# Regarding a consumption aggregator: How do I differentiate under the integral sign?

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".

• This answer has some related math literature, economics.stackexchange.com/a/231/61 Nov 5 '16 at 19:14
• @AlecosPapadopoulos Thanks for the link! I've been thinking about that problem also. Nov 5 '16 at 19:34

## 2 Answers

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.

• 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? Nov 5 '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. Nov 7 '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? Nov 8 '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. Nov 8 '16 at 20:58
• 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. Nov 11 '16 at 16:48

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 $C_t(i)$