Can someone show me how this integrates to 1?

$\displaystyle \int_{-\infty}^{\infty} \frac{\gamma \theta^{\gamma}}{x^{\gamma +1}}$

Also, can you show me how to derive the variance and quartiles? Thank you.

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- Feb 26th 2009, 07:34 PMVENIPareto PDF
Can someone show me how this integrates to 1?

$\displaystyle \int_{-\infty}^{\infty} \frac{\gamma \theta^{\gamma}}{x^{\gamma +1}}$

Also, can you show me how to derive the variance and quartiles? Thank you. - Feb 27th 2009, 02:09 AMmr fantastic
Since this is a Pareto distribution:

1. The support is $\displaystyle [\theta, + \infty)$. So you need to show that $\displaystyle \int_{\theta}^{+ \infty} \frac{\gamma \theta^{\gamma}}{x^{\gamma +1}} \, {\color{red} dx} = 1$.

(*Ahem .... note what is in red by the way).

2. You should realise that $\displaystyle \gamma > 0$ and $\displaystyle \theta > 0$. Knowing $\displaystyle \gamma > 0$ is of particular importance in solving the integral.

Now you should note that $\displaystyle \int_{\theta}^{+ \infty} \frac{\gamma \theta^{\gamma}}{x^{\gamma +1}} \, {\color{red} dx} = \gamma \theta^{\gamma} \int_{\theta}^{+ \infty} \frac{1}{x^{\gamma +1}} \, {\color{red} dx} = \gamma \theta^{\gamma} \int_{\theta}^{+ \infty} x^{-(\gamma +1)} \, {\color{red} dx}$.

So you have a simple improper integral to calculate.

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As for the variance, you should know that $\displaystyle Var(X) = E(X^2) - [E(X)]^2$. So set up the necessary integrals and calculate the required expectations. Again, you have simple improper integrals.

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As for the quartiles:

To find the median $\displaystyle Q_2$ you need to solve $\displaystyle \gamma \theta^{\gamma} \int_{\theta}^{Q_2} x^{-(\gamma +1)} \, {\color{red} dx} = \frac{1}{2}$ for $\displaystyle Q_2$. The integral is simple to do and the subsequent equation is simple to solve.

The other quartiles are found in a similar way.