<p>$$y = \frac {a(x)}{b+a(x)} \implies \frac{\partial y}{\partial x} = \frac{\partial y}{\partial a}\cdot \frac{\partial a}{\partial x}$$</p>

<p>Since</p>

<p>$$\frac{\partial y}{\partial a} = \frac {b+a -a}{(b+a)^2} = \frac {b}{(b+a)^2} &gt;0$$</p>

<p>because in your case "$b$"$&gt;0$,</p>

<p>the sign of $\frac{\partial y}{\partial x}$ will be the same as the sign of $\frac{\partial a}{\partial x}$, which is what Romer is asserting.</p>

<p>$\theta &gt;1$ would make $\frac{\partial a}{\partial x} &lt;0$, so </p>

<p>$$\theta &lt;1 \implies \frac{\partial a}{\partial x}&gt;0\implies \frac{\partial y}{\partial x}&gt;0$$</p>

CC BY-SA 3.0
2016-02-13T02:23:51.660Z
Yes thank you @Alecos Papadopoulos, I actually understood this part, my question .. I guess I should have put this first, sorry, was how to see this graphically.  Either:

1) to graph $s(r)$ as a function of $\theta$, that is a graph with axis, $(\theta, s(\theta,r))$, but when I did, I didn't see any change around $\theta=1$

or 2) graph with axis $(r, s(r))$ twice, once with $\theta<1$ and again with $\theta>1$ here I see a difference.

But I think we should be able to make this conclusion from observing either graph.
Sunhwa
CC BY-SA 3.0
2016-02-13T01:28:57.723Z
2016-02-13T01:28:57.723Z
Alecos Papadopoulos
<blockquote>
  <p>Romer equation 2.55 on the Diamond model, says:</p>
  
  <p>Let $s(r)$ being the fraction of income saved, $r$ the interest rate,$\rho$ the discount rate, and $\theta$ the household's willingness to shift consumption between periods, with</p>
  
  <p>$$ s(r) = \frac{(1+r)^{(1-\theta)/\theta}}{ (1+\rho)^{1/\theta} + (1+r)^{(1-\theta)/\theta} } $$
  implies that saving is increasing in $r \iff (1+r)^{(1 - \theta)/\theta)}$ is increasing in r; equivalently $ds / dr &gt; 0$ if $\theta &lt; 1$</p>
</blockquote>

<p>Is the first statement related to monoticity?  Also, how to show this graphically, that is to show, $s(r)$ a function of $\theta$?  Considering this, plotting $(\theta, s(\theta, r=.1))$ I do not see a change at $\theta=1$, but when I graphed $(r, s(r))$ with values $\theta=0.5,$ and $\theta=1.5$ I saw the desired change in orientation.  Which way is right?</p>

<macroeconomics>
Saving increasing in interest rate
Sunhwa
# Saving increasing in interest rate

(Asked by: username_0 on 2016-02-12)

> Romer equation 2.55 on the Diamond model, says:
>
> Let $s(r)$ being the fraction of income saved, $r$ the interest rate,$\rho$ the discount rate, and $\theta$ the household's willingness to shift consumption between periods, with
>
> $$ s(r) = \frac{(1+r)^{(1-\theta)/\theta}}{ (1+\rho)^{1/\theta} + (1+r)^{(1-\theta)/\theta} } $$
> implies that saving is increasing in $r \iff (1+r)^{(1 - \theta)/\theta)}$ is increasing in r; equivalently $ds / dr > 0$ if $\theta < 1$

Is the first statement related to monoticity? Also, how to show this graphically, that is to show, $s(r)$ a function of $\theta$? Considering this, plotting $(\theta, s(\theta, r=.1))$ I do not see a change at $\theta=1$, but when I graphed $(r, s(r))$ with values $\theta=0.5,$ and $\theta=1.5$ I saw the desired change in orientation. Which way is right?

**Tags:** macroeconomics

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## 1 answers

### Answer by username_1 on 2016-02-13. Score: 1

$$y = \frac {a(x)}{b+a(x)} \implies \frac{\partial y}{\partial x} = \frac{\partial y}{\partial a}\cdot \frac{\partial a}{\partial x}$$

Since

$$\frac{\partial y}{\partial a} = \frac {b+a -a}{(b+a)^2} = \frac {b}{(b+a)^2} >0$$

because in your case "$b$"$>0$,

the sign of $\frac{\partial y}{\partial x}$ will be the same as the sign of $\frac{\partial a}{\partial x}$, which is what Romer is asserting.

$\theta >1$ would make $\frac{\partial a}{\partial x} <0$, so

$$\theta <1 \implies \frac{\partial a}{\partial x}>0\implies \frac{\partial y}{\partial x}>0$$

> **Comments:**
> **username_0**: Yes thank you @Alecos Papadopoulos, I actually understood this part, my question .. I guess I should have put this first, sorry, was how to see this graphically.  Either:  1) to graph $s(r)$ as a function of $\theta$, that is a graph with axis, $(\theta, s(\theta,r))$, but when I did, I didn't see any change around $\theta=1$  or 2) graph with axis $(r, s(r))$ twice, once with $\theta<1$ and again with $\theta>1$ here I see a difference.  But I think we should be able to make this conclusion from observing either graph.  

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