<p>Decreasing returns to standard of living means that if you add a certain amount <span class="math-container">$\delta$</span> to the income of a low income person, then the increase in her standard of living is bigger than if you would give the same amount to a high income person.</p>
<p>Let <span class="math-container">$SL(x)$</span> be the standard of living for a person with income <span class="math-container">$x$</span>.</p>
<p>Take a person with income <span class="math-container">$x$</span> and a person with income <span class="math-container">$y &gt; x$</span>, and let us add an amount <span class="math-container">$\delta$</span> to the income of both persons. Then we should have that:
<span class="math-container">$$
SL(x + \delta) - SL(x) &gt; SL(y + \delta) - SL(\delta).
$$</span>
The left  hand side is the increase in the standard of living for the poor person. The right hand side is the increase in the standard of living for a rich person.</p>
<p>Dividing by <span class="math-container">$\delta$</span> and taking the limit for <span class="math-container">$\delta \to 0$</span> gives:
<span class="math-container">$$
\frac{d SL(x)}{dx} &gt; \frac{d SL(y)}{dy}.
$$</span>
The left hand side is the marginal return to the standard of living for a person with income <span class="math-container">$x$</span> and the right hand side is the marginal return to the standard of living for person with the higher income <span class="math-container">$y$</span>.</p>
<p>Decreasing marginal returns means that this marginal return decreases when income increases. So it is lower for the higher value <span class="math-container">$y$</span> compared to the lower value <span class="math-container">$x$</span>.</p>
<p>A function with a decreasing slope is called a concave function. So what you want in the end is a function that is</p>
<ul>
<li>Increasing, i.e. <span class="math-container">$SL(x)$</span> increases with <span class="math-container">$x$</span></li>
<li>concave, i.e. <span class="math-container">$dSL(x)/dx$</span> decreases with <span class="math-container">$x$</span>.</li>
</ul>
<p>One (popular) function that satisfies these two conditions is the natural log function. Note that <span class="math-container">$d \ln(x)/dx = 1/x$</span> which indeed decreases with <span class="math-container">$x$</span>.</p>
<p>The function that the world bank uses:
<span class="math-container">$$
SL(x) = \frac{\ln(x) - \ln(x_{min})}{\ln(x_{max}) - \ln(x_{min})},
$$</span>
is just a linear transformation that makes sure that:</p>
<ul>
<li><span class="math-container">$SL(x_{min}) = 0$</span></li>
<li><span class="math-container">$SL(x_{max}) = 1$</span>.</li>
</ul>
<p>So the function is:</p>
<ul>
<li>increasing</li>
<li>concave</li>
<li>between 0 and 1</li>
</ul>

CC BY-SA 4.0
2021-05-14T14:34:28.817Z
Why $\ln$ though? Seems like for any increasing concave $f$, the transformation $T(x) = \frac{f(x) - f(x_{min})}{f(x_{max}) - f(x_{min})}$ would have the same properties. Why not use $\sqrt{ \ \ }$, $\sqrt[5]{ \ \ }$ or $\ln \ln$ instead?
Giskard
CC BY-SA 4.0
2021-05-14T14:43:20.583Z
@Giskard, You probably have to ask the Word Bank for an answer -- as they don't offer more info on their website ;-). 
A "statistical reason could be that income is usually log-normally distributed. As such, the computed $SL$ will be (truncated) normally distribution. This makes it probably easier to use as inputs for statistical analysis.
tdm
CC BY-SA 4.0
2021-05-14T14:45:25.013Z
Yeah, I was surprised your answer focused so heavily on concavity and did not mention the distribution of income! Or perhaps I missed it?
Giskard
CC BY-SA 4.0
2021-05-14T13:49:34.667Z
2021-05-14T13:49:34.667Z
tdm
<p>For calculation of HDI, the Income Index used is <span class="math-container">$\mathrm{I_{income} = \dfrac{ln(Income) - ln(min Income)}{ln(max Income) - ln(min Income)}}$</span>. Why is the log transformed income used here? <a href="http://hdr.undp.org/en/content/why-hdi-using-logarithm-income-component" rel="nofollow noreferrer">World Bank has answered</a> that it is to emphasize the diminishing returns to the standard of living but I don't get how log-transform can reflect diminishing returns.</p>
<p>I know that log-transform is generally used to reduce skewness in the data and a log-transformed independent variable does reflect diminishing returns in a linear regression model but for calculating HDI, we take geometric mean and don't perform multiple linear regression.</p>

