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Above is an image of what I’m referring to. The top ‘rule of thumb’ is derived using calculus, and hence is only valid for infinitesimal changes in t. However, this equation is considered over the period of a year, which is a large unit of time. How can this equation be justified?

I've put a derivation of the 'rule of thumb' below (in case you weren't sure how it was derived). I've derived the product case, but it can be used to show the quotient case, which is what I put at the very bottom. Please excuse my naff handwriting!!

enter image description here

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$$\frac {\Delta d_{t+1}}{d_t} = \frac {\frac {D_{t+1}}{Y_{t+1}}-\frac {D_{t}}{Y_{t}}}{\frac {D_{t}}{Y_{t}}} = \frac {D_{t+1}Y_{t}-D_{t}Y_{t+1}}{D_{t}Y_{t+1}}$$

$$=\frac {D_{t+1}Y_{t}}{{D_{t}Y_{t+1}}}-1 = \frac{Y_{t}}{Y_{t+1}}\cdot \left[{\frac {D_{t+1}}{D_{t}} -\frac {Y_{t+1}}{{Y_{t}}}}\right]$$

$$=\frac{Y_{t}}{Y_{t+1}}\cdot \left[{\frac {\Delta D_{t+1}}{D_{t}} -\frac {\Delta Y_{t+1}}{{Y_{t}}}}\right]$$

So the "rule of thumb" deviates from the above exact expression by (multiplicatively) $Y_{t}/Y_{t+1}$.

Now, we are talking about the growth rate of the Debt/GDP ratio. Say the ratio was $120\text{%}$ and it went to what? in one year? Say it went to $D_{t+1} /Y_{t+1} = 130\text{%}$. Then its exact growth rate is $130/120 -1 = 8.33\text{%}$ Assume also that GDP grew at 5% yearly rate, so $Y_{t}/Y_{t+1} = 0.95238$.

In this numerical example, the "rule of thumb will give as Debt/GDP ratio growth rate $8.33\text{%} \times 0.95238 = 7.933 \text{%}$, or less than half percentage point underestimation. This is considered a small inaccuracy in the real world, when one discusses broadly these magnitudes. And in general the above numerical example exaggerates, in most cases the growth rate of the GDP and of the Debt/GDP ratio are smaller than above, so the inaccuracy will be even smaller. Hence, the "rule of thumb" becomes acceptable in general also in theoretical models.

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  • $\begingroup$ Thanks! It's kind of weird that they (+ another undergraduate macro textbook I was poring over) use a straight-up equality sign instead of an 'approximately equal' sign. $\endgroup$ – Jayden Watson Feb 15 '18 at 13:55
  • $\begingroup$ Also, do you know if there is a mathematical name for this approximation? (e.g. was this approximation named after a mathematician or does it have a name like, idk, the 'ratio approximation' etc.) $\endgroup$ – Jayden Watson Feb 15 '18 at 14:06
  • $\begingroup$ Discrete approximation of a continuous relation $\endgroup$ – Alecos Papadopoulos Feb 15 '18 at 14:20

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