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Given a distribution function $F$, the Theil index is defined to be:

$$\theta(F)=\int_0^\infty\frac{x}{\mu(F)} \ln\left(\frac{x}{ \mu(F) }\right) \, dF(x)$$

where $\mu(F)$ is the mean (expected value) of the distribution. Suppose that $X\sim F$ and $Y \sim G$, where $G$ is the distribution function of $Y=aX$ for some $a>0$. Show that $\theta(G)=\theta(F)$

I'm not quite sure how to go about evaluating this integral; it doesn't look as if integration by parts will get me very far.

Just to be clear, what is written in the problem text above is equivalent to this:

$$\theta(F)=\int_0^\infty\frac{x}{\mu(F)} \ln\left(\frac{x}{ \mu(F) } \right) f(x) \, dx$$

where $f(x)$ represents the probability density function. Obviously, by using the CDF technique I realize that $G(y)=F(y/a)$, but since I'm not sure how to comput this integral, I don't know how to progress to show equivalence.

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This is just a change of variable:

$$ \begin{align} \theta(F)&=\int_0^\infty\frac{x}{\mu(F)} \ln\left(\frac{x}{ \mu(F) } \right) f(x) \, dx\\ &=\int_0^\infty\frac{a x}{\mu(G)} \ln\left(\frac{a x}{ \mu(G) } \right) \frac1a f(x) \, d(ax)\\ &=\int_0^\infty\frac{y}{\mu(G)} \ln\left(\frac{y}{ \mu(G) } \right) \frac1a f(y/a) \, dy\\ &=\int_0^\infty\frac{y}{\mu(G)} \ln\left(\frac{y}{ \mu(G) } \right) g(y) \, dy = \theta(G), \end{align} $$ where I used $\mu(G) = a\mu(F)$ and $g(y) = \frac1a f(y/a)$.

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  • $\begingroup$ I wish I could see these things. Thanks so much. $\endgroup$ – ereHsaWyhsipS Apr 7 '18 at 6:50

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