# Verifying that a function is a cumulative distribution function

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?

• 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. – TSF Dec 2 '15 at 22:29
• If $F(x)$ is continuous then it is right-continuous. – Henry Dec 2 '15 at 23:51

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