For a wage as a function of profit: $w(\pi)$ and profit $\pi \in [\pi_{min},\pi_{max}]$, the owner of a company sets the minimum wage to satisfy the following condition:

obs $e = \{e_l, e_h\}$ but in this case effort is observable and contractible, and $f(\pi|e)$ is the pdf of $\pi$, thus:

$$\min_{w(\pi)} \int^{\pi_{max}}_{\pi_{min}}w(\pi)f(\pi|e)d\pi$$ subject to manager's participation condition: $$\int^{\pi_{max}}_{\pi_{min}}v(w(\pi))f(\pi|e)d\pi-g(e) = \bar{u}$$ after doing a lagrange optimization F.O.C $$-f(\pi|e) + \gamma v'(w(\pi))f(\pi|e)=0$$

What exactly I do not understand is how the author took the derivative of the integrand and disappeared with the integral signs... why does it work?

edit: Just in case, you can find this situation in MWG page 480-481, but it happens often in microeconomics.

  • $\begingroup$ derivative and integral cancel each other out $\endgroup$ – WilliamT Oct 19 at 10:34
  • $\begingroup$ I can see that, but the author took the derivative of the inside too... $\endgroup$ – Victor Dahan Oct 19 at 15:31

The minimization problem $$\min_{w(\cdot)} \int^{\pi_{max}}_{\pi_{min}}w(\pi)f(\pi|e)d\pi$$ s.t $$\int^{\pi_{max}}_{\pi_{min}}v(w(\pi))f(\pi|e)d\pi-g(e) = \bar{u},$$ is evidently an infinite dimensional optimization problem. The FOC, a la Lagrange, comes from standard consideration for such problems.

To make this more explicit, define the objective and constrain functionals $\Phi$ and $G$ (on an appropriate function space, say the Banach space $C[\pi_{min}, \pi_{max}]$) by $$ \Phi(w(\cdot)) = \int^{\pi_{max}}_{\pi_{min}}w(\pi)f(\pi|e)d\pi $$ and $$ G(w(\cdot)) = \bar{u} - \int^{\pi_{max}}_{\pi_{min}}v(w(\pi))f(\pi|e)d\pi + g(e). $$ Then the problem is simply $$ \min_{w(\cdot)} \Phi(w(\cdot)) \;\; s.t. \;\; G(w(\cdot)) = 0, $$ which is a standard sort of optimization problem over an infinite dimensional space.

The necessary Lagrangian FOC (as in the finite dimensional setting) would be $$ D_{w} \Phi + \lambda D_{w} G = 0, $$ where $\lambda$ is the Lagrange multiplier. Here, the appropriate notion of the derivative $D_{w}$ is the Frechet derivative.

In this specific case, $\Phi$ is a linear functional. Just as in the finite dimensional setting, the derivative of a linear functional is itself, i.e. $$ D_{w} \Phi = f, $$ and, an integration by parts calculation would tell us $$ D_{w} G = v'(w(\pi))f(\pi|e). $$ So one arrives at the FOC $$ -f(\pi|e) + \gamma v'(w(\pi))f(\pi|e)=0. $$

...this situation ...happens often in microeconomics.

By "this", you mean "solving" such problems by differentiating under the integral sign then setting the integrand equal to identically zero. Yes, this somewhat loose approach---with various accompanying hand-wavy justifications---often suffices in economic contexts.

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For taking a derivative under the integral, I find it helpful to consider the discrete analog, i.e. taking a derivative under summation.

Instead of having $\pi\in[\pi_{min},\pi_{max}]$, suppose $\pi$ takes value from a discrete set $\{\pi_1,\pi_2,\dots,\pi_n\}$. Then the problem becomes \begin{equation} \min_{w(\pi_i)}\sum_{i=1}^nw(\pi_i)f(\pi_i|e)\quad\text{s.t.}\quad \sum_{i=1}^nv(w(\pi_i))f(\pi_i|e)-g(e)\ge \bar u. \end{equation} Now if you expand the summation in a Lagrangian, you'll get \begin{multline} \min_{w(\pi_i)}\quad w(\pi_1)f(\pi_1|e)+\cdots+w(\pi_i)f(\pi_i|e)+\cdots+w(\pi_n)f(\pi_n|e)\\ -\gamma\biggl[v(w(\pi_1))f(\pi_1|e)+\cdots+v(w(\pi_i))f(\pi_i|e)+\cdots+v(w(\pi_n))f(\pi_n|e)\biggr.\\ \biggl.-g(e)-\bar u\biggr] \end{multline} From here, it should be obvious that the derivative with respect to $w(\pi_i)$ equals \begin{equation} f(\pi_i|e)-\gamma v'(w(\pi_i))f(\pi_i|e). \end{equation} The expression in MWG is obtained by setting the above expression to $0$ and multiplying both sides by $-1$.

