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

Question

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".

Answer 1

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.

Answer 2

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 $C_t(i)$