2

In a book I have found this example, where:

$$ \alpha (x) = \begin{cases} 0, & \mbox{if } x < 1, \\ x^2-2x+2, & \mbox{if } 1\le x < 2, \\ 3, & \mbox{if } x = 2, \\ x+2, & \mbox{if } x > 2 \end{cases} $$ Then after some computation: $$\int_{[0,3)}x^2 d\alpha = \dfrac{109}{6}$$

How $\alpha (x)$ is computed here and for what $f(x)$? Does $x^2$ mean some $f(x)^2$?

1 Answers1

1

The function $f(x) = x^2$ is continuous. Taking into account the jumps of $\alpha$, that occurs at $x=1$ and $x=2$, you get $$ \int_{[1,3)} f(x) d\alpha(x) = f(1) [\alpha(1+) - \alpha(1-)] +\int_1^2 f(x) (2x-2) dx + f(2) [\alpha(2+) - \alpha(2-)] + \int_2^3 f(x) dx $$ where, as usual, $\alpha(x\pm)$ denote the right and left limits of $\alpha$ at $x$.

Rigel
  • 14,434
  • Its like we're calculating second moment, $E[X^2]$, where $\alpha (x)$ is CDF, right? –  May 19 '17 at 05:53