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In a problem I was working on, I found it convenient to use the notation $ dX/dA_{i \rightarrow j} $ to represent the marginal change in $X$ from redistributing a marginal amount of $A_i$ to $A_j$. Is there a name for this, and can it even be called a "derivative"? Is there a better or more conventional way to write this?

Context: $\{ A_i \}$ is a finite sequence of loan payments and $X$ could be something like the associated internal rate of return. There are many valid payment sequences that satisfy a set of constraints, each having a different $X$.

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  • $\begingroup$ So basically you got yourself a function $X$ which depends on the variable $A$ if I understand correctly. That is you have $X(A)$. I think $dX/dA$ will suffice as a notation. $\endgroup$ – MathematicianByMistake Oct 12 '17 at 9:45
  • $\begingroup$ $A$ is a sequence, and there are many ways to move payments around, e.g., $A_1 \rightarrow A_4$, $A_3 \rightarrow A_2$, etc. So I need to convey $i$ and $j$ somehow $\endgroup$ – sirallen Oct 12 '17 at 9:49
  • $\begingroup$ I don't know enough about what's common in Economics/what you mean when you use the word "marginal" here, but it almost seems to me like this might be similar to a difference of "partial derivatives": If $X$ depends on all the money in the $A$s, then maybe you're looking for a quantity like $\dfrac{\partial X}{\partial A_{j}}-\dfrac{\partial X}{\partial A_{i}}$, but I'm really not certain. $\endgroup$ – Mark S. Oct 12 '17 at 10:02
  • $\begingroup$ "marginal" just means "differential". Hm, I don't think it makes sense to write $\partial X/\partial A_i$ on its own. Any increase in $A_i$ has to be offset by a decrease in $A_j$, otherwise constraints are violated. $\endgroup$ – sirallen Oct 12 '17 at 10:12
  • $\begingroup$ @sirallen If the sum of the $A_i$s is constant, then (ignoring any other constraints) that really just means the last is the sum of the others and all the others can vary essentially independently. I think your best bet would be to find an analogous situation in an economics paper/textbook and see what they do there. Alternatively, if you can formalize more precisely what you want to calculate, then someone might be able to tell you if there is a standard notation for it in math. $\endgroup$ – Mark S. Oct 12 '17 at 11:00
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If I understand the question correctly, I believe that this falls under the concept of directional derivative.

In particular, if we treat $X$ as a function of the $A_i$, e.g.: $X(A_1, A_2, \ldots, A_n)$, then the quantity described as $dX/dA_{i \rightarrow j}$ could be described equivalently as $\nabla_v X$ where $v$ is some vector that is positive in the $A_j$ direction, negative in the $A_i$ direction, and zero elsewhere.

EDIT:

To make it even more explicit in this case, a redistribution represented by $A_{i \rightarrow j}$ corresponds to a situation where a one unit decrease in $A_i$ is matched by a $(1+r)^{j−i}$ increase in $A_j$. Then, we can define $u$ to be a vector with $-1$ in the $i$th place, $(1+r)^{j−i}$ in the $j$th place, and zero elsewhere. Define $v = u/||u||$ (i.e.: $v$ is $u$ normalized to unit length). Then, $dX / dA_{i \rightarrow j} = \nabla_v X$.

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  • $\begingroup$ Hm, I'm not so sure. Directional derivative has the property $ \nabla_{-v} X = -\nabla_v X$, right? In my problem, I have $dA_i/dA_{i \rightarrow j} = -1 $ and $dA_i / dA_{j \rightarrow i} = (1 + r)^{i - j} $. $\endgroup$ – sirallen Oct 18 '17 at 3:10
  • $\begingroup$ I'm not sure I understand your concern. Specifically, I'm not sure how $dA_i/dA_{i \rightarrow j}$ and $dA_i/dA_{j \rightarrow i}$ are supposed to relate to $dX /dA_{i \rightarrow j}$. $\endgroup$ – mhum Oct 18 '17 at 3:22
  • $\begingroup$ Also, if we let $dX/dA_{i\rightarrow j} = \nabla_v X$, there's no need to assume that $dX/dA_{j\rightarrow i} = \nabla_{-v} X$. $\endgroup$ – mhum Oct 18 '17 at 3:28
  • $\begingroup$ (1) What do you mean? $X$ can be anything. I just gave an example in the question details. $X$ can be $A_i$. (2) Okay, I guess I misinterpreted your answer. $A_{j \rightarrow i}$ has the same direction as $-v$, different magnitude $\endgroup$ – sirallen Oct 18 '17 at 3:34
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    $\begingroup$ Ok, thanks. I've extended my answer to make explicit how to get the vector $v$ in your specific case. $\endgroup$ – mhum Oct 18 '17 at 3:54

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