<p>I'm working on a model involving choice probabilities in a discrete choice framework where the random utility component is assumed to be Gumbel. Below is the problem definition and my attempt at deriving the necessary probabilities and expectations. However, I'm encountering some difficulties in understanding the steps and ensuring that my derivations are correct. I'd greatly appreciate any help or insights you could provide!</p>
<p><strong>Problem Definition:</strong></p>
<p>I need to calculate the probability:
<span class="math-container">$$
\Pr\{c_j &lt; \min_{i \neq j} c_i\}
$$</span>
where the costs <span class="math-container">$(c_i)$</span> are given by:
<span class="math-container">$$
c_i = \exp(v_i + t_i)
$$</span>
with <span class="math-container">$t_i \sim F(t) = \exp(-\exp(-\theta t))$</span>.</p>
<p><strong>My Attempt at the Derivation:</strong></p>
<ol>
<li><p><strong>Starting with the distribution:</strong>
<span class="math-container">$$
   f(t) = \frac{d}{dt}F(t) = \theta \exp(-\theta t) \exp(-\exp(-\theta t))
   $$</span></p>
</li>
<li><p><strong>The required probability becomes:</strong>
<span class="math-container">$$
   \Pr\{c_j &lt; \min_{i \neq j} c_i\} = \Pr\{\exp(v_j + t_j) &lt; \min_{i \neq j} \exp(v_i + t_i)\}
   $$</span>
which simplifies to:
<span class="math-container">$$
   \Pr\{v_j + t_j &lt; \min_{i \neq j} (v_i + t_i)\}
   $$</span></p>
</li>
<li><p><strong>Further expansion leads to:</strong>
<span class="math-container">$$
   \Pr\{t_j &lt; \min_{i \neq j} (v_i - v_j + t_i)\} = \int_{-\infty}^{\infty} \Pr\{t_j &lt; \min_{i \neq j} (v_i - v_j + t_i)\} f(t_j) dt_j
   $$</span>
<span class="math-container">$$
   = \int_{-\infty}^{\infty} \prod_{i \neq j} F(v_i - v_j + t_i) f(t_j) dt_j
   $$</span></p>
</li>
<li><p><strong>Substituting the distribution:</strong>
<span class="math-container">$$
   = \int_{-\infty}^{\infty} \prod_{i \neq j} \exp(-\exp(-\theta(v_i - v_j + t_i))) \theta \exp(-\theta t_j) \exp(-\exp(-\theta t_j)) dt_j
   $$</span>
Substituting <span class="math-container">$ \exp(-\exp(-\theta t_j)) = \exp(-\exp(-\theta(t_j + v_j - v_j))) $</span>:
<span class="math-container">$$
   = \int_{-\infty}^{\infty} \prod_{i} \exp(-\exp(-\theta(v_i - v_j + t_j))) \theta \exp(-\theta t_j) dt_j
   $$</span>
<span class="math-container">$$
   = \int_{-\infty}^{\infty} \exp(-\exp(-\theta t_j) \sum_{i} \exp(-\theta(v_i - v_j))) \theta \exp(-\theta t_j) dt_j
   $$</span></p>
</li>
<li><p><strong>Integral transformation <span class="math-container">$u = -\exp(-\theta t_j)$</span>:</strong>
<span class="math-container">$$
   = \int_{-\infty}^{0} \exp(u \sum_{i} \exp(-\theta(v_i - v_j))) du
   $$</span>
<span class="math-container">$$
