1
$\begingroup$

I am reading papers about moral hazard. What is one dimensional, ordered type $\theta\in\Theta$?

What is one dimensional, not ordered type $\theta\in\Theta$?

Could you please give an example?


References:

Riley (1979)

Azevedo and Gottlieb (2016)

$\endgroup$
5
$\begingroup$

Judging from the reference you provide, this refers to whether the set $\Theta$ is ordered or not. For example, natural numbers or the alphabet are ordered sets. In the context of moral hazard problems, examples could be effort, or ability. Since this is a numerical variable, it is an ordered set.

In the paper you mention, the phrase "one dimensional, ordered type..." is not found. However, the paper states that:

... differences among sellers are assumed to be parameterized by a single unobservable characteristic $\theta \in \Theta$. Then the family of utility functions can be written in the alternative form,

$U_i=U(\theta_i;y,p) \text{, } i \in I$

If seller $i$ has $\theta_i$ units of the characteristic we may therefore refer to him as being of "type $\theta_i$.

The first bold statement refers to the one dimensionality of the unobserved type, whereas the second refers to it being a numerical variable, which means $\Theta$ is an ordered set.

Conversely, a non-ordered set contains elements which have no intrinsic order. For instance, you can think of a case where agents do not know the city where they are to be located. The set "cities" is unordered because cities do not necessarily hold a particular order in a one-dimension set. You can add a second dimension (e.g. population), where you could order them based on city size.

I would say that having ordered sets facilitates algebraic analysis, since it enables the introduction of monotonic strategies, whereby an important outcome variable is a monotonic function of the agent's type. In effect, this seems to be important in the paper, when the author states that:

A seller of type $\theta_i$ brings to market a product that is valued equally by each of the set of potential buyers, $J$. This valuation, $V_i$, measured in dollars, is an increasing function of $\theta$ and (possibly) also of the intensity of sales-related activity, $y$:

$V_i = V(\theta_i;y)$

The bold emphasis highlights precisely the monotonicity enabled by the order property of $\Theta$. Another example is given in Athey (2001):

$\hskip1in$enter image description here


In response to the second reference you give (here), in the phrase

We now show that Riley equilibria may not exist when types are not ordered.

the "ordered" adjective refers to the second interpretation given above (i.e. of monotonicity). First, they assume that

Consumers are uniformly distributed in the unit interval, $\theta in [0,1]$

which is an ordered set. However, because both the utility function and the cost function are quadratic, it results that

... the types in the extreme points have the strongest preference for one of the contracts and are the most costly to serve.

which means that the utility and cost functions are not monotonic over $\theta$.

$\endgroup$
  • $\begingroup$ So loosely speaking, let $U(\theta_i)$ denotes the utility of ordered type, then $dU(\theta)/d\theta>0$? If it not ordered type, $dU(\theta)/d\theta>0$ might not hold? $\endgroup$ – High GPA Jul 19 '17 at 15:10
  • 1
    $\begingroup$ @HighGPA Yes and no. A unidimensional, numerical variable is, I think, always ordered. My point is that an ordered set facilitates the algebraic treatment of monotonicity, but does not require it (as in your second example). Think in the variable "effort". We can imagine situations whereby exerting so much effort ends up being less productive than by having a bit of time off (to rest or etc). Here, the variable effort its still ordered, yet its outcome (productivity) might not be monotonic. Meanwhile, if the set is nor ordered, the derivative makes less (no?) sense (e.g. cities). $\endgroup$ – luchonacho Jul 19 '17 at 15:24

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.