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  • 2
    $\begingroup$ +1 but shouldn't there be minus in front of lagrangian multiplier? I think that could explain why the final equation has minus in front of it - authors just mulitiply the whole expression by -1. If my memory serves right if constraint is solved as $g(x,y,z...)−c=0$ the multiplier would be negative (i.e. the lagrangian function would be $\mathcal{L} = f(x,y,z...)-\lambda(g(x,y,z...)−c)$ So the solution would be $f(\pi_i|e)-\gamma v'(w(\pi_i))f(\pi_i|e) = 0$ which is the same as $-f(\pi_i|e)+\gamma v'(w(\pi_i))f(\pi_i|e)=0$ $\endgroup$ – 1muflon1 Oct 19 at 16:29
  • $\begingroup$ @1muflon1: Thanks. I edited my answer. $\endgroup$ – Herr K. Oct 19 at 17:41
  • $\begingroup$ thanks, Herr. that is what I was thinking about in terms of the discrete case. I was wondering why can we discretize the problem, but apparently, MWG noted it on a footnote. $\endgroup$ – Victor Dahan Oct 20 at 4:20
  • $\begingroup$ @VictorDahan: I think Michael's answer addresses your question in a rigorous manner. $\endgroup$ – Herr K. Oct 20 at 19:28

The answer by user Herr K. is very sensible and in fact is what MWG p. 481 footnote 6 suggest to do in order to obtain the f.o.c.

But this approach begs the question: Then why on earth did we use the integrals in the first place, only to abandon them for the discrete formulation?

If our problem is formulated in terms of continuous profits, then profits are a continuous random variable, and considering cases "at each level of $\pi$ separately" (as MWG write in their footnote), is not possible because there are uncountably infinite "levels of profit". MWG attempt to rectify this by writing in the same footnote

To be rigorous, we should add that when we have a continuum of possible levels of $\pi$, an optimal compensation scheme need only satisfy the f.o.c at a set of profit levels that is of full measure.

Now, one should tells us how we can obtain a "set of full measure" by including in it a finite number of points from a set that is uncountably infinite (the continuum, that is).

So, once more: why then formulate the problem in continuous terms, only to change the formulation to discrete in order to obtain the f.o.c? Why not formulate the problem in discrete terms from the beginning?

Moreover the description of the situation is

  1. $\pi$ is a random variable
  2. $w$ is a function of $\pi$
  3. We want to choose the optimal $w$

But 2. means that $w$ is a random variable, so the only meaning 3. may have is that what we are going to choose is $w$ as a function of $\pi$, not $w$ as a number. Because if we choose $w$ as a number, we essentially eliminate its dependence on the random variable $\pi$...

...but this is exactly what we can do in order to arrive at the f.o.c. So, treat $w$ as a decision variable independent of $\pi$. We want to

$$\min_w \int_{\pi_{min}}^{\pi_{max}} w f(\pi\mid e)d\pi\,-\,\gamma \int_{\pi_{min}}^{\pi_{max}} v(w) f(\pi\mid e)d\pi\,.$$

Take the derivative with respect to $w$ and set it equal to zero:

$$\int_{\pi_{min}}^{\pi_{max}} f(\pi\mid e)d\pi\,-\,\gamma \int_{\pi_{min}}^{\pi_{max}} v'(w) f(\pi\mid e)d\pi = 0.$$

Because we treat $w$ as a decision variable independent of $\pi$, we can take it out of the integral,

$$\int_{\pi_{min}}^{\pi_{max}} f(\pi\mid e)d\pi\,-\,\gamma v'(w)\int_{\pi_{min}}^{\pi_{max}} f(\pi\mid e)d\pi = 0.$$

Both integrals now equal unity, since $f(\pi\mid e)$ is a proper density over the specific domain, so we end up with

$$1\,-\,\gamma v'(w) = 0 \implies \gamma = \frac{1}{v'(w)},$$

...which is exactly the solution one can find in MWG p. 481. So this f.o.c corresponds also to describing an optimization problem where $w$ is presented initially as a function of $\pi$, and then to solving the problem by treating $w$ as not being a function of $\pi$.

To recapitulate:

  1. We formulated a problem over the continuum, and where the decision variable is a function of a random variable.

  2. In order to arrive at the f.o.c we either
    a) Abandon the continuum formulation and look at a discrete version or
    b) Abandon the assumption that the decision variable is a function of a random variable

This rather twisted situation deserves some contemplation from the side of any interested reader, and I will leave them to it. See also https://economics.stackexchange.com/a/231/61.

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  • $\begingroup$ Thanks, Alecos, this is a quite interesting point of view. Actually, it feels like the problem can't be analytically solved without dropping one of these two assumptions. $\endgroup$ – Victor Dahan Oct 20 at 4:24
  • $\begingroup$ @VictorDahan There is hope. See an old post of mine, here, economics.stackexchange.com/a/231/61 $\endgroup$ – Alecos Papadopoulos Oct 20 at 8:07

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