Problem 1¶

Below, we derive

$\Pr(d_i = j) = \frac{\exp(\delta_j)}{\sum_{k \in \mathcal{J}} \exp(\delta_k)}$

Let $f$ be the pdf of Extreme Value Type I random variable: $f(x) := F'(x) = \mathrm e^{-x} \cdot \exp(-\exp(-x))$ .

$\begin{align*} \Pr(d_i = j) &= \Pr(\forall k \neq j; \ u_{ij} \geq u_{ik}) \\ &= \Pr(\forall k \neq j; \ \epsilon_{ij} + \delta_j - \delta_k \geq \epsilon_{ik}) \\ &= \prod_{k \neq j} \Pr(\epsilon_{ij} + \delta_j - \delta_k \geq \epsilon_{ik}) \quad (\because \ \text{indep.}) \\ &= \int_{-\infty}^\infty F(\epsilon_{ij} + \delta_j - \delta_k) f(\epsilon_{ij}) \mathrm d \epsilon_{ij} \\ &= \int_{-\infty}^\infty \exp\left( \sum_{k \neq j} - \exp(-\epsilon_{ij} - \delta_j + \delta_k) \right) \mathrm e^{-\epsilon_{ij}} \cdot \exp(-\exp(-\epsilon_{ij})) \mathrm d \epsilon_{ij} \\ &= \int_{-\infty}^\infty \exp\left( \sum_{k} - \exp(-x - \delta_j + \delta_k) \right) \mathrm e^{-x} \mathrm d x \quad (x := \epsilon_{ij})\\ &= \int_{-\infty}^\infty \exp\left( - \mathrm e^{-x} \sum_{k} \exp(\delta_k - \delta_j) \right) \mathrm e^{-x} \mathrm d x \\ &= \int_{0}^\infty \exp\left( -t \sum_{k} \exp(\delta_k - \delta_j) \right) \mathrm d t \quad (t := \mathrm e^{-x}) \\ &= \left[- \left(\sum_{k} \exp(\delta_k - \delta_j)\right)^{-1} \exp\left( -t \sum_{k} \exp(\delta_k - \delta_j) \right) \right]_{0}^{\infty} \\ &= \left(\sum_{k} \exp(\delta_k - \delta_j)\right)^{-1} \\ &= \frac{\exp(\delta_j)}{\sum_k \exp (\delta_k)} \end{align*}$

Problem 2¶

The log-likelihood is:

$\sum_i \sum_j y_{ij} \cdot \log \left( \frac{\exp(X_{ij} \beta)}{1 + \sum_k \exp(X_{ik} \beta)} \right)$

In order to use scipy.optimize.minimize, we define a loss function that is the likelihood multiplied by $-1$ .

The estimated value is $\beta^* \approx (-1.88673942, 0.09717683, -1.02683967)$ .

Below is the code I used:

In [1]:

# import
import pandas as pd
import numpy as np
from scipy.optimize import minimize


# read csv
df = pd.read_csv('../data/DataPS201901.csv', header=None)
data = df.values

# pre-processing
x = [{'hp': 1.0, 'fe': 5.0}, {'hp': 1.2, 'fe': 3.5}, {'hp': 1.4, 'fe': 2.0}]
y = data[:,0]

age = data[:, 1]
gender = data[:, 2]
N = data.shape[0] # the number of agents
J = data.shape[1] # the number of goods

age = age[:, np.newaxis]/100 # rescaling
gender = gender[:, np.newaxis]

X = []
for j in range(data.shape[1]):
    temp = np.hstack([np.ones((N,1)), age * x[j]['hp'], gender * x[j]['fe']])
    X.append(temp)

    
# loss function
## use numpy to speed up
def loss(beta):
    # exp_Xij_beta(i,j) = exp(X_ij @ beta)
    for j in range(J):
        if j == 0:
            exp_Xij_beta = np.exp(X[j] @ beta)[:, np.newaxis]
        else:
            exp_Xij_beta = np.hstack([exp_Xij_beta, np.exp(X[j] @ beta)[:, np.newaxis]])

    exp_sum = np.sum(exp_Xij_beta, axis=1) + 1

    # z_ij は，(i,j)に対応するlogの中身
    z0 = np.ones((N,1)) / exp_sum[:, np.newaxis] # outside option
    z1 = exp_Xij_beta / exp_sum[:, np.newaxis]
    z = np.hstack([z0, z1])
    w = np.log(z)
    
    # Y_ij = 1 iff agent i chooses option j
    Y = np.zeros((N, J+1))
    for i in range(N):
        Y[i][y[i]] = 1
    
    return - (Y * w).sum()

res = minimize(loss, x0=[1, 1, 1])
beta = res.x
beta[1] = beta[1] / 100 # rescaling

print('The estimated beta is {}'.format(beta))

The estimated beta is [-1.88673942  0.09717683 -1.02683967]