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Let $\varepsilon>1$ and let $$C_t\equiv\left(\int_0^1C_t(i)^{(\varepsilon-1)/\varepsilon} \, di\right)^{\varepsilon/(\varepsilon-1)}$$ denote a consumption basket in time period $t$, where $C_t(i)$ is consumption of good $i\in [0,1]$. In e.g. new Keynesian models we want to differentiate $C_t$ with respect to $C_t(i)$ for some $i\in [0,1]$ so as to solve a utility optimization problem. In my lecture notes, and in many texts on this subject, it is said that $$\frac{\partial C_t}{\partial C_t(i)} = \frac{\varepsilon}{\varepsilon-1} C_t^{1/\varepsilon} \frac{\varepsilon-1}{\varepsilon} C_t(i)^{-1/\varepsilon}.$$ Does anyone know how this differentiation is accomplished? This is my question I want answered. Below I will outline how I have thought about this question.

I am prone to thinking that it is wrong. For using the chain rule I would say that the answer is the following. $$\frac{\partial C_t}{\partial C_t(i)} = \frac{\varepsilon}{\varepsilon-1} C_t^{1/\varepsilon} \left(\frac{\partial}{\partial C_t(i)} \int_0^1C_t(i)^{(\varepsilon-1)/\varepsilon} \, di\right),$$ which, when asssuming that the function is such that we may differentiate under the integral sign, I get $$\frac{\partial C_t}{\partial C_t(i)} = \frac\varepsilon {\varepsilon-1} C_t^{1/\varepsilon} \left(\int_0^1 \frac{\varepsilon-1} \varepsilon C_t(i)^{-1/\varepsilon} \, di\right).$$ Now, using the mean value theorem for integrals it would be possible to say that $$\int_0^1 C_t(i)^{-1/\varepsilon} \, di = C_t(j)^{-1/\varepsilon}(1-0)$$ for some $j\in (0,1)$, and insert this result above and then get a similar result to what was shown in my lecture notes. However, this would lead us to considering another good $j$ not necessarily equal to good $i$.

The reader may think that I am confusing the symbol '$i$' in the integral, for the same symbol used when differentiating with respect to $C_t(i)$, and that I should, when differentiating, consider a good $i_0$, and then perform the following differentiation: $$\frac{\partial C_t}{\partial C_t(i_0)} = \frac{\partial }{\partial C_t(i_0)} \left(\int_0^1 C_t(i)^{(\varepsilon-1)/\varepsilon} \, di\right)^{\varepsilon/(\varepsilon-1)}.$$ This may be so, but I do not know how to get the desired result from this, and if I take this approach, I would say that the derivative is equal to $0$ (!) as the integral is just a real constant if $t$ is fixed, which it is.

It is sometimes said that we may differentiate the integral just mentioned by looking at the integral as beeing a sum. What they mean by this, I do not know. Maybe they represent the integral as the limit of a Riemann sum, which it is, and write $$\frac{\partial }{\partial C_t(i_0)} \int_0^1 C_t(i)^{(\varepsilon-1/\varepsilon} \, di = \frac{\partial }{\partial C_t(i_0)} \lim_{n\to\infty} \sum_{k=1}^n C_t(\xi_k)^{(\varepsilon-1)/\varepsilon}(i_k-i_{k-1}),$$ with $i_0=0<i_1<\cdots < i_{n-1}<i_n=1$ and $i_{k-1}\leq\xi_k\leq i_k$ for each $k=1,2\ldots,n$. When the authors write that we should look at the integral as beeing a sum, this must be it. But differentiating this sum with respect to $C_t(i_0)$ would in the best cases (i.e., when we can do differentiation inside the limit) be equal to $\lim_{n\to\infty}\frac{\varepsilon-1} \varepsilon C_t(i_0)^{1/(\varepsilon-1)} \cdot (i_\alpha - i_{\alpha-1})$ for some $\alpha\in\{1,2,\ldots,n\}$ such that $i_{\alpha-1}\leq i_0\leq i_\alpha$; the problem now is that $\lim_{n\to\infty}\frac{\varepsilon-1} \varepsilon C_t(i_0)^{-1/\varepsilon}\cdot (i_\alpha - i_{\alpha-1})=0$, which is consistent with modern advanced real analysis (to my knowledge) in the sense that if we just increase or decrease the value of $C_t(i)$ at one $i=i_0$, then the value of the integral will not change, and hence the derivative should be $0$ (i.e., no change in the value of the integral for a change in $C_t(i_0)$).

Note: These problems occur when studying e.g. the so called "Dixit-Stiglitz aggregator".

