d \varepsilon d y \\
&amp; =-\int_{\mathcal{Y}} e^{-c+U(x, y)} d y \\
&amp; =-\exp \left[-c+\log \int_{\mathcal{Y}} \exp U(x, y) d y\right]
\end{aligned}
$$</span></p>
<p>Hence <span class="math-container">$Z$</span> is a <span class="math-container">$\left(\log \int_{\mathcal{Y}} \exp U(x, y) d y, 1\right)$</span> Gumbel. In particular,
<span class="math-container">$\mathbb{E}\left[\max_k \lbrace U\left(x, y_k^m\right)+\varepsilon_k^m \rbrace \right]=\log \int_{\mathcal{y}} \exp U(x, y) d y,$</span></p>
<p>and the choice probabilities are given by their density with respect to the Lebesgue measure</p>
<p><span class="math-container">$$\pi(y \mid x)=\exp [U(x, y)] /\left[\int_{\mathcal{Y}} \exp U\left(x, y^{\prime}\right) d y^{\prime}\right] .$$</span></p>
<p>The same logic also implies that <span class="math-container">$\lbrace \varepsilon_k: k \in \mathbb{N} \rbrace$</span> has a Gumbel distribution. Indeed, the probability that this Poisson point process has no element in the set <span class="math-container">$\{\varepsilon: \varepsilon&gt;c\}$</span> is equal to <span class="math-container">$$
\exp \left(-\int_c^{+\infty} e^{-\varepsilon} d \varepsilon\right)=\exp [-\exp (-c)]
$$</span>
which is equivalent to saying that <span class="math-container">$\operatorname{Pr}\left(\max _{k \in \mathbb{N}} \varepsilon_k \leq c\right)=\exp [-\exp (-c)]$</span>. Finally, note that a similar argument would show that <span class="math-container">$m$</span> has almost surely an infinite, though countable, number of acquaintances, as announced. &quot;</p>
<p>///</p>
<p>I think I fully understand the derivation towards &quot;<span class="math-container">$Z$</span> is a Gumbel&quot;. But then I stuck on deriving the perhaps most important equation of logit model: <span class="math-container">$$ \pi(y \mid x)=\exp [U(x, y)] /\left[\int_{\mathcal{Y}} \exp U\left(x, y^{\prime}\right) d y^{\prime}\right]$$</span>​​ . I don't see how it comes from the previous derivation.</p>
<p>///</p>
<p>I even checked one of the paper cited, Dagsvik 1994, and found in its appendix (PROOF OF THEOREM4) there is a similar derivation (A.8 to A.9) but again without any further explanation. In case anyone is interested, the equations there are &quot;(A.8)
<span class="math-container">$$
\begin{gathered}
P\left(\sup _{T(z) \in A,(T(z), E(z)) \in H, z \in Z}(\hat{v}(\hat{p}(T(z)), T(z), K)+E(z)) \leqslant y\right) \\
= \begin{cases}\exp \left\{-e^{-y} \mu \int_A \exp (\hat{v}(\hat{p}(t), t, K)) G(d t)\right\} &amp; \text { for } y \geqslant c, \\
0 &amp; \text { for } y&lt;c .\end{cases}
\end{gathered}
$$</span></p>
<p>From (A.8) we get
(A.9)
<span class="math-container">$$
\begin{gathered}
P\left(\sup _{T(z) \in A,(T(z), E(z)) \in H, z \in Z}(\hat{v}(\hat{p}(T(z)), T(z), K)+E(z))\right. \\
\left.&gt;\sup _{T(z) \in D-A,(T(z), E(z)) \in H, z \in Z}(\hat{v}(\hat{p}(T(z)), T(z), K)+E(z))\right) \\
\quad=\frac{\int_{u \leqslant t, u \in D} \exp (\hat{v}(\hat{p}(u), u, K)) G(d u)}{\int_D \exp (\hat{v}(\hat{p}(u), u, K)) G(d u)} \cdot\left(1-\exp \left(-\tilde{\Lambda}_c\right)\right),
\end{gathered}
$$</span>
where
<span class="math-container">$$
\tilde{\Lambda}_c \equiv \mu e^{-c} \int_D \exp (\hat{v}(\hat{p}(t), t, K)) G(d t) .
$$</span></p>
<p>Since <span class="math-container">$\Lambda_c$</span> is the expected number of Poisson points in <span class="math-container">$H \cap(D \times R)$</span> the probability that <span class="math-container">$H \cap(D \times R)$</span> is nonempty equals
<span class="math-container">$$
1-\exp \left(-\tilde{\Lambda}_c\right) \text {. }
$$</span>&quot;</p>

<mathematical-economics><structural-estimation><logit>
Continuous logit framework
Alalalalaki
# Continuous logit framework

(Asked by: username_0 on 2024-02-13)

I am reading [Dupuy & Galichon (2014)](https://www.journals.uchicago.edu/doi/full/10.1086/677191), which extend the estimation of matching model in Choo & Siow (2006) to continuous types. The way they build the continuous logit model is based on the insights of Cosslett (1988) and Dagsvik (1994) which "have independently suggested using max-stable processes to model continuous choice."

