1. ## approximation problem

A book I'm using makes heavy use of the approximation (X-Y)/Y = ln X - ln Y. (ln = natural log.) Can anybody tell me (a) what the intuition of this is, and (b) what the exact RHS should be? Many thanks in anticipation ...

2. Originally Posted by AndyDenis
A book I'm using makes heavy use of the approximation (X-Y)/Y = ln X - ln Y. (ln = natural log.) Can anybody tell me (a) what the intuition of this is, and (b) what the exact RHS should be? Many thanks in anticipation ...
Are you sure you are not mis-reading something? That's not close at all. For example, if X= 53 and Y= 25, then (X-Y)/Y= 1.12 while ln(X)- ln(Y)= 0.751. Taking X and Y very large, say X= 100000 and Y= 800000 makes it a bit better: (X-Y)/Y= 0.25 and ln(X)- ln(Y)= 0.22. Is this supposed to be for X and Y very large? Or perhaps your book is requiring that X and Y be very close- that is that X-Y is small compared to Y?

3. Originally Posted by HallsofIvy
Are you sure you are not mis-reading something? That's not close at all. For example, if X= 53 and Y= 25, then (X-Y)/Y= 1.12 while ln(X)- ln(Y)= 0.751. Taking X and Y very large, say X= 100000 and Y= 800000 makes it a bit better: (X-Y)/Y= 0.25 and ln(X)- ln(Y)= 0.22. Is this supposed to be for X and Y very large? Or perhaps your book is requiring that X and Y be very close- that is that X-Y is small compared to Y?
Sorry, I wasn't very clear. Yes, it's an approximation when X and Y are close. I tried it numerically and when the LHS is 5%, the right is 4.88% - so, as you say, not very close, but rule-of-thumb OK for small differences. In one case (I understand) it is exact, dX/X = d ln X. Since the dx is supposed to be very small indeed relative to X, this should work exactly. Anyway, this approximation is very widely used in economics. But I'm interested in the intuition behind it and what the exact equality would be. I'm guessing it will just be the RHS plus some cross-multiplication term, something small times something small, which can therefore be ignored. Apologies again for the lack of clarity.

4. If t is small (less than 1) then there is a power series expansion for ln(1+t), namely $\displaystyle \ln(1+t) = t - \tfrac12t^2+\ldots$. If x–y is small compared to y, then we can put $\displaystyle t = \tfrac{x-y}y$, so that $\displaystyle 1+t = \tfrac xy$. That gives $\displaystyle \ln x - \ln y = \ln\bigl(\tfrac xy\bigr) = \tfrac{x-y}y - \tfrac12\bigl(\tfrac{x-y}y\bigr)^2 + \ldots$.

So the error in using the approximation $\displaystyle \tfrac{x-y}y \approx \ln x - \ln y$ is at most $\displaystyle \tfrac12\bigl(\tfrac{x-y}y\bigr)^2$.

5. ## approximation problem

Originally Posted by Opalg
If t is small (less than 1) then there is a power series expansion for ln(1+t), namely $\displaystyle \ln(1+t) = t - \tfrac12t^2+\ldots$. If x–y is small compared to y, then we can put $\displaystyle t = \tfrac{x-y}y$, so that $\displaystyle 1+t = \tfrac xy$. That gives $\displaystyle \ln x - \ln y = \ln\bigl(\tfrac xy\bigr) = \tfrac{x-y}y - \tfrac12\bigl(\tfrac{x-y}y\bigr)^2 + \ldots$.

So the error in using the approximation $\displaystyle \tfrac{x-y}y \approx \ln x - \ln y$ is at most $\displaystyle \tfrac12\bigl(\tfrac{x-y}y\bigr)^2$.
This is extremely helpful - many thanks!