   = \frac{\exp(u \sum_{i} \exp(-\theta(v_i - v_j)))}{\sum_{i} \exp(-\theta(v_i - v_j))} \Bigg|_{-\infty}^{0}
   $$</span>
<span class="math-container">$$
   = \frac{1}{\sum_{i} \exp(-\theta(v_i - v_j))}
   $$</span>
<span class="math-container">$$
   = \frac{\exp(-\theta v_j)}{\sum_{i} \exp(-\theta v_i)}
   $$</span></p>
</li>
<li><p><strong>Expectation:</strong>
<span class="math-container">$$
   F(c) = \Pr(c_j \leq c, \forall j)
   $$</span>
<span class="math-container">$$
   = \Pr(\exp(v_j + t_j) \leq c, \forall j)
   $$</span>
<span class="math-container">$$
   = \Pr(v_j + t_j \leq \ln(c), \forall j)
   $$</span>
<span class="math-container">$$
   = \Pr(t_j \leq \ln(c) - v_j, \forall j)
   $$</span>
<span class="math-container">$$
   F(c) = \prod_{j=1}^{m} \Pr(t_j \leq \ln(c) - v_j)
   $$</span>
<span class="math-container">$$
   \Pr(t_j \leq \ln(c) - v_j) = F(\ln(c) - v_j) = \exp(-\exp(-\theta(\ln(c) - v_j)))
   $$</span>
<span class="math-container">$$
   F(c) = \prod_{j=1}^{m} \exp(-\exp(-\theta(\ln(c) - v_j))) = \exp(-\sum_{j=1}^{m} \exp(-\theta(\ln(c) - v_j)))
   $$</span>
<span class="math-container">$$
   F(c) = \exp(-c^{-\theta} \sum_{j} \exp(\theta v_j))
   $$</span>
<span class="math-container">$$
   f(c) = \frac{d}{dc} F(c) = \theta c^{-\theta-1} \left(\sum_{j} \exp(\theta v_j)\right) \exp(-c^{-\theta} \sum_{j} \exp(\theta v_j))
   $$</span>
<span class="math-container">$$
   E[c] = \int_{0}^{\infty} c \cdot f(c) dc
   $$</span>
<span class="math-container">$$
   = \int_{0}^{\infty} c \cdot \theta c^{-\theta-1} \left(\sum_{j} \exp(\theta v_j)\right) \exp(-c^{-\theta} \sum_{j} \exp(\theta v_j)) dc
   $$</span>
<span class="math-container">$$
   = \theta \left(\sum_{j} \exp(\theta v_j)\right) \int_{0}^{\infty} c^{-\theta} \exp(-c^{-\theta} \sum_{j} \exp(\theta v_j)) dc
   $$</span>
Let <span class="math-container">$ u = c^{\theta} \sum_{j} \exp(-\theta v_j), c = \left(\frac{u}{\sum_{j} \exp(-\theta v_j)}\right)^{1/\theta}, dc = \frac{1}{\theta} u^{1/\theta - 1} \left(\frac{1}{\sum_{j} \exp(-\theta v_j)}\right)^{1/\theta} du $</span>:
<span class="math-container">$$
   E[c] = \theta \left(\sum_{j} \exp(\theta v_j)\right) \frac{1}{\theta} \int_{0}^{\infty} \left(\frac{u}{\sum_{j} \exp(-\theta v_j)}\right)^{1/\theta} u^{1/\theta - 1} \exp(-u) \cdot \frac{1}{\sum_{j} \exp(-\theta v_j)} du
   $$</span>
<span class="math-container">$$
   = \left(\sum_{j} \exp(\theta v_j)\right) \cdot \left(\sum_{j} \exp(-\theta v_j)\right)^{-1/\theta} \int_{0}^{\infty} u^{1/\theta - 1} \exp(-u) du
   $$</span></p>
</li>
</ol>
<p><strong>Questions:</strong></p>
<p>Have I correctly derived the probability and expectation expressions? If not, where did I go wrong, and how can I correct it and finish the derivation?</p>
<p>Thank you in advance for your assistance!</p>

