# Obtaining cdf from pdf when pdf is defined on limited region/support

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 $$F(x) = -32x^{-5} + 1$$?

• Yes, you are correct. – jbowman Jan 30 at 2:49
• 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. – Glen_b -Reinstate Monica Jan 30 at 5:20
• 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 $0$ – 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 $$x$$.

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

• 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. – whuber Jan 30 at 14:24