The detail of the continuous logit model model is described in the Appendix A:

"In this paragraph, we expound the main ideas of Cosslett (1988) and Dagsvik (1994), who show how to obtain a continuous version of the multinomial logit model. Assume that $\lbrace\left(y\_k^m, \varepsilon\_k^m\right), k \in \mathbb{N} \rbrace$ are the points of a Poisson point process on $\mathcal{Y} \times \mathbb{R}$ of intensity $d y \times e^{-\varepsilon} d \varepsilon$. We recall that this implies that for $S$ a subset of $\mathcal{Y} \times \mathbb{R}$, the probability that man $m$ has no acquaintance in set $S$ is $\exp \left(-\int\_S e^{-\varepsilon} d y d \varepsilon\right)$. From (2), man $m$ chooses woman $k$ among his acquaintances such that his utility is maximized; that is, man $m$ solves
$$\max\_k \lbrace U\left(x,y\_k^m\right)+\varepsilon\_k^m \rbrace$$.

Letting $Z$ be the value of this maximum, one has for any $c \in \mathbb{R}$
$$
\operatorname{Pr}(Z \leq c)=\operatorname{Pr}\left(U\left(x, y\_k^m\right)+\varepsilon\_k^m \leq c \forall k\right),
$$
which is exactly the probability that the Poisson point process $\left(y\_k, \varepsilon\_k^m\right)$ has no point in $\{(y, \varepsilon): U(x, y)+\varepsilon>c\}$; thus
$$
\begin{aligned}
\log \operatorname{Pr}(Z \leq c) & =-\iint\_{\mathcal{Y} \times \mathbb{R}} 1(U(x, y)+\varepsilon>c) d y e^{-\varepsilon} d \varepsilon \\
& =-\int\_{\mathcal{Y}} \int\_{c-U(x, y)} e^{-\varepsilon} d \varepsilon d y \\
& =-\int\_{\mathcal{Y}} e^{-c+U(x, y)} d y \\
& =-\exp \left[-c+\log \int\_{\mathcal{Y}} \exp U(x, y) d y\right]
\end{aligned}
$$

Hence $Z$ is a $\left(\log \int\_{\mathcal{Y}} \exp U(x, y) d y, 1\right)$ Gumbel. In particular,
$\mathbb{E}\left[\max\_k \lbrace U\left(x, y\_k^m\right)+\varepsilon\_k^m \rbrace \right]=\log \int\_{\mathcal{y}} \exp U(x, y) d y,$

and the choice probabilities are given by their density with respect to the Lebesgue measure

$$\pi(y \mid x)=\exp [U(x, y)] /\left[\int\_{\mathcal{Y}} \exp U\left(x, y^{\prime}\right) d y^{\prime}\right] .$$

The same logic also implies that $\lbrace \varepsilon\_k: k \in \mathbb{N} \rbrace$ has a Gumbel distribution. Indeed, the probability that this Poisson point process has no element in the set $\{\varepsilon: \varepsilon>c\}$ is equal to $$
\exp \left(-\int\_c^{+\infty} e^{-\varepsilon} d \varepsilon\right)=\exp [-\exp (-c)]
$$
which is equivalent to saying that $\operatorname{Pr}\left(\max \_{k \in \mathbb{N}} \varepsilon\_k \leq c\right)=\exp [-\exp (-c)]$. Finally, note that a similar argument would show that $m$ has almost surely an infinite, though countable, number of acquaintances, as announced. "

///

I think I fully understand the derivation towards "$Z$ is a Gumbel". But then I stuck on deriving the perhaps most important equation of logit model: $$ \pi(y \mid x)=\exp [U(x, y)] /\left[\int\_{\mathcal{Y}} \exp U\left(x, y^{\prime}\right) d y^{\prime}\right]$$​​ . I don't see how it comes from the previous derivation.

///

I even checked one of the paper cited, Dagsvik 1994, and found in its appendix (PROOF OF THEOREM4) there is a similar derivation (A.8 to A.9) but again without any further explanation. In case anyone is interested, the equations there are "(A.8)
$$
\begin{gathered}
P\left(\sup \_{T(z) \in A,(T(z), E(z)) \in H, z \in Z}(\hat{v}(\hat{p}(T(z)), T(z), K)+E(z)) \leqslant y\right) \\
= \begin{cases}\exp \left\{-e^{-y} \mu \int\_A \exp (\hat{v}(\hat{p}(t), t, K)) G(d t)\right\} & \text { for } y \geqslant c, \\
0 & \text { for } y<c .\end{cases}
\end{gathered}
$$

