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Relearning cumulative distribution functions and their definitions, it says that $F_x$ is a CDF if it is non-decreasing and right continuous with value of $1$ for large $x$ and value of $0$ for negatively large $x$.

But this just says that it can increase to $1$, not that outside its domain, it is $1$.
Might be confusing if I don't give a function:
I'll give a PDF:

$f(x) = 2nx^{2n-1}$ for $0<x<1$.

How would I find the cumulative distribution function? I'd integrate it but is this correct?

$$F(X) = \begin{cases} x^{2n} \quad 0<x<1 \\ 1 \quad \text{otherwise}.\end{cases}$$
Logic tells me that it has to be $1$ (otherwise) because I want the integral from $0$ to $a$ where $a>1$ to be equal to 1?

A question that I'm stuck on that's related to this: How would I show a random variable with this distribution converges in probability to 1? I know the definition but I'm not sure how to apply it.

Natash1
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1 Answers1

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Let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space. The cumulative distribution function of a random variable $X$ is defined to be the function $F$ such that $$ F(x) = \mathbb{P}(X\le x). $$ Let $x \to -\infty$, do you think that $\lim_{x \to - \infty} F(x) = 1$ (your claim implies so)? On the other hand, $\lim_{x\to +\infty} F(x) = 1$ since it is $\lim_{x\to +\infty} \mathbb{P}(X\le x) = \mathbb{P}(X\le +\infty) = 1$.

In particular, one can prove that the cumulative distribution function has the following properties:

  • $F$ is non-decreasing, right-continuous,
  • $\lim_{x\to - \infty} F(x) = 0$ and $\lim_{x\to + \infty} F(x) = 1$.

For a proof, see, e.g., Proposition 3.9 of I. Florescu and C.A. Tudor's book Handbook of Probability or Theorem 7.2 of J. Jacod and P. Protter's book Probability Essentials.

EDIT (i.e., tentative of answer to the second question)

Let $\{X_n\}$ be the sequence of random variables with cumulative distribution function given by the one you have in the exercise, i.e., $$ F_n(x) = \begin{cases} 0, & \text{if } x \le 0, \\ x^{2n}, & \text{if } 0<x<1, \\ 1, & \text{if } x \ge 1. \end{cases} $$

Notice $$ \lim_{n \to +\infty} F_n(x) = \begin{cases} 0, & \text{if } x < 1, \\ 1, & \text{if } x \ge 1, \end{cases} $$ that is, $X_n$ converges in distribution to $1$ (i.e., a constant). What can you conclude?

Arthur11
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  • Thanks!

    I see now, it converges in distribution (to a constant) to it must converge in probability (because it's a constant)?

     – Natash1 Jun 10 '17 at 09:27
  • Another related question: If we had a PDF, such that its endpoint was the same $(0,1)$, except the PDF's maximum value is (say) $2$ (i.e. $f(1) = 2$), would it be safe to say that for $x\geq 1$, the CDF is defined to be $2$ instead of $1$ that we have in the example I first gave?
    I ask this because it seems the CDF is defined to be $1$ only because $f(1) = 1$.     – Natash1 Jun 10 '17 at 09:31
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    It converges in probability to $1$ because it converges in distribution to $1$, that is a constant. Notice that if it does not converge in distribution to a constant but to some other random variable, then you can not conclude that it converges in probability. Answer to the second question: in this case, $F(1) = \int_{-\infty}^1 f(x) , dx = 1$, hence $F$ has to be $1$ for $x \ge 1$ since $F$ is non-decreasing and always less than 1 (def. of probability measure). – Arthur11 Jun 10 '17 at 09:38
  • Thank you! I meant if there was a PDF such that $F(2) = 1 = \int_{-\infty}^{2} f(x)dx$ then can we conclude that $F(a)$ for $a\geq 2$ is equal to $2$ (since right continuous)? I didn't know that $F$ had to be less than 1 – Natash1 Jun 10 '17 at 09:41
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    It has to be less than $1$ since $F(x) = \mathbb{P}(X \le x)$ and, by definition, $\mathbb{P} \colon \mathcal{F} \to [0,1]$. – Arthur11 Jun 10 '17 at 09:44
  • Ah I see, thanks, cleared all my confusions! :D – Natash1 Jun 10 '17 at 09:50