I have a teacher of economics who likes to use mathematical proofs all the time. Specifically in one exercise he proposed to use the partial derivative of the total expenditure of a economy with respect to the interest rate to show that it is a decreasing function. I explain. The function of the total income / expenditure in a closed economy is defined as follows (is-lm model): $ Y = C(R , Y - T) + I(R) + G$. Where Y,the total expenditure, is equal to the total consumption of the agents (which depends of the interest rate R and the disposable income, that is income minus taxes); the total investment made by firms (which also depends of the interest rate R) and finally the government expending G (which is exogenus to the model). Now here's the part i get confused, how can I derive a function I do not know what looks like? How should I interpret the $C(R, Y-T)$ part mathematically? I know that is a function that depends on several variables, but how could I represent generically the derivative it will yield? I've tried to look at general forms of chain rule and product rule and then apply it to the case but it failed. Since i did not see this proof nowhere in textbooks and over the internet I'm starting to consider that his answer maybe wrong. Here it goes:
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That looks correct. What part is of concern? – Mark Viola Jul 09 '15 at 15:29
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I don't get the transition of the general formula of partial derivatives to what he done. Like, shouldn't the first part be $C_{1}$ TIMES $C(Y-T$, that is, the derivative of the first part times the second (fixing the second part)? – matt_zarro Jul 09 '15 at 15:45
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No. The development is correct. – Mark Viola Jul 09 '15 at 15:52
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If I interpret $C(R, Y-T)$ as derivative of $C(R)$ plus derivative of $C(Y-T)$, would that be correct? – matt_zarro Jul 09 '15 at 17:29
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No, that is not correct. I'll post an answer to help. – Mark Viola Jul 09 '15 at 17:34
1 Answers
Write $C=C(x,y)$ where $x=R$ and $y=Y(R)-T$. Then, we have
$$\begin{align} \frac{\partial C(x,y)}{\partial R}&=\frac{\partial C(x,y)}{\partial x}\frac{\partial x}{\partial R}+\frac{\partial C(x,y)}{\partial y}\frac{\partial y}{\partial R}\\\\ &=\left(\left.\frac{\partial C(x,y)}{\partial x}\right|_{x=R,y=Y(R)-T}\right)\frac{\partial R}{\partial R}+\left(\left.\frac{\partial C(x,y)}{\partial y}\right|_{x=R,y=Y(R)-T}\right)\frac{\partial Y(R)-T}{\partial R}\\\\ &=C_1(R,Y(R)-T)\times 1+C_2(R,Y(R)-T)\times \frac{\partial Y(R)}{\partial R}\\\\ &=C_1(R,Y(R)-T)+C_2(R,Y(R)-T) \frac{\partial Y(R)}{\partial R} \end{align}$$
where
$$C_1(R,Y(R)-T)\equiv\left.\frac{\partial C(x,y)}{\partial x}\right|_{x=R,y=Y(R)-T}$$
$$C_2(R,Y(R)-T)\equiv\left.\frac{\partial C(x,y)}{\partial y}\right|_{x=R,y=Y(R)-T}$$
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Wow.. i was blind and now I see.. but just before giving you the answer, wouldn't this be true? I'm well aware that you can't just cancel as it were a mere fraction, I'm just using the exact definition of the chain rule http://www.tiikoni.com/tis/view/?id=9c4d500 – matt_zarro Jul 09 '15 at 19:17
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sorry for using a link image, but the
\cancelmethod seems to don't work inside partial derivatives $\require{cancel} \cancel{x}$ only with single variables – matt_zarro Jul 09 '15 at 19:18 -
You cannot cancel those notations and assign any meaning thereafter. – Mark Viola Jul 09 '15 at 19:34
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