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Let there be a unit measure of inputs. Define by $C_i = \left[\int_0^i c(i)^{1-\alpha}\right]^\frac{1}{1-\alpha}$ a cost index of using up to $i$ as input.

I discretize $c(i)$ by $c_0, \dots c_i, \dots c_I$. Let for example $c_i = 1 \,\forall i$, and $\alpha = 0.5$. Let $I = 5$. To approximate the integral, I simply sum up terms and divide by $I$.

Then we have $C_1 = (\frac{2}{5}\sqrt 1)^2$, $C_2 = (\frac{3}{5}\sqrt 1)^2$ etc. So this is an increasing sequence, and $C_1 = 4/25=0.16$.

Now I need the corner case, $C_0$. $C_0$ only uses $c(0)$ as input, so I'd think that

$$ C_0 = (c(0)^{1-\alpha})^\frac{1}{1-\alpha} = 1$$

But that clearly can't be the limit of the discretized sequence, since $\lim\limits_{i\to 0} C_i \to 0$.

So, as the variety of inputs decreases, the cost index decreases - until you only have one variety, and then it explodes? Could someone please shed some light on this?

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  • $\begingroup$ @denesp fixed, thx $\endgroup$ – FooBar Jan 21 '17 at 21:19
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I think the appropriate discretisation should be something like $$C_i=\left[\int_0^i\!c(i)^{1-\alpha}\,di\right]^{\frac{1}{1-\alpha}}=\lim_{I\rightarrow\infty}\left[\sum_{j=0}^{(I-1)i}\frac{1}{I}c\left(\frac{j}{I}\right)^{1-\alpha}\right]^{\frac{1}{1-\alpha}}$$

The below figure illustrates where each term of the sum comes from (for the case of $I=5$)

enter image description here

(The reason the sum goes up to $(I-1)i$ is that we start counting from $j=0$ and want to have $I$ bins in total.)

In your example with $c(i)=1$ and $\alpha=1/2$, this would yield $$\left[\sum_{j=0}^{(I-1) i}\frac{1}{I}\sqrt{1}\right]^2=\left[I i\frac{1}{I}\sqrt{1}\right]^2=i^2.$$ We can compare this to the true value of $C$: $(\int_0^i\!\sqrt{1}di)^2=i^2$. Thus, this particular example is special because the discrete approximation gives an exact expression for the integral even without taking the limit. The reason for this is that $c(i)=1$ is a constant function so a rectangular approximation of the areas beneath it is perfect for any number/size of Rieman rectangles.

But suppose we take $c(i)=i^2$. We then have $(\int_0^i\!\sqrt{i^2}di)^2=i^4/4$ and $$\left[\sum_{j=0}^{(I-1) i}\frac{1}{I}\sqrt{\frac{j^2}{I^2}}\right]^2=\frac{(I-1)^2 i^2 (1-i+I i)^2}{4 I^4}.$$ In the limit this approximation converges on the true value: $$\lim_{I\rightarrow\infty}\frac{(I-1)^2 i^2 (1-i+I i)^2}{4 I^4}=\frac{i^4}{4},$$ but for finite $I$ the approximation is imperfect.

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  • $\begingroup$ So I suppose in my discretization instead of computing the value at the left end of the bins, I should compute it at the right end of the bins to prevent a firm of having literally zero costs... $\endgroup$ – FooBar Jan 21 '17 at 22:37
  • $\begingroup$ Right. So the reason I invoked the $I-1$ contrivance was so that the sum could be evaluated at $i=0$ to give an approximation for $C_i$. But one could indeed evaluate at the right end on the bins (essentially by replacing $I-1$ with $I$ and $j=0$ with $j=1$) to get an equivalent formulation. The difference is that the first term in this sum would be evaluated at $c(1/I)$ rather than $c(0)$. As $I\rightarrow\infty$, for course, the two formulations are equivalent. $\endgroup$ – Ubiquitous Jan 22 '17 at 10:23

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