Suppose that we have the following probability density function \begin{equation} f(x)=\frac{\alpha x_m^{\alpha}}{x^{\alpha+1}}, \quad x>x_m \end{equation} and its cumulative density function \begin{equation} F(x)=1-\left(\frac{x_m}{x}\right)^{\alpha}, \quad x>x_m \end{equation} I calculate expected value of $X$ in two ways.
First way:
\begin{eqnarray} \int_{x_m}^{\infty}xf(x)dx&=&\int_{x_m}^{\infty}\frac{\alpha x_m^{\alpha}}{x^{\alpha}}dx\\ &=&\frac{\alpha x_m}{\alpha-1},\quad \alpha>1 \end{eqnarray} Second way: \begin{eqnarray} \int_{x_m}^{\infty}(1-F(x))dx&=&\int_{x_m}^{\infty} \left(\frac{x_m}{x}\right)^{\alpha} dx\\ &=&\frac{x_m}{\alpha-1},\quad \alpha>1 \end{eqnarray} why the second way gives an incorrect answer? I do not know where I am doing wrong.
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- 2$\begingroup$ How is this distribution-theory? $\endgroup$md2perpe– md2perpe2021-08-17 12:13:15 +00:00Commented Aug 17, 2021 at 12:13
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In the second method you have to add $\int_0^{x_m} (1-0)dx=x_m$ (since $f(x)=0$ for $0<x<x_m$). Now you get the same answer as in the first method.
[The general formula for the mean of a positive random variable is $\int_0^{\infty} (1-F(x))dx$].
answered Aug 17, 2021 at 12:11
Kavi Rama Murthy
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