We are given three observations on binary choice with \(y_{1}=1, y_{2}=1, y_{3}=0\). Consider a logit model with only an intercept, \(P(y=1)=\Lambda\left(\gamma_{1}\right)\), where \(\Lambda(\bullet)\) is the logistic \(c d f\).

a. Show that the log-likelihood function is \(\ln L\left(\gamma_{1}\right)=2 \ln \Lambda\left(\gamma_{1}\right)+\ln \left[1-\Lambda\left(\gamma_{1}\right)\right]\).

b. Show that \(d \ln L\left(\gamma_{1}\right) / d \gamma_{1}=2 \lambda\left(\gamma_{1}\right) / \Lambda\left(\gamma_{1}\right)-\lambda\left(\gamma_{1}\right) /\left[1-\Lambda\left(\gamma_{1}\right)\right]\), where \(\lambda(\cdot)\) is the logistic \(p d f\) in (16.14). 

c. The value of \(\gamma_{1}\) such that \(d \ln L\left(\gamma_{1}\right) / d \gamma_{1}=0\) is the maximum likelihood estimator \(\tilde{\gamma}_{1}\). True, false, or maybe?

d. It can be shown that for the logit model \(\ln L\left(\gamma_{1}\right)\) is strictly concave, meaning that the second derivative is negative for all values of \(\gamma_{1}\) or \(d^{2} \ln L\left(\gamma_{1}\right) / d \gamma_{1}{ }^{2}

e. Setting the derivative in (c) to zero and solving, show that \(\Lambda\left(\tilde{\gamma}_{1}\right)=2 / 3\). [Note: This does not require you to first solve for \(\tilde{\gamma}_{1}\).]

f. Now, solve the condition in (c) to show that \(\tilde{\gamma}_{1}=-\ln (1 / 2)\).

Data From Equation 16.14:-

Transcribed Image Text:

P(y = 1, y = 1, y3 = 0x1, x2, x3) = [ + (1.5)] [ + B(0.6)] = x{1-B, +B(0.7)]} L(B1, Bly, x) (16.14)