<p>First you need the value <span class="math-container">$w_0$</span> of <span class="math-container">$w$</span> for <span class="math-container">$s=0$</span>:</p>
<p><span class="math-container">$$
0=\alpha(1-w_0)^{\zeta-1}-(1-\alpha)w_0^{\zeta-1}\;,\\
(1-\alpha)w_0^{\zeta-1}=\alpha(1-w_0)^{\zeta-1}\;,\\
(1-\alpha)^\frac1{\zeta-1}w_0=\alpha^\frac1{\zeta-1}(1-w_0)\;;\\
w_0=\frac{\alpha^\frac1{\zeta-1}}{\alpha^\frac1{\zeta-1}+(1-\alpha)^\frac1{\zeta-1}}=\frac1{1+\left(\alpha^{-1}-1\right)^\frac1{\zeta-1}}\;.
$$</span></p>
<p>Then you can consider small deviations <span class="math-container">$\delta$</span> from <span class="math-container">$w_0$</span> with <span class="math-container">$w=w_0+\delta$</span> and linearize by keeping only terms of first order in <span class="math-container">$\delta$</span>. Since the numerator is <span class="math-container">$0$</span> at <span class="math-container">$w_0$</span>, terms of first order in the denominator don’t contribute, so</p>
<p><span class="math-container">\begin{eqnarray*}
s&amp;\approx&amp;-\beta\frac{\alpha(1-w_0-\delta)^{\zeta-1}-(1-\alpha)(w_0+\delta)^{\zeta-1}}{\alpha(1-w_0)^\zeta+(1-\alpha)w_0^\zeta}
\\
&amp;\approx&amp;
-\beta\frac{\alpha\left((1-w_0)^{\zeta-1}-(\zeta-1)\delta(1-w_0)^{-1}\right)-(1-\alpha)\left(w_0^{\zeta-1}+(\zeta-1)\delta w_0^{-1}\right)}{\alpha(1-w_0)^\zeta+(1-\alpha)w_0^\zeta}
\\
&amp;=&amp;
\beta(\zeta-1)\delta\frac{\alpha(1-w_0)^{-1}+(1-\alpha)w_0^{-1}}{\alpha(1-w_0)^\zeta+(1-\alpha)w_0^\zeta}\;.
\end{eqnarray*}</span></p>
<p>So</p>
<p><span class="math-container">$$
w\approx w_0+\frac1{\beta(\zeta-1)}\frac{\alpha(1-w_0)^\zeta+(1-\alpha)w_0^\zeta}{\alpha(1-w_0)^{-1}+(1-\alpha)w_0^{-1}}s\;.
$$</span></p>

CC BY-SA 4.0
2022-08-22T06:16:45.693Z
2022-08-22T06:16:45.693Z
joriki
<p>I have the following equation for a variable <span class="math-container">$s$</span>:</p>
<p><span class="math-container">$$
s = -\beta\frac{\alpha(1-w)^{\zeta-1}-(1-\alpha)w^{\zeta-1}}{\alpha(1-w)^{\zeta}+(1-\alpha)w^{\zeta}}
$$</span></p>
<p>where <span class="math-container">$\zeta$</span>, <span class="math-container">$\alpha$</span>, and <span class="math-container">$\beta$</span> are parameters. As far as I am aware there is not a way to explicitly write <span class="math-container">$w$</span> as a function of <span class="math-container">$s$</span>. My goal is to obtain a linear (or log-linear) approximation <span class="math-container">$f(s)=w$</span> for <span class="math-container">$s$</span> near zero. When <span class="math-container">$\zeta=0$</span> and <span class="math-container">$\beta=1$</span>, this comes out to have a nice parabolic slope and simple intercept:</p>
<p><span class="math-container">$$
w \approx (1-\alpha) + \left(\left(\alpha - \frac{1}{2}\right)^2-\frac{1}{4}\right)s
$$</span></p>
<p>I have yet to work out a linear approximation that is general for <span class="math-container">$\zeta$</span>, <span class="math-container">$\beta$</span>, and <span class="math-container">$\alpha$</span>. I have been trying to use log-linearization but the problems I run into are the following:</p>
<p>(1) When log-linearizing we have <span class="math-container">$s=\overline{s}e^{\tilde{s}}$</span> where <span class="math-container">$\tilde{s}$</span> is the deviation <span class="math-container">$\ln\left(\frac{s}{\overline{s}}\right)$</span> and <span class="math-container">$\overline{s}$</span> is our value around which we are approximating. In this case, <span class="math-container">$\overline{s}=0$</span>. We can see that we have an issue of division by zero or having <span class="math-container">$s=0$</span> always. This I can fix by approximating around some small <span class="math-container">$\epsilon&gt;0$</span>, though maybe there is a better way to do this.</p>
<p>(2) When I make substitutions to bring in the log-deviation <span class="math-container">$\tilde{s}$</span> then try to simplify, I end up with a logarithm of a sum since on the right hand side of the equation we have summations of exponents in both the denominator and numerator. I haven't figured out how to get rid of the sums in the logarithms. This is my main issue.</p>
<p>Any and all hints / tips will be much appreciated.</p>

