This is a very simple question, but I want to make sure I am doing it correctly.

I have the pdf from a Pareto distribution:

$$f(x) = 160 x^{-6}, \ \ 2 \leq x < \infty$$

and want to obtain the cdf

$$F(x) = \int_{- \infty}^x f(t) \mathrm{d}t$$

In this case, is it the same if I substitue the lower bound of the integral to $2$ since the pdf is specifically defined for $x \in [2, \infty)$ such that $F(x) = -32x^{-5} + 1$?

  • 1
    $\begingroup$ Yes, you are correct. $\endgroup$ – jbowman Jan 30 at 2:49
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    $\begingroup$ The confusion only arises because you failed to define what $f$ was to the left of $2$. If you correct the omission, the difficulty immediately disappears. $\endgroup$ – Glen_b -Reinstate Monica Jan 30 at 5:20
  • $\begingroup$ I have seen the problems defined this way almost all of the time. I was unsure if it always was implied that everything outside this interval was $0$ $\endgroup$ – The Bosco Jan 30 at 9:48

Let's write the calculus formally.

$$f(x) = \begin{cases} 160x^{-6} & x\ge2\\ 0 & x< 2 \end{cases} $$

To get from $f(x)$ to $F(x)$, integrate:

$$F(x) = \int_{-\infty}^{x}f(t)dt.$$

But we can break up the integral into an integral from $-\infty$ to $2$ and another from $2$ to $x$.

$$ F(x) = \int \limits_{-\infty}^{x} f(t) \ dt = \int \limits_{-\infty}^2 0 \ dt + \int \limits_2^{x} 160t^{-6} \ dt = \int \limits_2^{x} 160t^{-6} \ dt.$$

  • $\begingroup$ For the answer to be completely accurate, you need to specify $x\ge 2.$ Obviously $F(x)=0$ for all $x \lt 2,$ but that differs from what your formula asserts. $\endgroup$ – whuber Jan 30 at 14:24

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