This question pertains to the paper "Optimal Tax Administration" by Slemrod and Keen (2017). The IMF working paper is freely available on SSRN, however, it is not necessary to know the paper in order to understand the question.

I am confused on how to arrive at equation (5) in the working paper version.

Suppose there is a welfare function:

$W(t,\alpha)= wl - tz(t,\alpha) -c(e,\alpha) - \psi(l) + v(tz(t,\alpha) -a(\alpha)) $

Here, $v$ is an increasing and concave function, $\psi$ is an incresing and convex function, $t$ is the tax rate, $\alpha$ is tax enforcement, $c$ is evasion/compliance cost and $z$ is declared income, which depends on $t$ and $\alpha$.

Declared income is given by $z=wl(t,w) - e(t,\alpha)$, where $w$ is the wage rate, $l$ is hours worked and $e$ is concealed income. Both $l$ and $e$ are derived optimally.

The first order conditions for $l$ and $e$ are given by:

$(1-t)w - \psi'(l) = 0$

$t - c_e(e,\alpha) = 0$

Here $c_e$ denotes the derivative of $c$ with respect to $e$ and $\psi'$ is the derivative of $\psi$.

I am interested in $\frac{dW}{dt}$.

The author's invoke the envelope property to arrive at:

$\frac{dW}{dt}= -z + v'*(z + tz_t)$

Here $z_t$ is the derivative of $z$ with resprect to t and $v'$ is the derivative of $v$.

How can I arrive at this expression?

  • $\begingroup$ Why the first order condiiton w.r.t. to $e$ is not $t-c_e -v'\cdot t=0$? $e$ is inside $z$ which is inside $v$ also. $\endgroup$ Nov 8 '17 at 14:56
  • $\begingroup$ The same quastion holds for $l$. Why we do not differentiate $v()$ with respect to $l$? $\endgroup$ Nov 8 '17 at 15:12
  • $\begingroup$ Government spending $g$ is exogenous to the consumer. The consumer des not take into account that if she works more and avoids taxes less then she will receive more public goods. This may not be very realistic, but it is an assumption often made in public finance, I believe. $\endgroup$
    – BB King
    Nov 8 '17 at 15:20
  • $\begingroup$ @Certainly. So I understand that it is in the assumptions of the model that agents treat the $v$ function as exogenous. $\endgroup$ Nov 8 '17 at 15:42
  • $\begingroup$ Yes, exactly. $v$ is exogenous to the consumer (but naturally not to the social planer). Sorry for any confusion. $\endgroup$
    – BB King
    Nov 8 '17 at 15:54

The answer to this "puzzle" has been clarified in the comments to the original OP's post.

The issue is that agents treat the $v$ function as exogenous when optimizing with respect to $l$ and $e$, while the social planner naturally takes it into account.

But this is a behavioral assumption with mathematical consequences. The Envelope Theorem does not allow for such asymmetries.

Let's apply the Envelope Theorem, which says that if we have a function $f(x;a)$ and we optimize it over, say, $x$, then the total derivative of the optimized function $f(x^*,a)$ with respect to what previously was treated as a parameter equals the partial derivative of the non-optimized function with respect to that parameter, keeping the $x$s fixed:

$$\frac {df(x^*,a)}{da} = \frac {f(x,a)}{\partial a}$$

Now, note that $z$ depends on $t$ only through $l$ and $e$. So, for the purpose of applying the theorem, we have that the partial derivative $\partial z/\partial t =0$.

Keeping that in mind let's write

$$W = H + v$$

where $H$ contains all the other terms. As the OP showed in his answer we have

$$\frac{dW}{dt}= l_t[w(1-t)-\psi'] + e'[t-c_e] - z + \frac{dv}{dt}$$

and taking into account the f.o.c for $l$ and $e$ given the behavioral assumptions related to $v$, we are left with

$$\frac{dW}{dt}= \frac{dH}{dt} + \frac{dv}{dt} = -z + \frac{dv}{dt}$$

$$\implies \frac{dH}{dt} = -z$$

Does this conform with the Envelope Theorem? It does, with respect to $H$ only, because this is the part of $W$ over which we maximize with respect to $l$ and $e$. And we have

$$\frac{\partial H}{\partial t} = \frac{\partial (-tz)}{\partial t} = -z$$

since as we have said, $\partial z/\partial t =0$.

So we should not see the partial of $z$ in the first position in any case, while the appearance of $dz/dt$ in relation to $v$ appears only because we have ignored it in the maximization with respect to $l$ and $e$.

If the agents optimized taking into account $v$ also we would have obtained

$$\frac{dW}{dt} = -z + v'\cdot z$$

and the Envelope Theorem would hold for the full $W$ function.

  • $\begingroup$ This was precisely my initial question. I will post the answer in a couple of days if no one else does. I can assure you, however, that the authors are not wrong. $\endgroup$
    – BB King
    Nov 4 '17 at 2:11
  • $\begingroup$ I have now posted the answer. Do you agree with it or have any comments? $\endgroup$
    – BB King
    Nov 8 '17 at 11:20
  • $\begingroup$ I will examine it in detail and respond later. Please verify the following: agents optimize by choosing $l,e$ treating as exogenous $t, \alpha, w$. The social planner maximizes the welfare function by choosing $t, \alpha$, given the optimizing behavior of the agents, and given the market-determined $w$. Also clarify please: we are interested at the derivative of the optimized welfare function with respect to the tax rate? $\endgroup$ Nov 8 '17 at 12:34
  • $\begingroup$ We are interested in optimizing the welfare function w.r.t. $t$ taking into account that agents optimize to choose $l$ and $e$. $\endgroup$
    – BB King
    Nov 8 '17 at 13:34

At first glance, it appears the answer should be $\frac{dW}{dt}= -(z + tz_t) + v'*(z + tz_t)$, as $l$ and $e$ are chosen optimally and the envelope theorem cancels these terms out. Furthermore, one could expect both $z_t$ terms in the final derivative to drop out due to the envelope theorem. However, one and only one $z_t$ term remains in the authors' calculations.

The envelope theorem is just an application of first order conditions. In this case, it is better to simply use the first order conditions we know to derive the results, instead of using the theorem directly. I believe the authors invoke the term "envelope property" to concisely convey that they are using the previously found first order conditions.

Taking the total derivative we arrive at:

$\frac{dW}{dt}= wl_t -z - tz_t -c_ee' - \psi'l_t + v'*(z + tz_t)$

Inserting the expression for $z_t$ and grouping some terms we have:

$\frac{dW}{dt}= l_t[w(1-t)-\psi'] + e'[t-c_e] - z + v'*(z + tz_t)$

Note that the terms in square brackets are the same as the first order conditions for $l$ and $e$ and hence equal to zero. This leaves us with:

$\frac{dW}{dt}= - z + v'*(z + tz_t)$

  • $\begingroup$ Glad you worked out your question. It'd be nice if you explained why "naively" applying the envelope theorem didn't work here, though. $\endgroup$ Nov 6 '17 at 22:03
  • $\begingroup$ I thought I already did that in the first paragraph, but I can expand upon it. $\endgroup$
    – BB King
    Nov 7 '17 at 10:27
  • $\begingroup$ Thanks for the edit! I’ll have to admit that I still don’t quite get it. The partial derivatives are set equal to $0$ even in a multi-variable optimisation problem (assuming an interior solution). I guess I’ll just have to take a closer look at the paper to find out... $\endgroup$ Nov 7 '17 at 14:55

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