Computing Gini coefficient for a 2 parameters density function

Question

I have a random variable $X$ defined by the following the density function,

\begin{equation} f_{\theta_1, \theta_2}(x) = \begin{cases} \frac{\theta_1 \theta_2^{\theta_1}}{x^{\theta_1 + 1}}, &\quad x \geq \theta_2 \\ 0, &\quad \text{otherwise} \end{cases} \end{equation}

with $\theta_1 > 2$ and $\theta_2 > 0$.

I tried to compute the Gini coefficient of $X$ but I'm not sure if my computation is right. In particular there is an integral that seems to diverge and if I try to ignore that, in the end I obtain a pretty big expression.

Anyway, here's my computation :

The Gini coefficient is defined as, \begin{equation} G_{\theta_1, \theta_2} := 2 \int_{0}^{1} \left( p - \frac{\int_{0}^{p} Q(t) dt}{E(X)} \right) dp \end{equation}

where $Q(t)$ is the quantile function of $X$ that we can compute the following way,

\begin{align*} P(X \leq x_t) &= \int_{-\infty}^{x_t} f_{\theta_1, \theta_2}(x) dx \\ &= \int_{\theta_2}^{x_t} \theta_1 \theta_2^{\theta_1} x^{-(\theta_1 + 1)} dx \\ &= - \frac{\theta_1 \theta_2^{\theta_1}}{\theta_1} \left[ x^{-\theta_1} \right]_{x=\theta_2}^{x=x_t} \\ &= - \frac{\theta_1 \theta_2^{\theta_1}}{\theta_1} \left( x_t^{-\theta_1} - \theta_2^{-\theta_1} \right) \\ \end{align*}

Let's solve $P(X \leq x_t) = t$ for $x_t$, \begin{align*} - \frac{\theta_1 \theta_2^{\theta_1}}{\theta_1} \left( x_t^{-\theta_1} - \theta_2^{-\theta_1} \right) = t &\iff x_t^{\theta_1} = \frac{t \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \\ &\iff x_t = \left( \frac{t \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \right)^{1/\theta_1} \equiv Q(t) \end{align*}

Then we compute the mean of $X$, \begin{align*} E(X) &= \int_{-\infty}^{+\infty} x f(x) dx \\ &= \int_{\theta_2}^{+\infty} x \frac{\theta_1 \theta_2^{\theta_1}}{x^{\theta_1 + 1}} dx \\ &= \theta_1 \theta_2^{\theta_1} \int_{\theta_2}^{+\infty} x^{- \theta_1} dx \\ &= - \frac{\theta_1 \theta_2^{\theta_1}}{(\theta_1 - 1)} \left[ x^{-(\theta_1 - 1)} \right]_{\theta_2}^{+\infty} \end{align*}

I guess I made a mistake here as this integral seems to diverge. I can continue showing you the computation ignoring the divergent part.

Let's compute the Gini coefficient, \begin{align*} G_{\theta_1, \theta_2} &:= 2 \left( \int_{0}^{1} p dp - \int_{0}^{1} \frac{\int_{0}^{p} Q(t) dt}{E(X)} dp \right) \end{align*}

We compute each integral separately, \begin{align*} \int_0^p Q(t) dt &= \int_0^p \left( \frac{t \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \right)^{1/\theta_1} dt \end{align*}

We use the change of variable $u = \frac{t \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1}$, $du = \frac{\theta_1}{\theta_1 \theta_2^{\theta_1}} dt$

The boundaries becomes, \begin{align*} \begin{cases} t = 0 &\implies u_1 \equiv - \theta_2^{-\theta_1} \\ t = p &\implies u_2 \equiv \frac{p \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \end{cases} \end{align*}

Then, \begin{align*} \int_0^p Q(t) dt &= \int_{u_1}^{u_2} u^{(1/\theta_1)} \frac{\theta_1 \theta_2^{\theta_1}}{\theta_1} du \\ &= \frac{\theta_1}{\theta_1 \theta_2^{\theta_1}} \left[ \frac{u^{(1/\theta_1) + 1}}{(1/\theta_1) + 1} \right]_{u_1}^{u_2} \\ &= \frac{\theta_1}{\theta_1 \theta_2^{\theta_1} ((1/\theta_1) + 1)} \left( \left( \frac{p \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} - \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} \right) \end{align*}

Therefore, \begin{align*} \frac{\int_{0}^{p} Q(t) dt}{E(X)} &= \frac{\theta_1(\theta_1 - 1)}{\theta_2^{(1 - \theta_1)} ((1/\theta_1) + 1)} \left( \left( \frac{p \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} - \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} \right) \end{align*}

Then, \begin{align*} \int_{0}^{1} \frac{\int_{0}^{p} Q(t) dt}{E(X)} dp &= \frac{\theta_1(\theta_1 - 1)}{\theta_2^{1 - \theta_1}} \left( \underbrace{\int_0^1 \left( \frac{p \theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} dp}_{\equiv A} - \underbrace{\int_0^1 \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} dp}_{\equiv B} \right) \end{align*}

Computing integral A and B. For A we use the same change of variable as before, \begin{align*} A &= \int_{u_1}^{u_2} u^{(1/\theta_1) + 1} \frac{\theta_1 \theta_2^{\theta_1}}{\theta_1} du \\ &= \frac{\theta_1 \theta_2^{\theta_1}}{\theta_1} \left( \left( \frac{\theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 2} - \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 2} \right) \end{align*}

\begin{align*} B &= \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} \int_0^1 dp \\ &= \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} \end{align*}

Then, \begin{align*} \int_0^1 pdp &= \frac{1}{2} \end{align*}

Eventually, \begin{align*} G_{\theta_1, \theta_2} &= 2 \left( \frac{1}{2} - \frac{\theta_1(\theta_1 - 1)}{\theta_2^{(1 - \theta_1)} ((1/\theta_1) + 1)} \left[ \frac{\theta_1 \theta_2^{\theta_1}}{\theta_1} \frac{1}{(1/\theta_1) + 2} \left( \left( \frac{\theta_1}{\theta_1 \theta_2^{\theta_1}} - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 2} - \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 2} \right) - \left( - \theta_2^{-\theta_1} \right)^{(1/\theta_1) + 1} \right] \right) \end{align*}