Elias
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2 Answers2

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Using your formalism above, you can think (heuristically) of the integral as $$ \sum_{i=1}^nC_t(i)^{(\epsilon-1)/\epsilon} $$

If we differentiate this with respect to $C_t(j)$, we get

$$ \frac{\epsilon - 1}{\epsilon} C_t(j)^{-1/\epsilon} $$

Which is exactly what we needed. To do this rigorously, you need a notion of taking derivatives on function spaces. Look up the Gâteaux and Fréchet derivatives.

Theoretical Economist
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  • Thanks for your answer! I am looking for a rigorous answer, how would you use the concept of a Gâteaux derivative to show the above result? As I said below, I think the right way to understand this is through an understaning of functional derivatives (e.g., Gâteaux derivatives), what do you think about my answer below? Is this how you think about it? – Elias Nov 05 '16 at 18:02
  • Yes, I am basically using the same notion of the functional derivative you give there. Here's something that I think should work; let me know if it doesn't.

    Set $\eta_t (i) = 1$ for $i = i_0$ and $0$ otherwise. The integral sign disappears, and we can differentiate $f(x)=x^{(\epsilon-1)/\epsilon}$ as we would normally.

     – Theoretical Economist Nov 07 '16 at 13:23
  • Hmmm, I've tried to understand your answer, but I don't get it yet. From my perspective, your approach leads just to a vanishing integral in the sense that it equals 0. Can you explain your answer in more detail? – Elias Nov 08 '16 at 20:50
  • As I see it, if $F(C_t):=\int_0^1C_t(i)^{(\epsilon-1)/\epsilon}$, then $\lim_{\epsilon\to 0^+}\frac{F(C_t+\epsilon\eta_t)-F(C_t)}{\epsilon}=\lim_{\epsilon\to 0^+}\int_0^1 \frac{(C_t(i)+\epsilon\eta_t(i))^{(\epsilon-1)/\epsilon}-C_t(i)^{(\epsilon-1)/\epsilon}}{\epsilon}, di$, then this becomes $\int_0^1\frac{\epsilon-1}{\epsilon}C_t(i)^{-1/\epsilon}\eta_t(i), di$. Hence, $\frac{\delta F}{\delta C_t(i)}=\frac{\epsilon-1}{\epsilon}C_t(i)^{-1/\epsilon}$. This argument is based on the strong assumption of taking the limit inside the integral, and my weak understanding of Gâteaux derivatives. – Elias Nov 08 '16 at 20:58
  • Remember, we don't need to find the derivative in an arbitrary direction, but only the derivative in the direction of, say $C_t(j)$. Thus, set $\eta_t(i) = 1$ if $i=j$ and $0$ for $i \neq j$. Hence: $\int_0^1\frac{(C_t(i) + \epsilon \eta_t (i))^{(\epsilon-1)/\epsilon}-C_t(i)^{(\epsilon-1)/\epsilon}}{\epsilon}di$ simplifies to $\frac{(C_t(j) + \epsilon)^{(\epsilon-1)/\epsilon}-C_t(j)^{(\epsilon-1)/\epsilon}}{\epsilon}$. You can then take limits as usual, without worrying about differentiating under the integral sign, etc. – Theoretical Economist Nov 10 '16 at 17:48
  • Sorry for confusing $\epsilon$ with $\varepsilon$, I should have written "if $F(C_t):=\int_0^1C_t(i)^{(\varepsilon-1)/\varepsilon}$, $\lim_{\epsilon\to 0^+}\frac{F(C_t+\epsilon\eta_t)-F(C_t)}{\epsilon}=\lim_{\epsilon\to 0^+}\int_0^1 \frac{(C_t(i)+\epsilon\eta_t(i))^{(\varepsilon-1)/\varepsilon}-C_t(i)^{(\varepsilon-1)/\varepsilon}}{\epsilon}, di$ which becomes $\int_0^1\frac{\varepsilon-1}{\varepsilon}C_t(i)^{-1/\varepsilon}\eta_t(i), di$. Hence, $\frac{\delta F}{\delta C_t(i)}=\frac{\varepsilon-1}{\varepsilon}C_t(i)^{-1/\varepsilon}$.". – Elias Nov 10 '16 at 21:01
  • I'm sorry, but I can't see how the integral sign just vanishes. If you let $\eta_t(i)=0$ for $i\neq j$, $i\in [0,1]$, for fixed $j\in (0,1)$ and $\eta_t(i)=0$ for $i=j$, then the numerator is zero for all $i$ but $i=j$. Thus, the integral is just the integral $\int_j^j \frac{(C_t(i)+\epsilon\eta_t(i))^{(\varepsilon-1)/\varepsilon}-C_t(i)^{(\varepsilon-1)/\varepsilon}}{\epsilon}=0$. What am I missing? I agree that we may just find the derivative in one direction, if we assume that the functional is Gâteaux differentiable to begin with. – Elias Nov 10 '16 at 21:05
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    Ah, yes, you're right. My mistake. I suspect a slightly sneakier choice for $eta_t$ should work. Alternatively, assume that $C_t(i)$ is uniformly bounded (for all $i$ -- I suspect you need this in equilibrium anyway, if not for integrability). You can then appeal to the dominated convergence theorem. – Theoretical Economist Nov 11 '16 at 16:48
  • Okay, I have not tried so much to find a sneaky choice for $\eta_t$. How does the dominated convergence theorem lead me to the conclusion? In this case, as I see it, the collection of integrands is ${\frac{C_t(i)+\epsilon\eta_t(i))^{(\varepsilon-1)/\varepsilon}-C_t(i)^{(\varepsilon-1)/\varepsilon}}{\epsilon}}{\epsilon\in\mathbb{R}+}$, which is not even a sequence as it is uncountable, and Arzela's dominated convergence theorem for Riemann integrals, is given for a countable sequence. This worries me. I can not see how you thought that the dominated convergence theorem solved this problem. – Elias Nov 19 '16 at 13:52
  • To answer myself, maybe one may just construct a sequence ${f_n(i)}={\frac{C_t(i)+\epsilon_n\eta_t(i))^{(\varepsilon-1)/\varepsilon}-C_t(i)^{(\varepsilon-1)/\varepsilon}}{\epsilon_n}}$ where $\epsilon_n\geq 0$ is some sequence such that $\epsilon_n\to 0$, and then apply the dominated convergence theorem. – Elias Nov 19 '16 at 13:58
  • Yes, exactly. Take a sequence of $\epsilon$s converging to zero. Note that the Dominated Convergence Theorem can be used to prove, for instance, Leibniz's rule for differentiating under the integral sign, and the fact that you are taking limits over sequences there (instead of over an uncountable net) does not matter. – Theoretical Economist Nov 19 '16 at 16:56
  • Okay, thanks! This answers my question. Yeah, Leibniz's rule is a nice corollary! – Elias Nov 19 '16 at 18:10
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I think I can answer my own question, so I will answer it here so as to mark it as an answer.

