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I have set up and solved an optimization problem with time $t$ endogenous state variables, $\alpha_t$ and $\beta_t$ and choice variable $s_t$. After some manipulation, the first-order condition for $s_t$ is of the form:

$f(\alpha_t,\beta_t,s_t,s_{t+1})=0$

where $f(\cdot)$ is a non-linear and contains expectations over future realizations of shocks. In some specifications there is no explicit solution for $s_t$.

I want to derive testable implications from the underlying theory. In particular, I am interested in the signs of:

$\dfrac{\partial s_t}{\partial \alpha_t}$ and $\dfrac{\partial s_t}{\partial \beta_t}$.

How should I treat $s_{t+1}$ when I take the total derivative? Should I treat it as a constant, or include it? What's the rationale? If more elaboration is needed, do let me know and I'll be happy to explain in more detail the problem.

Thanks!

John
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2 Answers2

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Try writing your equation explicitly with a conditional expectation operator:

$$ \mathbb{E}_t f(\alpha_t, \beta_t, s_t, s_{t+1}) = 0$$

Hopefully this clarifies things. You can take the total derivative as you normally would with any function, however $s_{t+1}$ is a random variable since it is not predetermined. Ultimately you are considering all cases of $s_{t+1}$ weighted by their probability.

Since $\mathbb{E}_t$ integrates over shocks, you can differentiate under the integral sign by Leibniz's integration rule.

Bryce
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  • Thanks, @Bryce.

    So, in order to obtain the partial $\dfrac{\partial s_t}{\partial \alpha_t}$, should I be totally differentiating the equation holding $\beta_t$ constant, but not $s_{t+1}$?

    If so, this will yield and expression for $\dfrac{\partial s_t}{\partial \alpha_t}$ that depends on $\dfrac{\partial s_{t+1}}{\partial \alpha_t}$. How then do I determine the sign of the latter partial derivative in order to determine the sign of the former (which is what I am interested in)?

     – John Jan 04 '15 at 17:40
  • For that, I recommend using the implicit function theorem. As for determining signs, we need a lot more information about the primitives of the problem. – Bryce Jan 04 '15 at 17:44
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The moment an unknown and unknowable in advance future quantity enters an optimal solution, we have no other option than to insert in its place some estimation of it. The more widely used such estimation (but not the only one) is the conditional expectation. Conditional on $t$, everything except $s_{t+1}$ will be treated as a constant. Since we do not know how exactly $s_{t+1}$ enters the $f$-function, we write in abstract notation as a function:

$$E\big[f(\alpha_t,\beta_t,s_t,h(s_{t+1},...)\mid t\big]=0 \Rightarrow f(\alpha_t,\beta_t,s_t,E[h(s_{t+1},...)\mid t]) =0$$

Remember also that the conditional expectation is a function, and not a constant (as is the unconditional expected value).

Since this is a solution, to study comparative statics one has to use the implicit function theorem which states that, starting at the solution,

$$\frac{{\rm d}s_t}{{\rm d}\alpha_t} = -\frac {\partial f(\alpha_t,\beta_t,s_t,E[h(s_{t+1},...)\mid t]) / \partial \alpha_t}{\partial f(\alpha_t,\beta_t,s_t,E[h(s_{t+1},...)\mid t]) / \partial s_t}$$

The partial derivative symbol in the right-hand side conveys the message that in the numerator, $s_t$ is not to be differentiated with respect to $\alpha_t$, and in the denominator, $\alpha_t$ is not to be differentiated with respect to $s_t$ (and the same holds with respect to $\beta_t$).
But $E[h(s_{t+1},...)\mid t]$ is fair game in both cases. What will the operations $\partial E[h(s_{t+1},...)\mid t] /\partial \alpha_t$ and $\partial E[h(s_{t+1},...)\mid t] /\partial s_t$ yield, will depend of course on what is the actual expression for the conditional expectation.

Alecos Papadopoulos
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    The first thing you are saying is not necessarily true. There is no way to tell if that holds without knowing more about OP's model. The expectation of non-linear function is not the same as taking an expectation, and then applying the non-linear function. – Bryce Jan 04 '15 at 20:41
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    @Bryce Oops, you're right, regarding how $s_{t+1}$ appears in $f$. Fixed it. – Alecos Papadopoulos Jan 04 '15 at 21:10
  • Thanks @AlecosPapadopoulos for all of the help. A follow-up question. It's still not entirely clear to me whether I should be taking the partial of $s_{t+1}$ with respect to $\alpha_t$. Since the evolution of $\alpha$ from $t$ to $t+1$ depends on the choice of $s_t$, and $s_{t+1}$ depends on $\alpha_{t+1}$, should the partials of $f$ with respect to $\alpha_t$ for example not account for this dependence? – John Jan 05 '15 at 18:31
  • Ie. should I write

    $\dfrac{\partial f(\alpha_t,\beta_t,s_t,E[h(s_{t+1},...)])}{\partial \alpha_t} = f_1 + f_4 \dfrac{\partial E[h(s_{t+1},...)]}{\partial s_{t+1}} \dfrac{\partial s_{t+1}}{\partial \alpha_t}$?

     – John Jan 05 '15 at 18:33
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    In essence yes, but note that the symbol $s_{t+1}$ will no longer be present in the conditional expectation. – Alecos Papadopoulos Jan 05 '15 at 18:54