<p>You could use the <span class="math-container">$\delta$</span>-method to get the standard deviation.</p>
<p>Let <span class="math-container">$X_0$</span> and <span class="math-container">$X_1$</span> be the random variabled denoting income in periods <span class="math-container">$0$</span> and period <span class="math-container">$1$</span>. Set <span class="math-container">$Y_0 = X_0 \ln(X_0)$</span> and <span class="math-container">$Y_1 = X_1 \ln(X_1)$</span>.</p>
<p>Let <span class="math-container">$\mu_0 = \mathbb{E}X_0$</span>, <span class="math-container">$\mu_1 = \mathbb{E} X_1$</span>, <span class="math-container">$\nu_0 = \mathbb{E}Y_0$</span> and <span class="math-container">$\nu_1 = \mathbb{E}Y_1$</span>.</p>
<p>The Theil index in periods 0 and 1 are given by:
<span class="math-container">$$
T(\mu_0, \nu_0) = \frac{\nu_0}{\mu_0} - \ln(\mu_0) \text{ and } T(\mu_1, \nu_1) = \frac{\nu_1}{\mu_1} - \ln(\mu_1).
$$</span></p>
<p>Given two finite independent samples <span class="math-container">$\{x_{0,1},\ldots, x_{0,N_0}\}$</span> and <span class="math-container">$\{x_{1,1},\ldots, x_{1,N_1}\}$</span> of period 0 and period 1 income levels, we can construct <span class="math-container">$y_{0,i} = x_{0,i} \ln(x_{0,i})$</span> and <span class="math-container">$y_{1,i} = x_{1,i} \ln(x_{(1,i)})$</span> and the finite sample analogues:
<span class="math-container">$$
\hat \mu_i = \frac{1}{N_i} \sum_{j=1}^{N_i} x_{i,j}
\text{ and } \hat \nu_i = \frac{1}{N_i} \sum_{j=1}^{N_i} y_{i,j}.
$$</span></p>
<p>By the central limit theorem:
<span class="math-container">$$
\begin{align*}
\sqrt{N_0} \begin{pmatrix} \hat \mu_0 - \mu_0\\
\hat \nu_0 - \nu_0,\\
\end{pmatrix} \sim^a {\cal N}(0, \Sigma_0)\\
\sqrt{N_1} \begin{pmatrix} \hat \mu_1 - \mu_1\\
\hat \nu_1 - \nu_1
\end{pmatrix} \sim^a {\cal N}(0, \Sigma_1)
\end{align*}
$$</span></p>
<p>Where <span class="math-container">$\Sigma_i$</span> is the Variance-Convariance matrix,
<span class="math-container">$$
\Sigma_i = \begin{bmatrix} \sigma^2_{\mu_i} &amp; \sigma_{\mu_i, \nu_i}\\ \sigma_{\mu_i, \nu_i} &amp; \sigma^2_{\nu_i}\end{bmatrix}
$$</span>
We can estimate this matrix by <span class="math-container">$\widehat{\Sigma}_i$</span> where:
<span class="math-container">$$
\hat \Sigma_i = \begin{bmatrix} \frac{1}{N_i} \sum_{j = 1}^{N_i} (x_{i,j} - \hat \mu_i)^2
&amp; \frac{1}{N_i} \sum_{j = 1}^{N_i} (x_{i,j} - \hat \mu_i)(y_{i,j} - \hat \nu_i)\\
\frac{1}{N_i}\sum_{j = 1}^{N_i} (x_{i,j} - \hat \mu_i)(y_{i,j} - \hat \nu_i) &amp; \frac{1}{N_i} \sum_{j = 1}^{N_i}(y_{i,j} - \hat \nu_i)^2\end{bmatrix}
$$</span>
Next, we can use the <span class="math-container">$\delta$</span>-method to derive the asymptotic distribution of <span class="math-container">$T(\hat \mu_i, \hat \nu_i) = \frac{\hat \nu_i}{\hat \mu_i} - \ln(\hat \mu_i).
$</span></p>
<p>Note that for <span class="math-container">$i = 0,1$</span>:
<span class="math-container">$$
\frac{\partial T(\mu, \nu)}{\partial \nu} = \frac{1}{\mu} \text{ and } \frac{\partial T(\mu,\nu)}{\partial \mu} = -\frac{\nu+ \mu}{(\mu)^2}.
$$</span>
Consider the gradient vector:
<span class="math-container">$$
\partial T_i = \begin{bmatrix}\dfrac{1}{\mu_i} &amp; -\dfrac{\mu_i + \nu_i}{(\mu_i)^2} \end{bmatrix} 
$$</span>
with estimator:
<span class="math-container">$$
\partial \widehat{T}_i = \begin{bmatrix} \dfrac{1}{\hat \mu_i} &amp; - \dfrac{\hat \mu_i + \hat \nu_i}{(\hat \mu_i)^2}\end{bmatrix}
$$</span>
Then the <span class="math-container">$\delta$</span>-method gives:
<span class="math-container">$$
(T(\hat \mu_i, \hat \nu_i) - T(\mu, \nu)) \sim^a {\cal N}\left(0,\frac{\partial T_i \, \Sigma_i \, (\partial T_i)'}{N_i}\right)
$$</span></p>
<p>The Null hypothesis is
<span class="math-container">$$H_0: \underbrace{T(\mu_0, \nu_0) - T(\mu_1, \nu_1)}_{G T(\mu_0, \nu_0, \mu_1, \nu_1)} = 0.
$$</span></p>
<p>Then under the null:
<span class="math-container">$$
\begin{align*}
(T(\hat \mu_1, \hat \nu_1) - T(\mu_1, \nu_1)) - (T(\hat \mu_0, \hat \nu_0) - T(\mu_0, \nu_0))
=T(\hat \mu_1, \hat \nu_1) - T(\hat \mu_0, \hat \nu_0)
\end{align*}
$$</span>
So under the null (if the two samples are independent):
<span class="math-container">$$
T(\hat \mu_1, \nu_1) - T(\hat \mu_0, \nu_0) \sim^a {\cal N}\left(0, \frac{\partial T_1 \, \Sigma_1 \, (\partial T_1)'}{N_1} + \frac{\partial T_0 \, \Sigma_0 \, (\partial T_0)'}{N_0}\right)
$$</span></p>
<p>Using the estimators for the variance, we have:
<span class="math-container">$$
\dfrac{T(\hat \mu_1, \nu_1) - T(\hat \mu_0, \nu_0)}{\left(\dfrac{\partial \hat T_1 \, \hat \Sigma_1 \, (\partial \hat T_1)'}{N_1} + \dfrac{\partial \hat T_0 \, \hat \Sigma_0 \, (\partial \hat T_0)}{N_0}\right)^{1/2}} \sim^a {\cal N}(0,1).
$$</span>
You can use this (asymptotic) distribution for a hypothesis test for <span class="math-container">$H_0$</span>.</p>

