I am reading Dupuy & Galichon (2014), 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 $$ 1-\exp \left(-\tilde{\Lambda}_c\right) \text {. } $$"


2 Answers 2


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.


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 $\dfrac{d \left(\int_Y exp(U(x,y')) dy'\right)}{dy} = \exp(U(x,y))$.

  • 1
    $\begingroup$ 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? $\endgroup$ Feb 14 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.