<economic-growth><development-economics><national-income><development>
Why does HDI use log-transformed data for income index?
reasonStore
# Why does HDI use log-transformed data for income index?

(Asked by: username_0 on 2021-05-14)

For calculation of HDI, the Income Index used is $\mathrm{I\_{income} = \dfrac{ln(Income) - ln(min Income)}{ln(max Income) - ln(min Income)}}$. Why is the log transformed income used here? [World Bank has answered](http://hdr.undp.org/en/content/why-hdi-using-logarithm-income-component) that it is to emphasize the diminishing returns to the standard of living but I don't get how log-transform can reflect diminishing returns.

I know that log-transform is generally used to reduce skewness in the data and a log-transformed independent variable does reflect diminishing returns in a linear regression model but for calculating HDI, we take geometric mean and don't perform multiple linear regression.

**Tags:** economic-growth, development-economics, national-income, development


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## 1 answers

### Answer by username_1 on 2021-05-14. Score: 1

Decreasing returns to standard of living means that if you add a certain amount $\delta$ to the income of a low income person, then the increase in her standard of living is bigger than if you would give the same amount to a high income person.

Let $SL(x)$ be the standard of living for a person with income $x$.

Take a person with income $x$ and a person with income $y > x$, and let us add an amount $\delta$ to the income of both persons. Then we should have that:
$$
SL(x + \delta) - SL(x) > SL(y + \delta) - SL(\delta).
$$
The left hand side is the increase in the standard of living for the poor person. The right hand side is the increase in the standard of living for a rich person.

Dividing by $\delta$ and taking the limit for $\delta \to 0$ gives:
$$
\frac{d SL(x)}{dx} > \frac{d SL(y)}{dy}.
$$
The left hand side is the marginal return to the standard of living for a person with income $x$ and the right hand side is the marginal return to the standard of living for person with the higher income $y$.

Decreasing marginal returns means that this marginal return decreases when income increases. So it is lower for the higher value $y$ compared to the lower value $x$.

A function with a decreasing slope is called a concave function. So what you want in the end is a function that is

* Increasing, i.e. $SL(x)$ increases with $x$
* concave, i.e. $dSL(x)/dx$ decreases with $x$.

One (popular) function that satisfies these two conditions is the natural log function. Note that $d \ln(x)/dx = 1/x$ which indeed decreases with $x$.

The function that the world bank uses:
$$
SL(x) = \frac{\ln(x) - \ln(x\_{min})}{\ln(x\_{max}) - \ln(x\_{min})},
$$
is just a linear transformation that makes sure that:

* $SL(x\_{min}) = 0$
* $SL(x\_{max}) = 1$.

So the function is:

* increasing
* concave
* between 0 and 1

> **Comments:**
> **username_2**: Why $\ln$ though? Seems like for any increasing concave $f$, the transformation $T(x) = \frac{f(x) - f(x_{min})}{f(x_{max}) - f(x_{min})}$ would have the same properties. Why not use $\sqrt{ \ \ }$, $\sqrt[5]{ \ \ }$ or $\ln \ln$ instead?  
> **username_1**: @username_2, You probably have to ask the Word Bank for an answer -- as they don't offer more info on their website ;-).  A "statistical reason could be that income is usually log-normally distributed. As such, the computed $SL$ will be (truncated) normally distribution. This makes it probably easier to use as inputs for statistical analysis.  
> **username_2**: Yeah, I was surprised your answer focused so heavily on concavity and did not mention the distribution of income! Or perhaps I missed it?  

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