Can the empirical Hessian of an M-estimator be indefinite? Jeffrey Wooldridge in his Econometric Analysis of Cross Section and Panel Data (page 357) says that the empirical Hessian "is not guaranteed to be positive definite, or even positive semidefinite, for the particular sample we are working with.".
This seems wrong to me as (numerical problems apart) the Hessian must be positive semidefinite as a result of the definition of the M-estimator as the value of the parameter which minimizes the objective function for the given sample and the well-known fact that at a (local) minimum the Hessian is positive semidefinite.
Is my argument right?
[EDIT: The statement has been removed in the 2nd ed. of the book. See comment.]
BACKGROUND
Suppose that $\widehat \theta_N$ is an estimator obtained by minimizing
$${1 \over N}\sum_{i=1}^N q(w_i,\theta),$$
where $w_i$ denotes the $i$-th observation.
Let's denote the Hessian of $q$ by $H$,
$$H(q,\theta)_{ij}=\frac{\partial^2 q}{\partial \theta_i \partial \theta_j}$$
The asymptotic covariance of $\widehat \theta_n$ involves $E[H(q,\theta_0)]$ where $\theta_0$ is the true parameter value. One way to estimate it is to use the empirical Hesssian
$$\widehat H=\frac{1}{N}\sum_{i=1}^N H(w_i,\widehat \theta_n)$$
It is the definiteness of $\widehat H$ which is in question.
 A: The hessian is indefinite at a saddle point.  It’s possible that this may be the only stationary point in the interior of the parameter space.
Update: Let me elaborate.  First, let’s assume that the empirical Hessian exists everywhere.
If $\hat{\theta}_n$ is a local (or even global) minimum of $\sum_i q(w_i, \cdot)$ and in the interior of the parameter space (assumed to be an open set) then necessarily the Hessian $(1/N) \sum_i H(w_i, \hat{\theta}_n)$ is positive semidefinite.  If not, then $\hat{\theta}_n$ is not a local minimum.  This follows from second order optimality conditions — locally $\sum_i q(w_i, \cdot)$ must not decrease in any directions away from $\hat{\theta}_n$.
One source of the confusion might the "working" definition of an M-estimator.  Although in principle an M-estimator should be defined as $\arg\min_\theta \sum_i q(w_i, \theta)$, it might also be defined as a solution to the equation $$0 = \sum_i \dot{q}(w_i, \theta)\,,$$ where $\dot{q}$ is the gradient of $q(w, \theta)$ with respect to $\theta$.  This is sometimes called the $\Psi$-type.  In the latter case a solution of that equation need not be a local minimum.  It can be a saddle point and in this case the Hessian would be indefinite.
Practically speaking, even a positive definite Hessian that is nearly singular or ill-conditioned would suggest that the estimator is poor and you have more to worry about than estimating its variance.
A: There's been a lot of beating around the bush in this thread regarding whether the Hessian has to be positive (semi)definite at a local minimum.  So I will make a clear statement on that.
Presuming the objective function and all constraint functions are twice continuously differentiable, then at any local minimum, the Hessian of the Lagrangian projected into the null space of the Jacobian of active constraints must be positive semidefinite.  I.e., if $Z$ is a basis for the null space of the Jacobian of active constraints, then $Z^T*(\text{Hessian of Lagrangian})*Z$ must be positive semidefinite.  This must be positive definite for a strict local minimum. 
So the Hessian of the objective function in a constrained problem having active constraint(s) need not be positive semidefinite if there are active constraints.
Notes:
1) Active constraints consist of all equality constraints, plus inequality constraints which are satisfied with equality. 
2) See the definition of the Lagrangian at https://www.encyclopediaofmath.org/index.php/Karush-Kuhn-Tucker_conditions . 
3) If all constraints are linear, then the Hessian of the Lagrangian  = Hessian of the objective function because the 2nd derivatives of linear functions are zero. But you still need to do the projection jazz if any of these constraints are active.  Note that lower or upper bound constraints are particular cases of linear inequality constraints.  If the only constraints which are active are bound constraints, the projection of the Hessian into the null space of the Jacobian of active constraints amounts to eliminating the rows and columns of the Hessian corresponding to those components on their bounds.
4) Because Lagrange multipliers of inactive constraints are zero, if there are no active constraints, the Hessian of the Lagrangian = the Hessian of the  objective function, and the Identity matrix is a basis for the null space of the Jacobian of active constraints, which results in the simplification of the criterion being the familiar condition that the Hessian of the objective function be positive semidefinite at a local minimum (positive definite if a strict local minimum).
A: The positive answers above are true but they leave out the crucial identification assumption - if your model is not identified (or if it is only set identified) you might indeed, as Wooldridge correctly indicated, find yourself with a non-PSD empirical Hessian. Just run some non-toy psychometric / econometric model and see for yourself. 