<log-linearization>
Log linearization with sums
daydaybroskii
# Log linearization with sums

(Asked by: username_0 on 2022-02-25)

I have the following equation for a variable $s$:

$$
s = -\beta\frac{\alpha(1-w)^{\zeta-1}-(1-\alpha)w^{\zeta-1}}{\alpha(1-w)^{\zeta}+(1-\alpha)w^{\zeta}}
$$

where $\zeta$, $\alpha$, and $\beta$ are parameters. As far as I am aware there is not a way to explicitly write $w$ as a function of $s$. My goal is to obtain a linear (or log-linear) approximation $f(s)=w$ for $s$ near zero. When $\zeta=0$ and $\beta=1$, this comes out to have a nice parabolic slope and simple intercept:

$$
w \approx (1-\alpha) + \left(\left(\alpha - \frac{1}{2}\right)^2-\frac{1}{4}\right)s
$$

I have yet to work out a linear approximation that is general for $\zeta$, $\beta$, and $\alpha$. I have been trying to use log-linearization but the problems I run into are the following:

(1) When log-linearizing we have $s=\overline{s}e^{\tilde{s}}$ where $\tilde{s}$ is the deviation $\ln\left(\frac{s}{\overline{s}}\right)$ and $\overline{s}$ is our value around which we are approximating. In this case, $\overline{s}=0$. We can see that we have an issue of division by zero or having $s=0$ always. This I can fix by approximating around some small $\epsilon>0$, though maybe there is a better way to do this.

(2) When I make substitutions to bring in the log-deviation $\tilde{s}$ then try to simplify, I end up with a logarithm of a sum since on the right hand side of the equation we have summations of exponents in both the denominator and numerator. I haven't figured out how to get rid of the sums in the logarithms. This is my main issue.

Any and all hints / tips will be much appreciated.

**Tags:** log-linearization

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## 1 answers

### Answer by username_1 on 2022-08-22. Score: 3

First you need the value $w\_0$ of $w$ for $s=0$:

$$
0=\alpha(1-w\_0)^{\zeta-1}-(1-\alpha)w\_0^{\zeta-1}\;,\\
(1-\alpha)w\_0^{\zeta-1}=\alpha(1-w\_0)^{\zeta-1}\;,\\
(1-\alpha)^\frac1{\zeta-1}w\_0=\alpha^\frac1{\zeta-1}(1-w\_0)\;;\\
w\_0=\frac{\alpha^\frac1{\zeta-1}}{\alpha^\frac1{\zeta-1}+(1-\alpha)^\frac1{\zeta-1}}=\frac1{1+\left(\alpha^{-1}-1\right)^\frac1{\zeta-1}}\;.
$$

Then you can consider small deviations $\delta$ from $w\_0$ with $w=w\_0+\delta$ and linearize by keeping only terms of first order in $\delta$. Since the numerator is $0$ at $w\_0$, terms of first order in the denominator don’t contribute, so

\begin{eqnarray\*}
s&\approx&-\beta\frac{\alpha(1-w\_0-\delta)^{\zeta-1}-(1-\alpha)(w\_0+\delta)^{\zeta-1}}{\alpha(1-w\_0)^\zeta+(1-\alpha)w\_0^\zeta}
\\
&\approx&
-\beta\frac{\alpha\left((1-w\_0)^{\zeta-1}-(\zeta-1)\delta(1-w\_0)^{-1}\right)-(1-\alpha)\left(w\_0^{\zeta-1}+(\zeta-1)\delta w\_0^{-1}\right)}{\alpha(1-w\_0)^\zeta+(1-\alpha)w\_0^\zeta}
\\
&=&
\beta(\zeta-1)\delta\frac{\alpha(1-w\_0)^{-1}+(1-\alpha)w\_0^{-1}}{\alpha(1-w\_0)^\zeta+(1-\alpha)w\_0^\zeta}\;.
\end{eqnarray\*}

So

$$
w\approx w\_0+\frac1{\beta(\zeta-1)}\frac{\alpha(1-w\_0)^\zeta+(1-\alpha)w\_0^\zeta}{\alpha(1-w\_0)^{-1}+(1-\alpha)w\_0^{-1}}s\;.
$$


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