Consider the simplest version of the Spence signalling model. There are two types of worker, with either productivity $\theta_H$ or instead productivity $\theta_H < \theta_L$. The proportions of the workers are $p_H$ (share of high types) and $1 - p_H$ respectively. Risk-neutral and competitive firms pay wages that may depend on a worker's education level $e \in \{0, 1 \}$. Finally, getting education ($e = 1$) costs $c_H$ for the high types and $c_L > c_H$ for the low types.

My question is simple:

What are the possible wage schedules that sustain a pooling equilibrium? In particular, can a pooling equilibrium wage schedule specify a wage of zero following $e = 1$?

I will briefly elaborate on my understanding of the problem and what is unclear to me. In any pooling equilibrium, workers must be paid their expected productivity. So following $e = 0$, the specified wage must be $$ w(e = 0) = \mathbb{E}(\theta) = p_H \theta_H + (1 - p_H)\theta_L $$ Furthermore, to ensure that nobody wants to deviate (by getting education), we only need to ensure that the high types won't deviate. (If they won't want to deviate, then neither will the low types). This requires that $$ w(e = 1) - c_H \leq w(e = 0) = p_H \theta_H + (1 - p_H)\theta_L $$ So the wage following $e = 0$ is uniquely determined in the pooling equilibrium (the equation). Moreover, the wage following $e = 1$ must be sufficiently low to ensure that nobody gets education (the inequality). I am wondering if there are any additional constraints that the wage schedule in a pooling equilibrium must satisfy, or whether any wage schedule satisfying these constraints will sustain pooling (as a PBE).

To make the issue a bit more salient, suppose that we propose some wage $w(e = 1) < \theta_L$. For example, suppose you make the wage for those who get education zero. Clearly, nobody will choose education, i.e. the inequality holds. Moreover, wages don't need to be set optimally following $e = 1$ since this happens with zero probability in equilibrium. On the other hand, no possible type could have a productivity of zero, so this wage schedule does not follow from any possible firm beliefs. This suggests that the wage following zero education needs to also satisfy $$ w(e = 1) \in [\theta_L, \theta_H] $$ Is this indeed a further constraint (in any pooling equilibrium where both types choose $e = 0$)?

  • $\begingroup$ It seems to me all you need is $w(e=1)\in[\theta_L,\, \mathbb E(\theta)+c_H]$. Not sure if this is what you're trying to get at. $\endgroup$
    – Herr K.
    Commented Feb 24, 2021 at 14:22
  • $\begingroup$ So we can't have, for example, $w(e = 1) = 0 < \theta_L$? This would definitely prevent any deviations from the pooling equilibrium! (But one could never find employer beliefs which lead to this wage...) $\endgroup$
    – afreelunch
    Commented Feb 24, 2021 at 14:35
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    $\begingroup$ What is a PBE here? With competitive firms, this is not really a game. $\endgroup$ Commented Feb 24, 2021 at 17:00
  • $\begingroup$ @MichaelGreinecker I think the idea is that firms play Bertrand competition over wages. However, I do agree that this is not modelled rigorously at least in the original Spence paper $\endgroup$
    – afreelunch
    Commented Feb 25, 2021 at 15:15

1 Answer 1


In my interpretation of the model, competitive firms imply that the wage always equals the expected productivity, which depends on the beliefs. Clearly, in any pooling equilibrium the on-path beliefs are equal to the prior such that the wage is simply $E[\theta]$. If, in your example, a worker sends the off-path messags $e=1$, the wage must be $w(e=1)=w_1=b_1 \theta_H + (1-b_1) \theta_L$, where $b_1$ is the belief in the information set of $(e=1)$ that the type is high. Since $b_1 \in [0,1]$, it must be that $w_1 \in [\theta_L,\theta_H]$. This information set is not reached on equilibrium path such that we are free to determine the out-of-equilibrium belief $b_{1}$. Only $b_{0}=p_H$ is a PBE-consistency requirement.

The set of all no-education pooling PBE is then given by the following:
$w_{1}=b_{1}\theta_H + (1-b_{1})\theta_L$,
$ b_{0}=p_H$,
$b_{1}: b_{1}\theta_H + (1-b_{1})\theta_L -c_H\leq E[\theta].$

Note that the beliefs are part of the equilibrium. To get rid of equilibrium multiplicity, have a look at the numerous refinements such as the intuitive criterion that kills unreasonable out-of-equilibrium beliefs (and thereby unreasonable equilibria). The restriction on belief $b_1$ originates from the incentive constraint of the high type. We want to make sure that $\theta_H$ does not want to deviate get eductation, and because the wages are competitive they are implied by the off-path belief which we can set freely.

The idea to set $w_1<\theta_L$ would violate the competitive nature of firms. This sentence can be interpreted in two ways.

  1. Wages are just exogenously set this way.
  2. There are two firms (having the same beliefs) who both set a wage menu $(w_1,w_2)$ and a la Bertrand the worker always goes to the firm setting the higher wage. For some off-path beliefs, one firm would deviate from $w_1<\theta_L$ and may attract workers by setting a wage $w_1=b_{1}\theta_H + (1-b_{1})\theta_L - \varepsilon$ and thereby make a profit. Because you did not specify the off-path belief, we have no idea whether this deviation on the wage would be profitable.
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    $\begingroup$ Ooops, I forgot about the cost of education $c_H$. $\endgroup$
    – Bayesian
    Commented Feb 24, 2021 at 16:22
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    $\begingroup$ "Your idea in the comments": this was in the main text and in fact the central question. To quote from the main text: "In particular, can a pooling equilibrium wage schedule specify a wage of zero following $e=0$?" $\endgroup$
    – afreelunch
    Commented Feb 25, 2021 at 10:28
  • $\begingroup$ I thought in the comments you referred to $w_1=w(e=1)=0$ and the main text has $w_0=w(e=0)=0$. However, there also cannot be $w_0=0$ with $\theta_L\geq0$ as a firm can always make a profit by setting a wage $w_0=E[\theta]-\varepsilon$. $\endgroup$
    – Bayesian
    Commented Feb 25, 2021 at 11:18
  • $\begingroup$ Many apologies, there was a typo in the question! Now fixed. $\endgroup$
    – afreelunch
    Commented Feb 25, 2021 at 11:41
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    $\begingroup$ On a separate note, I don't see why $w(e = 1) \leq \mathbb{E}[\theta]$. On seeing somebody get education (an impossible event, given by prior beliefs) why can't I conclude that the person getting education was a high type? $\endgroup$
    – afreelunch
    Commented Feb 25, 2021 at 11:45

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