I think the problem boils down to differentiating a functional $$F(C_t)=\int_0^1C_t(i) \, di$$ where I purposefully have ignored the exponent $\frac{\varepsilon-1}{\varepsilon}$ stated in my question. To give meaning to the notion of a partial derivate, see Functionals and the Functional Derivative. Basically, when studying the differential of a functional w.r.t to its argument, we study the directional derivative $$\lim_{\varepsilon\to 0⁺}\frac{F(C_t+\varepsilon\eta_t)-F(C_t)}{\varepsilon},$$ where $\eta$ is some continuous test function, and define the first partial derivative of the functional w.r.t. $C_t(i)$ for some given good $i\in [0,1]$, which I write as $\frac{\delta F(C_t)}{\delta C_t(i)}$, in such a way that $$\lim_{\varepsilon\to 0⁺}\frac{F(C_t+\varepsilon\eta_t)-F(C_t)}{\varepsilon}=:\int_0^1 \, di \frac{\delta F(C_t)}{\delta C_t(i)}\eta_t(i).$$

In our case, the directional derivative is $$\int_0^1 \eta_t(i) \, di$$ and so $\frac{\delta F(C_t)}{\delta C_t(i)}=1$. For the functional $G(C_t)=\int_0^1C_t(i)p_t(i) \, di$, where $p_t(i)$ is the price of good $i$ in time period $t$, we have the directional derivative $$\int_0^1\eta_t(i)p_t(i) \, di$$ and thus $\frac{\delta G(C_t)}{\delta C_t(i)}=p_t(i)$.

To generalize, I conjecture that, under some not so strong conditions, we have that for a functional $\int_0^1 H(C_t(i),i) \, di$, where $H$ is some continuously differentiable function, we get that $\frac{\partial H}{\partial C_t(i)}$ will be the partial derivative of the functional with respect to $C_t(i)$

Michael Hardy
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Elias
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  • You are correct! See the lecture notes on calculus of variations here https://pmbbrito.github.io/cursos/phd/ame/ame2122/ame_2122.html – RAGMS Jun 13 '23 at 15:13