CC BY-SA 4.0
2024-08-29T10:34:32.430Z
The $Pr(c_j < \min_{i\not=j} c_i) = Pr(c_j = \min_i c_i) = Pr(-c_j = \max_i -c_i)$ for which you derive the wellknown result. It is unclear exactly what you are trying to calculate the expectation of. It would be easier to do that derivation if you start with defining the expectation you are trying to derive. In particular, it is unclear whether you are trying to calculate $E[min(c_i)]$ or simply $E[c_i]$. You are assuming independence of $t_i$'s so why would the expectation of $c_i$ depend on the values $v_j$ with $j\not=i$? There seems to me to be some mistake here.
Jesper Hybel
CC BY-SA 4.0
2024-09-01T17:09:18.433Z
Thank you for your suggestion! I followed your approach and attempted the derivation, but I encountered an issue where introducing − altered the distribution function. This change made the integral quite complex and challenging to solve. @Jesper Hybel
XJ.C
<statistics><probability><expectations>
Derivation of Choice Probabilities and Expected Values in Discrete Choice Models with minimal values
XJ.C
# Derivation of Choice Probabilities and Expected Values in Discrete Choice Models with minimal values

(Asked by: username_0 on 2024-08-29)

I'm working on a model involving choice probabilities in a discrete choice framework where the random utility component is assumed to be Gumbel. Below is the problem definition and my attempt at deriving the necessary probabilities and expectations. However, I'm encountering some difficulties in understanding the steps and ensuring that my derivations are correct. I'd greatly appreciate any help or insights you could provide!

**Problem Definition:**

I need to calculate the probability:
$$
\Pr\{c\_j < \min\_{i \neq j} c\_i\}
$$
where the costs $(c\_i)$ are given by:
$$
c\_i = \exp(v\_i + t\_i)
$$
with $t\_i \sim F(t) = \exp(-\exp(-\theta t))$.

