Problem 3. The groundhog problem. Each of $n$ blood samples is sent to a lab with two
technicians. One technician is highly skilled, and the other is a groundhog. Each blood sample is
independently assigned to the skilled technician with probability 1/2 and to the groundhog with
probability 1/2. Both the skilled technician's and the groundhog's measurements are normally
distributed around the true value, $\theta$, but the variance of the measurement produced by the skilled
technician, $\sigma_0^2$, is much less than the variance of the groundhog's measurement, $\sigma_1^2$. We observe
$(Y_1, A_1), ..., (Y_n, A_n)$ iid, where $A_i \sim Bern(1/2)$ is the indicator of the groundhog analyzing the
$i$th sample and $Y_i|A_i \sim N(0, \sigma_{A_i}^2)$. Assume $\sigma_0^2$ and $\sigma_1^2$ are known and $\theta$ is unknown.
(a) Show that the MLE is of the form $\hat{\theta}_{MLE} = c_0Y_0 + c_1Y_1$, where
$Y_0 = \frac{\sum_{i=1}^n (1-A_i)Y_i}{\sum_{i=1}^n (1-A_i)}$
is the mean of the skilled technician's measurements and
$Y_1 = \frac{\sum_{i=1}^n A_iY_i}{\sum_{i=1}^n A_i}$
is the mean of the groundhog's measurements, and $c_0$ and $c_1$ are functions of $A = (A_1,..., A_n)$.
(b) Compute the Fisher information of the sample. That is, compute $-E_0[l''(\theta)]$, where $l''(\theta)$ is
the second derivative of the log-likelihood based on $(Y_1, A_1),..., (Y_n, A_n)$.
(c) The observed information of the sample is defined as $i = -l''(\hat{\theta}_{MLE})$. Compute the observed
information. What does it equal when all of the measurements happen to be made by the skilled
technician? What does it equal when all of the measurements happen to be made by the ground-
hog? Why might we consider the observed information to be a more relevant measure of uncertainty
than the Fisher information?
This problem illustrates that while the Fisher information represents the expected or average cur-
vature of the log likelihood, the curvature of the actual, realized log-likelihood function may be
quite different. The observed information accounts for this.
