本文从决策者效用最大化，利用 Gumbel 分布的假设，推出各选项的 logit 概率，并分析了其隐含的 IIA 问题。
效用最大化与 Logit
链接到标题
一个决策者 n n n J J J j j j U n j = V n j + ε n j
U_{nj} = V_{nj} + \varepsilon_{nj}
U nj = V nj + ε nj V n j V_{nj} V nj ε n j \varepsilon_{nj} ε nj
决策者选择效用最高的选项，那么决策者 n n n i i i P n i = P r o b ( V n i + ε n i > V n j + ε n j ∀ j ≠ i ) = P r o b ( ε n j < ε n i + V n i − V n j ∀ j ≠ i ) .
\begin{align*}
P_{ni} & = Prob(V_{ni} + \varepsilon_{ni} > V_{nj} +
\varepsilon_{nj} \quad \forall j \neq i) \newline
& = Prob(\varepsilon_{nj} < \varepsilon_{ni} + V_{ni} - V_{nj} \quad \forall j \neq i).
\end{align*}
P ni = P ro b ( V ni + ε ni > V nj + ε nj ∀ j = i ) = P ro b ( ε nj < ε ni + V ni − V nj ∀ j = i ) . ε n j \varepsilon_{nj} ε nj P n i ∣ ε n i = ∏ j ≠ i e − e − ( ε n i + V n i − V n j ) .
P_{ni} \mid \varepsilon_{ni} = \prod_{j \neq i} e^{-e^{-(\varepsilon_{ni} + V_{ni} - V_{nj})}}.
P ni ∣ ε ni = j = i ∏ e − e − ( ε ni + V ni − V nj ) . P n i = ∫ ( ∏ j ≠ i e − e − ( ε n i + V n i − V n j ) ) e − ε n i e − e − ε n i d ε n i .
P_{ni} = \int \left( \prod_{j \neq i} e^{-e^{-(\varepsilon_{ni} + V_{ni} - V_{nj})}} \right) e^{-\varepsilon_{ni}} e^{-e^{-\varepsilon_{ni}}} d \varepsilon_{ni}.
P ni = ∫ j = i ∏ e − e − ( ε ni + V ni − V nj ) e − ε ni e − e − ε ni d ε ni . P n i = e V n i ∑ j e V n j ,
P_{ni} = \frac{
e^{V_{ni}}
}{
\sum_j e^{V_{nj}}
},
P ni = ∑ j e V nj e V ni , CB495-03Drv.tex (berkeley.edu) 第 3.10 小节。
当效用函数是线性形式时，例如 V n j = β ’ x n j V_{nj} = \beta’ x_{nj} V nj = β ’ x nj P n i = e β ’ x n j ∑ j e β ’ x n j ,
P_{ni} = \frac{
e^{\beta’ x_{nj}}
}{
\sum_j e^{\beta’ x_{nj}}
},
P ni = ∑ j e β ’ x nj e β ’ x nj , x n j x_{nj} x nj j j j
Independence from Irrelevant Alternatives
链接到标题
Logit 模型对选项间的替代模式有特殊限制，对于任意两个选项 i i i k k k P n i P n k = e V n i / ∑ j e V n j e V n k / ∑ j e V n j = e V n i e V n k = e V n i − V n k .
\frac{P_{ni}}{P_{nk}} =
\frac{e^{V_{ni}}/ \sum_j e^{V_{nj}}}
{e^{V_{nk}}/ \sum_j e^{V_{nj}}} =
\frac{e^{V_{ni}}}{e^{V_{nk}}} =
e^{V_{ni} - V_{nk}}.
P nk P ni = e V nk / ∑ j e V nj e V ni / ∑ j e V nj = e V nk e V ni = e V ni − V nk . i i i k k k
这一特性有时并不符合现实，一个经典的例子是红蓝巴士问题：假设一个人选择开车 (c) 或乘蓝色巴士 (bb) 上班，并且概率各为一半：P c = P b b = 1 2 P_c = P_{bb} = \frac{1}{2} P c = P bb = 2 1 P r b / P b b = 1 P_{rb} / P_{bb} = 1 P r b / P bb = 1 P c / P b b P_c / P_{bb} P c / P bb P c / P b b = 1 和 P r b / P b b = 1 ⟹ P c = P b b = P r b = 1 3 .
P_c / P_{bb} = 1 \text{ 和 } P_{rb} / P_{bb} = 1
\implies P_c = P_{bb} = P_{rb} = \frac{1}{3}.
P c / P bb = 1 和 P r b / P bb = 1 ⟹ P c = P bb = P r b = 3 1 . 1 2 \frac{1}{2} 2 1 P c = 1 2 和 P b b = R r b = 1 4 .
P_c = \frac{1}{2} \text{ 和 } P_{bb} = R_{rb} = \frac{1}{4}.
P c = 2 1 和 P bb = R r b = 4 1 .