<blockquote>
<p>How <span class="math-container">$\frac{\triangle G_t}{Y_t}=\triangle \left(\frac{G_t}{Y_t}\right)$</span>
relationship can be established?</p>
</blockquote>
<p>If <span class="math-container">$\Delta G_t={G_t-G_{t-1}}$</span> and <span class="math-container">$Y_t$</span> is fixed:</p>
<p><span class="math-container">$$
\frac{\Delta G_t}{Y_t}=\frac {G_t-G_{t-1}}{Y_t}=\frac {G_t}{Y_t}-\frac {G_{t-1}}{Y_t}=\Delta \left(\frac{G_t}{Y_t}\right)
$$</span>.</p>
<p>It is just a different way of writing the same thing.</p>

CC BY-SA 4.0
2024-01-03T13:07:29.923Z
2024-01-05T18:18:36.050Z
2024-01-05T18:18:36.050Z
BakerStreet
<p>The calculation below is from a NBER paper.</p>
<p>If <span class="math-container">$Y_t$</span> grows at the constant rate <span class="math-container">$g=r$</span>, then
<span class="math-container">$$
\triangle G_t+\frac{\triangle G_{t+1}}{1+r}+\frac{\triangle G_{t+2}}{(1+r)^2}+\dots$$</span>
can be rewritten as
<span class="math-container">$$
Y_t \left(\triangle \frac{G_t}{Y_t}+\triangle \frac{G_{t+1}}{Y_{t+1}}+\triangle \frac{G_{t+2}}{Y_{t+2}}+\dots\right)
$$</span>
I think this calculation is from <span class="math-container">$Y_{t+1}=Y_t (1+r)$</span> and <span class="math-container">$\frac{\triangle G_t}{Y_t}=\triangle \left(\frac{G_t}{Y_t}\right)$</span>.<br />
How <span class="math-container">$\frac{\triangle G_t}{Y_t}=\triangle \left(\frac{G_t}{Y_t}\right)$</span> relationship can be established?<br />
Or are there any other missing clues I didn't noticed?</p>

<macroeconomics><mathematical-economics>
How can the relationship $\frac{\triangle G_t}{Y_t}=\triangle \left(\frac{G_t}{Y_t}\right)$ be established?
guest
# How can the relationship $\frac{\triangle G_t}{Y_t}=\triangle \left(\frac{G_t}{Y_t}\right)$ be established?

(Asked by: username_0 on 2024-01-02)

The calculation below is from a NBER paper.

If $Y\_t$ grows at the constant rate $g=r$, then
$$
\triangle G\_t+\frac{\triangle G\_{t+1}}{1+r}+\frac{\triangle G\_{t+2}}{(1+r)^2}+\dots$$
can be rewritten as
$$
Y\_t \left(\triangle \frac{G\_t}{Y\_t}+\triangle \frac{G\_{t+1}}{Y\_{t+1}}+\triangle \frac{G\_{t+2}}{Y\_{t+2}}+\dots\right)
$$
I think this calculation is from $Y\_{t+1}=Y\_t (1+r)$ and $\frac{\triangle G\_t}{Y\_t}=\triangle \left(\frac{G\_t}{Y\_t}\right)$.  
How $\frac{\triangle G\_t}{Y\_t}=\triangle \left(\frac{G\_t}{Y\_t}\right)$ relationship can be established?  
Or are there any other missing clues I didn't noticed?

**Tags:** macroeconomics, mathematical-economics

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## 1 answers

### Answer by username_1 on 2024-01-03. Score: 2 | Accepted answer

> How $\frac{\triangle G\_t}{Y\_t}=\triangle \left(\frac{G\_t}{Y\_t}\right)$
> relationship can be established?

If $\Delta G\_t={G\_t-G\_{t-1}}$ and $Y\_t$ is fixed:

$$
\frac{\Delta G\_t}{Y\_t}=\frac {G\_t-G\_{t-1}}{Y\_t}=\frac {G\_t}{Y\_t}-\frac {G\_{t-1}}{Y\_t}=\Delta \left(\frac{G\_t}{Y\_t}\right)
$$.

It is just a different way of writing the same thing.


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