From (A.8) we get
(A.9)
$$
\begin{gathered}
P\left(\sup \_{T(z) \in A,(T(z), E(z)) \in H, z \in Z}(\hat{v}(\hat{p}(T(z)), T(z), K)+E(z))\right. \\
\left.>\sup \_{T(z) \in D-A,(T(z), E(z)) \in H, z \in Z}(\hat{v}(\hat{p}(T(z)), T(z), K)+E(z))\right) \\
\quad=\frac{\int\_{u \leqslant t, u \in D} \exp (\hat{v}(\hat{p}(u), u, K)) G(d u)}{\int\_D \exp (\hat{v}(\hat{p}(u), u, K)) G(d u)} \cdot\left(1-\exp \left(-\tilde{\Lambda}\_c\right)\right),
\end{gathered}
$$
where
$$
\tilde{\Lambda}\_c \equiv \mu e^{-c} \int\_D \exp (\hat{v}(\hat{p}(t), t, K)) G(d t) .
$$

Since $\Lambda\_c$ is the expected number of Poisson points in $H \cap(D \times R)$ the probability that $H \cap(D \times R)$ is nonempty equals
$$
1-\exp \left(-\tilde{\Lambda}\_c\right) \text {. }
$$"

**Tags:** mathematical-economics, structural-estimation, logit


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## 2 answers

### Answer by username_0 on 2024-02-14. Score: 3

This derivation follows the same procedure as the one used in discrete logit model.

Let's denote the index of the maximum given $y$ is selected is $(y\_i^m,\varepsilon\_i^m)$, where $y\_i^m = y$. We have $$\operatorname{Pr}\_i = \operatorname{Pr}\left(U\left(x, y\_k^m\right)+\varepsilon\_k^m \leq U\left(x, y\_i^m\right)+\varepsilon\_i^m \ \forall k \neq i \right) \\ = \exp \left( -\exp \left[-U\left(x, y\_i^m\right)-\varepsilon\_i^m+\log \int\_{\mathcal{Y}} \exp U(x, y) d y\right] \right)$$​.

Then we can write the choice probabilities as $$\pi(y \mid x) = \int \operatorname{Pr}\_i e^{-\varepsilon\_i} d \varepsilon\_i =\int \exp \left( -\exp \left[-U\left(x, y \right)-\varepsilon\_i +\log \int\_{\mathcal{Y}} \exp U(x, y') d y'\right] \right) e^{-\varepsilon\_i} d \varepsilon\_i \\ = \int \exp \left( -\exp \left[-\varepsilon\_i\right] \exp \left[-U\left(x, y \right)\right] \int\_{\mathcal{Y}} \exp U(x, y') d y' \right) e^{-\varepsilon\_i} d \varepsilon\_i \\ = \int \exp \left( -\exp \left[-\varepsilon\_i\right] \int\_{\mathcal{Y}} \exp \left[ U(x, y')-U\left(x, y \right) \right] d y \right) e^{-\varepsilon\_i} d \varepsilon\_i$$

Then define $t=\exp(-\varepsilon\_i)$ such that $-\exp (-\varepsilon\_i) d \varepsilon\_i=d t$ , we have $$\pi(y \mid x) = \int\_{\infty}^0 \exp \left( - t \int\_{\mathcal{Y}} \exp \left[ U(x, y')-U\left(x, y \right) \right] d y' \right) (-dt) \\ = \int\_0^{\infty} \exp \left( - t \int\_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right]d y' \right) dt \\ = \frac{\exp \left( - t \int\_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right] d y' \right)}{ - \int\_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right] d y' } \left.\right|\_0 ^{\infty} \\ = \frac{1}{ \int\_{\mathcal{Y}} \exp \left[U(x, y')-U\left(x, y \right) \right] d y' } =\frac{\exp U\left(x, y \right)}{ \int\_{\mathcal{Y}} \exp U(x, y') d y' }$$

Any comments are welcome.


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### Answer by username_1 on 2024-02-14. Score: 2

As they state, $\pi(y|x)$ is the density of $\ln\left(\int\_Y \exp(U(x,y) dy\right)$ (the density being the derivative of the cdf).

If we take the derivative of $\ln\left(\int\_Y \exp(U(x,y')) dy' \right)$ with respect to $y$, we obtain:

$$
\pi(y|x) = \frac{\exp(U(x,y))}{\int\_Y \exp(U(x,y') dy'}
$$
The denominator follows from the $\ln$ operator. The numerator follows from the chain rule given that $\dfrac{d \left(\int\_Y exp(U(x,y')) dy'\right)}{dy} = \exp(U(x,y))$.

> **Comments:**
> **username_0**: Thanks very much for your answer as always. The derivation is clear but can you explain why $\log \int_Y \exp U(x, y) d y$ is the cdf? And their statement is "choice probabilities are given by their density with respect to the Lebesgue measure"  but not the choice probabilities are the density function itself?  

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