CC BY-SA 4.0
2023-12-28T08:44:07.940Z
2023-12-28T08:44:07.940Z
tdm
<p>In principle there is nothing that makes <span class="math-container">$t$</span>-test for difference in means not possible to apply to inequality measures.</p>
<p>However, you cannot apply <span class="math-container">$t$</span>-test just to two numbers. The test can be only used to test means of two groups.</p>
<p>For example, if by 2016 Theil's T index you mean average Theil's T index for whole world and 2017 is the mean average Theil's T index for whole world, and you have standard errors for both samples you can calculate whether the difference between the two years is statistically significant.</p>
<p>Of course, the group does not need to be all countries, it could be regions in a country etc.</p>
<p>An important caveat is that for <span class="math-container">$t$</span>-test comparing two means you should have at least approximately 25-30 observations since its parametric test. If you don't have that you can use some non-parametric test like Wilcoxon test.</p>

CC BY-SA 4.0
2023-12-24T18:42:43.357Z
2023-12-24T18:42:43.357Z
1muflon1
<p>I am curious, is it logical to implement the significance test for an inequality index, in order to interpret it. Consider Theil's T index is for the consumption inequality for year 2016 is <span class="math-container">$T_{2016}=1.3$</span> and for year 2017 is <span class="math-container">$T_{2017} = 0.7$</span>.</p>
<p>Our main purpose is to test whether the difference between two values is different than 0. So we employ t test, since they are found through samples:</p>
<p><span class="math-container">$$H_0:T_{2017}-T_{2016}=0\;\;\;\;H_1 : T_{2017}-T_{2016} \neq 0$$</span></p>
<p>Is it reasonable, and statistically possible to perform this test. I assume the answer is yes since we have obtained these values from sample data, it can be considered a sample parameter as sample mean or standard deviation. Am I right?</p>