**My Attempt at the Derivation:**

1. **Starting with the distribution:**
   $$
   f(t) = \frac{d}{dt}F(t) = \theta \exp(-\theta t) \exp(-\exp(-\theta t))
   $$
2. **The required probability becomes:**
   $$
   \Pr\{c\_j < \min\_{i \neq j} c\_i\} = \Pr\{\exp(v\_j + t\_j) < \min\_{i \neq j} \exp(v\_i + t\_i)\}
   $$
   which simplifies to:
   $$
   \Pr\{v\_j + t\_j < \min\_{i \neq j} (v\_i + t\_i)\}
   $$
3. **Further expansion leads to:**
   $$
   \Pr\{t\_j < \min\_{i \neq j} (v\_i - v\_j + t\_i)\} = \int\_{-\infty}^{\infty} \Pr\{t\_j < \min\_{i \neq j} (v\_i - v\_j + t\_i)\} f(t\_j) dt\_j
   $$
   $$
   = \int\_{-\infty}^{\infty} \prod\_{i \neq j} F(v\_i - v\_j + t\_i) f(t\_j) dt\_j
   $$
4. **Substituting the distribution:**
   $$
   = \int\_{-\infty}^{\infty} \prod\_{i \neq j} \exp(-\exp(-\theta(v\_i - v\_j + t\_i))) \theta \exp(-\theta t\_j) \exp(-\exp(-\theta t\_j)) dt\_j
   $$
   Substituting $ \exp(-\exp(-\theta t\_j)) = \exp(-\exp(-\theta(t\_j + v\_j - v\_j))) $:
   $$
   = \int\_{-\infty}^{\infty} \prod\_{i} \exp(-\exp(-\theta(v\_i - v\_j + t\_j))) \theta \exp(-\theta t\_j) dt\_j
   $$
   $$
   = \int\_{-\infty}^{\infty} \exp(-\exp(-\theta t\_j) \sum\_{i} \exp(-\theta(v\_i - v\_j))) \theta \exp(-\theta t\_j) dt\_j
   $$
5. **Integral transformation $u = -\exp(-\theta t\_j)$:**
   $$
   = \int\_{-\infty}^{0} \exp(u \sum\_{i} \exp(-\theta(v\_i - v\_j))) du
   $$
   $$
   = \frac{\exp(u \sum\_{i} \exp(-\theta(v\_i - v\_j)))}{\sum\_{i} \exp(-\theta(v\_i - v\_j))} \Bigg|\_{-\infty}^{0}
   $$
   $$
   = \frac{1}{\sum\_{i} \exp(-\theta(v\_i - v\_j))}
   $$
   $$
   = \frac{\exp(-\theta v\_j)}{\sum\_{i} \exp(-\theta v\_i)}
   $$
6. **Expectation:**
   $$
   F(c) = \Pr(c\_j \leq c, \forall j)
   $$
   $$
   = \Pr(\exp(v\_j + t\_j) \leq c, \forall j)
   $$
   $$
   = \Pr(v\_j + t\_j \leq \ln(c), \forall j)
   $$
   $$
   = \Pr(t\_j \leq \ln(c) - v\_j, \forall j)
   $$
   $$
   F(c) = \prod\_{j=1}^{m} \Pr(t\_j \leq \ln(c) - v\_j)
   $$
   $$
   \Pr(t\_j \leq \ln(c) - v\_j) = F(\ln(c) - v\_j) = \exp(-\exp(-\theta(\ln(c) - v\_j)))
   $$
   $$
   F(c) = \prod\_{j=1}^{m} \exp(-\exp(-\theta(\ln(c) - v\_j))) = \exp(-\sum\_{j=1}^{m} \exp(-\theta(\ln(c) - v\_j)))
   $$
   $$
   F(c) = \exp(-c^{-\theta} \sum\_{j} \exp(\theta v\_j))
   $$
   $$
   f(c) = \frac{d}{dc} F(c) = \theta c^{-\theta-1} \left(\sum\_{j} \exp(\theta v\_j)\right) \exp(-c^{-\theta} \sum\_{j} \exp(\theta v\_j))
   $$
   $$
   E[c] = \int\_{0}^{\infty} c \cdot f(c) dc
   $$
   $$
   = \int\_{0}^{\infty} c \cdot \theta c^{-\theta-1} \left(\sum\_{j} \exp(\theta v\_j)\right) \exp(-c^{-\theta} \sum\_{j} \exp(\theta v\_j)) dc
   $$
   $$
   = \theta \left(\sum\_{j} \exp(\theta v\_j)\right) \int\_{0}^{\infty} c^{-\theta} \exp(-c^{-\theta} \sum\_{j} \exp(\theta v\_j)) dc
   $$
   Let $ u = c^{\theta} \sum\_{j} \exp(-\theta v\_j), c = \left(\frac{u}{\sum\_{j} \exp(-\theta v\_j)}\right)^{1/\theta}, dc = \frac{1}{\theta} u^{1/\theta - 1} \left(\frac{1}{\sum\_{j} \exp(-\theta v\_j)}\right)^{1/\theta} du $:
   $$
   E[c] = \theta \left(\sum\_{j} \exp(\theta v\_j)\right) \frac{1}{\theta} \int\_{0}^{\infty} \left(\frac{u}{\sum\_{j} \exp(-\theta v\_j)}\right)^{1/\theta} u^{1/\theta - 1} \exp(-u) \cdot \frac{1}{\sum\_{j} \exp(-\theta v\_j)} du
   $$
   $$
   = \left(\sum\_{j} \exp(\theta v\_j)\right) \cdot \left(\sum\_{j} \exp(-\theta v\_j)\right)^{-1/\theta} \int\_{0}^{\infty} u^{1/\theta - 1} \exp(-u) du
   $$

**Questions:**

Have I correctly derived the probability and expectation expressions? If not, where did I go wrong, and how can I correct it and finish the derivation?

Thank you in advance for your assistance!

**Tags:** statistics, probability, expectations
> **Comments:**
> **username_1**: The $Pr(c_j < \min_{i\not=j} c_i) = Pr(c_j = \min_i c_i) = Pr(-c_j = \max_i -c_i)$ for which you derive the wellknown result. It is unclear exactly what you are trying to calculate the expectation of. It would be easier to do that derivation if you start with defining the expectation you are trying to derive. In particular, it is unclear whether you are trying to calculate $E[min(c_i)]$ or simply $E[c_i]$. You are assuming independence of $t_i$'s so why would the expectation of $c_i$ depend on the values $v_j$ with $j\not=i$? There seems to me to be some mistake here.  
> **username_0**: Thank you for your suggestion! I followed your approach and attempted the derivation, but I encountered an issue where introducing − altered the distribution function. This change made the integral quite complex and challenging to solve. @Jesper Hybel  

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