I'm working out a sum out that asks to find P(a < x < b), and the usual way to do that is to calculate F(b) - F(a). This particular sum however adds up F(a) and F(b).

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The PDF is:

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  • 1
    $\begingroup$ Can you specify the pdf with the range of values that $x$ can take? Please also add the self-study tag. $\endgroup$
    – user64106
    Feb 13, 2018 at 13:30
  • $\begingroup$ Thanks for bringing that to my notice. I've added more details to the question and added the self-study tag. $\endgroup$
    – WorldGov
    Feb 13, 2018 at 13:33
  • 2
    $\begingroup$ Please explain what you mean by "$F$". According to the formula you have displayed, it is not the (usual) distribution function. Look at the limits of the integrals. $\endgroup$
    – whuber
    Feb 13, 2018 at 13:39
  • $\begingroup$ By F(x), I mean the cumulative distribution function of x. From the small hint in your comment, I'm guessing we can't simply use F(1/2) - F(-1/2) because the function is broken into two parts and so we are integrating over the two different limits separately? $\endgroup$
    – WorldGov
    Feb 13, 2018 at 13:46
  • 1
    $\begingroup$ Why do you think the sum adds up $F(a)$ and $F(b)$? It doesn't. Look at the limits of the integrals, as @whuber has already indicated you should. $\endgroup$
    – jbowman
    Feb 16, 2018 at 2:37

2 Answers 2


I've computed that $F(1/2) = 23/32$ and $F(-1/2) = 8/32$.

This is how I approached the 2 integrations:

  • $F(1/2) = P(X \le 1/2) = 1-P(X > 1/2) = 1-\int_{1/2}^{2} f(x)dx = 23/32$
  • $F(-1/2) = \int_{-1}^{-1/2} f(x) dx = 8/32$

Therefore $P(-1/2 \le X \le 1/2) = 15/32 = F(1/2)-F(-1/2)$.


The reason is that there is one step missing:

$P(-1/2 \le X \le 1/2) = P(X\le 1/2) - P(X\le -1/2)$

$= F(1/2) - F(-1/2) = \int_{-1}^{1/2} f(x)dx - \int_{-1}^{-1/2} f(x)dx$

$ = \int_{-1/2}^{1/2} f(x)dx = \int_{-1/2}^{0}1/2dx + \int_{0}^{1/2} 1/4(2-x)dx$


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