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Given a function

$$F(x)=(1/3)x^2, \text{$0 \le x \le 3$}$$

Is the following function valid CDF?

I have two ways to check it. First, I used three properties of CDF. Second, I will take derivative of $F(x)$ , and it will become PDF. After that, I will take integration of PDF from $-\infty$ to $\infty$, if the result is $1$, then the above function is CDF. So, what is correct way to do with above function? If it is possible, could you show $\lim_{x \to -\infty}F(x)=0$?

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  • $\begingroup$ You need to define the function on the whole of $\mathbb{R}$ first. $\endgroup$ – ajr Oct 3 '15 at 8:57
  • $\begingroup$ What does it mean of define the function? $\endgroup$ – user3051460 Oct 3 '15 at 9:02
  • $\begingroup$ You have provided us only with the values of $F$ in the interval $[0,3]$. What about, for example, $F(5)$? Should we assume that $F(x) = 0$ for $x<0$ and $F(x) = F(3) = 9$ for $x>3$? $\endgroup$ – ajr Oct 3 '15 at 9:05
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The four necessary properties of a CDF are:

  1. non decreasing.
  2. right continuous.
  3. $\lim\limits_{x\to-\infty} F(x) = 0$
  4. $\lim\limits_{x\to+\infty} F(x) = 1$

You have a suspected-to-be cumulative distribution function: $$F(x) = \begin{cases} 0 & : x \leq 0 \\ x^2/3 & : 0\leq x\leq 3 \\ 1& : 3 \leq x\end{cases}$$

Although you've only provided the middle piece.   However the other pieces are actually important.   With them included, are you satisfied that the four properties hold ?

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  • $\begingroup$ Should I also check that the PDF integrates to 1? Or does that follow from these 4 conditions? $\endgroup$ – Undertherainbow Sep 23 '19 at 11:41
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    $\begingroup$ The existence of a CDF does not require the existence of a PDF. But yes, if the proposed CDF does satisfy the four properties and may be derived over the support (the region where it is non-zero), then the integral of that derivative over the support will equal one. No need to check. $\endgroup$ – Graham Kemp Sep 24 '19 at 1:48
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Nonetheless, the info that you give, already lets us resolve the problem, if by CDF we mean a CDF of a probability distribution. If a real function $F$ is a CDF, then in particular $0\leq F(x) \leq 1$ for every real $x$ (because $F(x) = P(Z \leq x)$ for some random variable $Z$). Here $F(3) > 1$, hence $F$ cannot be a CDF of a probability distribution.

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  • $\begingroup$ I checked three properties of CDF and it is satisfied, I think it is a CDF $\endgroup$ – user3051460 Oct 3 '15 at 9:20
  • $\begingroup$ Your reasoning is incorrect though. First thing is the fact that $F$ is not differentiable at 3, and the second thing is that the mentioned integral is not 1: $\int\limits_0^3 \frac 13 x^2 dx = \frac {x^3}9 |_0^3 = 3$. $\endgroup$ – ajr Oct 3 '15 at 9:25
  • $\begingroup$ You are wrong. The integration of derivative of $F(x)$ is 1. $\endgroup$ – user3051460 Oct 3 '15 at 9:45
  • $\begingroup$ @ajr you're confusing Probability Density Function for Cumulative Distribution Function. $\endgroup$ – Graham Kemp Oct 3 '15 at 11:18
  • $\begingroup$ Indeed, that previous comment is pretty flawed. The derivative of $F$, i.e. the PDF, is $\frac 23 x \chi_{[0,3]}$. Integrating it over $\mathbb{R}$ gives $\frac{x^2}3|_0^3 = \frac {3^2}3 = 3$. $\endgroup$ – ajr Oct 3 '15 at 12:25

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