How to arrive at a specific formulation of the relative median deviation?

Question

I am an economist currently working with this book: Frank Cowell - Measuring Inequality

On page 25 a formulation of the relative mean deviation is given as follows: $$ M = 2 \left[ F\left(\bar{y}\right) - \Phi(\bar{y}) \right] $$

$F$ is the CDF, $\Phi$ is the proportion of total income received by persons who have an income less than or equal to $y$ ( per the book's definition: $\Phi=\frac{1}{\bar{y}} \int_0^y zdF(z)$), and $\bar{y}$ is the mean. All this is also defined on page 152 in the appendix. The appendix also gives a definition of $M$:

$$ M = \int \left| \frac{y}{\bar{y}} -1\right|dF $$

The book says that the former formulation can be derived from the latter, but I have no idea how to begin with this. How do I perform the integration here and get to the first formulation?

Answer 1

The relationship is universal and does not depend on the income distribution. The only assumption to be made is that F is a CDF and $F(-\infty)=0$ (or $F(0)=0$ in this special case) and $F(\infty)=1$. Then: \begin{aligned} M &= \int_0^\infty \left|\frac{y}{\bar y}-1 \right|dF(y) \\ &= \int_0^{\bar y} \left(1 - \frac{y}{\bar y}\right)dF(y) + \int_{\bar y}^\infty \left(\frac{y}{\bar y}-1 \right)dF(y)\\ &= \left(F(\bar y)-0-\Phi(\bar y)\right) + \left( 1 - \Phi(\bar y) -1 +F(\bar y)\right) \\ &= 2 \left( F(\bar y) - \Phi(\bar y) \right) \end{aligned}

Step 1 is possible because of the monotone dependency of F on y. Step 2 uses the properties of a CDF and the definition of $\Phi$. More explicitely for the first summand: $$ \int_0^{\bar y} dF(y) = F(\bar y) - F(0) = F(\bar y) $$ since $F(0)=0$ and with definition of $\Phi$ $$ \int_0^{\bar y} \frac{y}{\bar y}dF(y) = \Phi(\bar y). $$ And for the second summand: $$ \int_{\bar y}^\infty dF(y) = F(\infty)-F(\bar y)=1-F(\bar y) $$ since $F(\infty)=1$ and \begin{aligned} \int_{\bar y}^\infty \frac{y}{\bar y} dF(y) &= \int_0^\infty \frac{y}{\bar y} dF(y) - \int_0^{\bar y} \frac{y}{\bar y} dF(y) \\ &= 1 - \Phi({\bar y}) \end{aligned} because $\int_0^{\infty} y dF(y)$ is the expectation value $\bar y$ and $\int_0^{\bar y} \frac{y}{\bar y} dF(y)$ is again the definition of $\Phi$. Note that I assumed that the income $y$ has to be positive (as in the definition of $\Phi$). Otherwise the lower integration limit $0$ has to be replaced by $-\infty$.