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If I have a function

$$ F(x) = \left\{\begin{array}{} 0, & \text{if } x \leq -1\\ \frac{1}{2} - \frac{x^2}{2}, & \text{if } -1 \leq x \leq 0\\ \frac{1}{2} + \frac{x^2}{2}, & \text{if } 0 \leq x \leq 1\\ 1, & \text{if } x \geq 1 \end{array}\right. $$

how do I verify that it is a cumulative distribution function?

I know that to be a cumulative distribution function, $F$ must

  • have $\lim_{x \to - \infty} = 0$,
  • have $\lim_{x \to + \infty} = 1$,
  • be non-decreasing,
  • and be right-continuous.

Clearly $F$ satisfies the first two conditions by definition.

For $F$ to be non-decreasing, is it enough to show that $F'$ is always positive?

And I'm not sure even how to show that a function is right-continuous.

Would it be enough to sketch $F$, and then explain that it "looks like" a CDF?

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  • $\begingroup$ Sketching will be good to add to your rigorous argument, but it is not a substitute alone. You are on the right track, you should be able to use to the derivative to show it is an increasing function. Alternatively, just assume $x_1>x_2$ and plug it in brute force. $\endgroup$ – TSF Dec 2 '15 at 22:29
  • $\begingroup$ If $F(x)$ is continuous then it is right-continuous. $\endgroup$ – Henry Dec 2 '15 at 23:51
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Yes, showing that the derivative is non-negative everywhere is sufficient to show the function is non-decreasing.   (When the derivative exists.)

To demonstrate right continuity of a piecewise function, show continuity within each piece and that the value at the left limit point of each piece is the limit of the piece from the right. Ie:

$$f(-1) = \lim\limits_{x\to -1^+} f(x) \\ f(0) = \lim\limits_{x\to 0^+} f(x) \\ f(1) = \lim\limits_{x\to 1^+} f(x)$$

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