<econometrics><statistics><inequality>
Significance Test for Inequality Index
Tunay Sabri Yüksel
# Significance Test for Inequality Index

(Asked by: username_0 on 2023-12-24)

I am curious, is it logical to implement the significance test for an inequality index, in order to interpret it. Consider Theil's T index is for the consumption inequality for year 2016 is $T\_{2016}=1.3$ and for year 2017 is $T\_{2017} = 0.7$.

Our main purpose is to test whether the difference between two values is different than 0. So we employ t test, since they are found through samples:

$$H\_0:T\_{2017}-T\_{2016}=0\;\;\;\;H\_1 : T\_{2017}-T\_{2016} \neq 0$$

Is it reasonable, and statistically possible to perform this test. I assume the answer is yes since we have obtained these values from sample data, it can be considered a sample parameter as sample mean or standard deviation. Am I right?

**Tags:** econometrics, statistics, inequality

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## 2 answers

### Answer by username_1 on 2023-12-28. Score: 1

You could use the $\delta$-method to get the standard deviation.

Let $X\_0$ and $X\_1$ be the random variabled denoting income in periods $0$ and period $1$. Set $Y\_0 = X\_0 \ln(X\_0)$ and $Y\_1 = X\_1 \ln(X\_1)$.

Let $\mu\_0 = \mathbb{E}X\_0$, $\mu\_1 = \mathbb{E} X\_1$, $\nu\_0 = \mathbb{E}Y\_0$ and $\nu\_1 = \mathbb{E}Y\_1$.

The Theil index in periods 0 and 1 are given by:
$$
T(\mu\_0, \nu\_0) = \frac{\nu\_0}{\mu\_0} - \ln(\mu\_0) \text{ and } T(\mu\_1, \nu\_1) = \frac{\nu\_1}{\mu\_1} - \ln(\mu\_1).
$$

Given two finite independent samples $\{x\_{0,1},\ldots, x\_{0,N\_0}\}$ and $\{x\_{1,1},\ldots, x\_{1,N\_1}\}$ of period 0 and period 1 income levels, we can construct $y\_{0,i} = x\_{0,i} \ln(x\_{0,i})$ and $y\_{1,i} = x\_{1,i} \ln(x\_{(1,i)})$ and the finite sample analogues:
$$
\hat \mu\_i = \frac{1}{N\_i} \sum\_{j=1}^{N\_i} x\_{i,j}
\text{ and } \hat \nu\_i = \frac{1}{N\_i} \sum\_{j=1}^{N\_i} y\_{i,j}.
$$

By the central limit theorem:
$$
\begin{align\*}
\sqrt{N\_0} \begin{pmatrix} \hat \mu\_0 - \mu\_0\\
\hat \nu\_0 - \nu\_0,\\
\end{pmatrix} \sim^a {\cal N}(0, \Sigma\_0)\\
\sqrt{N\_1} \begin{pmatrix} \hat \mu\_1 - \mu\_1\\
\hat \nu\_1 - \nu\_1
\end{pmatrix} \sim^a {\cal N}(0, \Sigma\_1)
\end{align\*}
$$

