<p><strong>How to derive elasticity of substitution</strong></p>
<p>The first step is to recall the definition of a differential. If you have a function <span class="math-container">$f: \Bbb R^n \to \Bbb R$</span>, say, <span class="math-container">$f(x_1,\cdots,x_n)$</span>, then: <span class="math-container">$${\rm d}f = \frac{\partial f}{\partial x_1}{\rm d}x_1 + \cdots + \frac{\partial f}{\partial x_n}\,{\rm d}x_n. $$</span></p>
<p>For example,  <span class="math-container">$$d\log v = \frac{1}{v}dv$$</span></p>
<p>Now suppose <span class="math-container">$v = \tfrac{y}{x}$</span>, then we have <span class="math-container">$$ d\log(y/x)=\frac{d(y/x)}{(y/x)}$$</span></p>
<p>and for <span class="math-container">$v = \tfrac{U_x}{U_y}$</span></p>
<p><span class="math-container">$$ d\log(U_x/U_y)=\frac{d(U_x/U_y)}{(U_x/U_y)}$$</span></p>
<p>In other words, <strong>if you reduce the problem to (1) understanding the definition of a differential and (2) use a simple change of variable</strong>, the problem becomes very straightforward.</p>
<p>You then get</p>
<p><span class="math-container">$$\sigma \equiv \frac{d\log\left(\frac{y}{x}\right)}{ d\log\left(\frac{U_x}{U_y}\right) }= \frac{ \frac{d(y/x)}{(y/x)} }{ \frac{d(U_x/U_y)}{(U_x/U_y)} } $$</span></p>
<p><strong>ASIDE:</strong></p>
<p>Note, it is important to recognize that <span class="math-container">$ d(y/x)$</span> is a meaningful concept. You simply apply quotient rule and you find</p>
<p><span class="math-container">$$ d(y/x)= \frac{xdy-ydx}{x^2}$$</span></p>
<p>This makes sense because</p>
<p><span class="math-container">$$ d\log(y/x) = d\log(y) - d\log(x) = \frac{dy}{y}-\frac{dx}{x}$$</span></p>
<p>And if you compute</p>
<p><span class="math-container">$$ d\log(y/x)=\frac{d(y/x)}{(y/x)}=\frac{ \frac{xdy-ydx}{x^2}}{y/x} =  \frac{xdy-ydx}{xy} = \frac{dy}{y}-\frac{dx}{x}$$</span></p>
<p>Same logic applies to <span class="math-container">$d(U_x/U_y)$</span>.</p>
<p>Thus, all of <span class="math-container">$\sigma$</span> is well-defined in the sense we are using the calculus tools correctly / legally.</p>
<hr />
<p><strong>What is elasticity of substitution?</strong></p>
<p>Elasticity is by how much % one thing changes relative to a % change in another. Therefore, in this case, it is % change in ratio of two goods  relative to a single % change in the <span class="math-container">$MRS$</span> for those two goods.</p>

CC BY-SA 4.0
2015-05-06T03:35:00.193Z
2020-07-31T15:30:15.573Z
2020-07-31T15:30:15.573Z
Stan Shunpike
<p>For two goods $x$ and $y$, the elasticity of substitution is defined as $$\sigma \equiv \frac{d\log\left(\frac{y}{x}\right)}{ d\log\left(\frac{U_x}{U_y}\right) }= \frac{\frac{d\left(\frac{y}{x}\right)}{\frac{y}{x}}}{ \frac{d\left(\frac{U_x}{U_y}\right)}{\frac{U_x}{U_y}}} $$</p>

<p>I am confused by two things: </p>

<ol>
<li>Why do we just write $d\log\left(\frac{y}{x}\right)$? What are we differentiating with respect to? </li>
<li>How do I use that that to show the above relation? </li>
</ol>

<p>Can someone explain?  </p>

<elasticity>
How does one derive the elasticity of substitution?
Stan Shunpike
# How does one derive the elasticity of substitution?

(Asked by: username_0 on 2015-05-06)

For two goods $x$ and $y$, the elasticity of substitution is defined as $$\sigma \equiv \frac{d\log\left(\frac{y}{x}\right)}{ d\log\left(\frac{U\_x}{U\_y}\right) }= \frac{\frac{d\left(\frac{y}{x}\right)}{\frac{y}{x}}}{ \frac{d\left(\frac{U\_x}{U\_y}\right)}{\frac{U\_x}{U\_y}}} $$

I am confused by two things:

1. Why do we just write $d\log\left(\frac{y}{x}\right)$? What are we differentiating with respect to?
2. How do I use that that to show the above relation?

Can someone explain?

**Tags:** elasticity


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## 1 answers

### Answer by username_0 on 2015-05-06. Score: 21 | Accepted answer

**How to derive elasticity of substitution**

The first step is to recall the definition of a differential. If you have a function $f: \Bbb R^n \to \Bbb R$, say, $f(x\_1,\cdots,x\_n)$, then: $${\rm d}f = \frac{\partial f}{\partial x\_1}{\rm d}x\_1 + \cdots + \frac{\partial f}{\partial x\_n}\,{\rm d}x\_n. $$

For example, $$d\log v = \frac{1}{v}dv$$

Now suppose $v = \tfrac{y}{x}$, then we have $$ d\log(y/x)=\frac{d(y/x)}{(y/x)}$$

and for $v = \tfrac{U\_x}{U\_y}$

$$ d\log(U\_x/U\_y)=\frac{d(U\_x/U\_y)}{(U\_x/U\_y)}$$

In other words, **if you reduce the problem to (1) understanding the definition of a differential and (2) use a simple change of variable**, the problem becomes very straightforward.

You then get

$$\sigma \equiv \frac{d\log\left(\frac{y}{x}\right)}{ d\log\left(\frac{U\_x}{U\_y}\right) }= \frac{ \frac{d(y/x)}{(y/x)} }{ \frac{d(U\_x/U\_y)}{(U\_x/U\_y)} } $$

**ASIDE:**

Note, it is important to recognize that $ d(y/x)$ is a meaningful concept. You simply apply quotient rule and you find

$$ d(y/x)= \frac{xdy-ydx}{x^2}$$

This makes sense because

$$ d\log(y/x) = d\log(y) - d\log(x) = \frac{dy}{y}-\frac{dx}{x}$$

And if you compute

$$ d\log(y/x)=\frac{d(y/x)}{(y/x)}=\frac{ \frac{xdy-ydx}{x^2}}{y/x} = \frac{xdy-ydx}{xy} = \frac{dy}{y}-\frac{dx}{x}$$

Same logic applies to $d(U\_x/U\_y)$.

Thus, all of $\sigma$ is well-defined in the sense we are using the calculus tools correctly / legally.

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**What is elasticity of substitution?**

Elasticity is by how much % one thing changes relative to a % change in another. Therefore, in this case, it is % change in ratio of two goods relative to a single % change in the $MRS$ for those two goods.


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