Hi I was trying to understand hazard rate and got stuck in the middle. Any suggestions are welcome. Below is the problem.

Consider the following distribution for the duration of an unemployment spell $T_{i}$:

$f(t|\theta_{i})=\theta_{i}e^{-\theta_{i}t}$ if $t\geq 0$ and 0 otherwise.

for an individual specific hazard rate $\theta_{i}$

(a) Argue that the random variable $F_{X}(X)\sim U[0,1]$ when $F_{X}$ is a strictly increasing cdf.

(b) Give the cdf of $T_{i}$ and by using (a) show that $T_{i}=-\frac{1}{\theta_{i}}\ln U_{i}$ where $U_{i}\sim U[0,1]$

Now specify individual-specific hazard rate $\theta_{i}=\lambda e^{x_{i}'\beta}$

(c) Demonstrate that you can represent a transformation of $T_{i}$ as a linear regression model with an error term $e_{i}$ that is independent of $x_{i}$

(d) Derive the c.d.f. of the error term $e_{i}$. Is $E[e_{i}] = 0$? If not, what is the interpretation of the coefficients from a least-squares regression for your model in (c)?

Here is my attempt.

(a) $P(F_{X}(X)\leq m)=P(X\leq F_{X}^{-1}(m))=F(F^{-1}(m))=m$ since $F$ is strictly increasing, the inverse function is well-defined. Thus F is uniformly distributed

(b) Clearly, $F(T_{i}|\theta_{i})=1-e^{-\theta_{i}T_{i}}$. This cdf is strictly increasing. By (a), $F(T_{i}|\theta_{i})=1-e^{-\theta_{i}t}=U_{i}$ where $U_{i}\sim U[0,1]$. Taking log on the both sides we have $T_{i}=-\frac{1}{\theta_{i}}\ln U_{i}$ as desired.

I am getting stuck at (c)

If we take log again, we may have

$\ln T_{i}=\ln \theta_{i}+\ln(\ln U_{i})$, but $\ln(\ln U_{i})$ is not defined. How can I proceed from here? I don’t get what transformation of $T_{i}$ means

  • $\begingroup$ Thanks! I think you are right $\endgroup$ – Shady Nov 5 '19 at 20:14
  • $\begingroup$ Use the self-study tag. $\endgroup$ – Michael R. Chernick Nov 8 '19 at 22:21

The broad approach is correct but the "$-$" needs to be paired with the $\ln U_i$ (making it positive) and then it's possible to take logs. The sign of the other term resolves itself if you continue by making a suitable substitution.

You end up with an "error term" corresponding to $\ln(-\ln(U_i))$ which has a non-zero mean, just as the question suggested (indeed its mean is the negative of a well-known constant).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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