Where $\Sigma\_i$ is the Variance-Convariance matrix,
$$
\Sigma\_i = \begin{bmatrix} \sigma^2\_{\mu\_i} & \sigma\_{\mu\_i, \nu\_i}\\ \sigma\_{\mu\_i, \nu\_i} & \sigma^2\_{\nu\_i}\end{bmatrix}
$$
We can estimate this matrix by $\widehat{\Sigma}\_i$ where:
$$
\hat \Sigma\_i = \begin{bmatrix} \frac{1}{N\_i} \sum\_{j = 1}^{N\_i} (x\_{i,j} - \hat \mu\_i)^2
& \frac{1}{N\_i} \sum\_{j = 1}^{N\_i} (x\_{i,j} - \hat \mu\_i)(y\_{i,j} - \hat \nu\_i)\\
\frac{1}{N\_i}\sum\_{j = 1}^{N\_i} (x\_{i,j} - \hat \mu\_i)(y\_{i,j} - \hat \nu\_i) & \frac{1}{N\_i} \sum\_{j = 1}^{N\_i}(y\_{i,j} - \hat \nu\_i)^2\end{bmatrix}
$$
Next, we can use the $\delta$-method to derive the asymptotic distribution of $T(\hat \mu\_i, \hat \nu\_i) = \frac{\hat \nu\_i}{\hat \mu\_i} - \ln(\hat \mu\_i).
$

Note that for $i = 0,1$:
$$
\frac{\partial T(\mu, \nu)}{\partial \nu} = \frac{1}{\mu} \text{ and } \frac{\partial T(\mu,\nu)}{\partial \mu} = -\frac{\nu+ \mu}{(\mu)^2}.
$$
Consider the gradient vector:
$$
\partial T\_i = \begin{bmatrix}\dfrac{1}{\mu\_i} & -\dfrac{\mu\_i + \nu\_i}{(\mu\_i)^2} \end{bmatrix}
$$
with estimator:
$$
\partial \widehat{T}\_i = \begin{bmatrix} \dfrac{1}{\hat \mu\_i} & - \dfrac{\hat \mu\_i + \hat \nu\_i}{(\hat \mu\_i)^2}\end{bmatrix}
$$
Then the $\delta$-method gives:
$$
(T(\hat \mu\_i, \hat \nu\_i) - T(\mu, \nu)) \sim^a {\cal N}\left(0,\frac{\partial T\_i \, \Sigma\_i \, (\partial T\_i)'}{N\_i}\right)
$$

The Null hypothesis is
$$H\_0: \underbrace{T(\mu\_0, \nu\_0) - T(\mu\_1, \nu\_1)}\_{G T(\mu\_0, \nu\_0, \mu\_1, \nu\_1)} = 0.
$$

Then under the null:
$$
\begin{align\*}
(T(\hat \mu\_1, \hat \nu\_1) - T(\mu\_1, \nu\_1)) - (T(\hat \mu\_0, \hat \nu\_0) - T(\mu\_0, \nu\_0))
=T(\hat \mu\_1, \hat \nu\_1) - T(\hat \mu\_0, \hat \nu\_0)
\end{align\*}
$$
So under the null (if the two samples are independent):
$$
T(\hat \mu\_1, \nu\_1) - T(\hat \mu\_0, \nu\_0) \sim^a {\cal N}\left(0, \frac{\partial T\_1 \, \Sigma\_1 \, (\partial T\_1)'}{N\_1} + \frac{\partial T\_0 \, \Sigma\_0 \, (\partial T\_0)'}{N\_0}\right)
$$

Using the estimators for the variance, we have:
$$
\dfrac{T(\hat \mu\_1, \nu\_1) - T(\hat \mu\_0, \nu\_0)}{\left(\dfrac{\partial \hat T\_1 \, \hat \Sigma\_1 \, (\partial \hat T\_1)'}{N\_1} + \dfrac{\partial \hat T\_0 \, \hat \Sigma\_0 \, (\partial \hat T\_0)}{N\_0}\right)^{1/2}} \sim^a {\cal N}(0,1).
$$
You can use this (asymptotic) distribution for a hypothesis test for $H\_0$.


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### Answer by username_2 on 2023-12-24. Score: 0

In principle there is nothing that makes $t$-test for difference in means not possible to apply to inequality measures.

However, you cannot apply $t$-test just to two numbers. The test can be only used to test means of two groups.

For example, if by 2016 Theil's T index you mean average Theil's T index for whole world and 2017 is the mean average Theil's T index for whole world, and you have standard errors for both samples you can calculate whether the difference between the two years is statistically significant.

Of course, the group does not need to be all countries, it could be regions in a country etc.

An important caveat is that for $t$-test comparing two means you should have at least approximately 25-30 observations since its parametric test. If you don't have that you can use some non-parametric test like Wilcoxon test.


---

