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Let $F$ be a c.d.f for a random variable X and let $$x_1<x_2<\cdots<x_n<x_{n+1}<\cdots$$ be a monotonic increasing sequence that converges to $+\infty$. I've managed to do a tight proof of the fact that: $$\lim_{n\to\infty}F(x_n)=1$$ But now I am a bit worried that this implies that $\lim_{x\to\infty}F(x)=1$. Here is my attempt at proving this second fact. Suppose $\lim_{x\to\infty}F(x)\ne1$. Then there exists an $\epsilon_0$ such that:

  1. There exists an $x_1^*>1$ so that $|F(x_1^*)-1|\ge \epsilon_0$
  2. There exists an $x_2^*>\text{max}\{x_1^*,2\}$ so that $|F(x_2^*)-1|\ge \epsilon_0$
  3. There exists an $x_3^*>\text{max}\{x_2^*,3\}$ so that $|F(x_3^*)-1|\ge \epsilon_0$
  4. Etc, etc, etc, ...

Thus, I've created a sequence $$x_1^*<x_2^*<\cdots<x_n^*<x_{n+1}^*<\cdots$$ that converges to $+\infty$, but $\lim_{n\to\infty}F(x_n^*)\ne 1$, which contradicts my first proof. Hence, $\lim_{x\to\infty}F(x)=1$.

Does this look like a valid proof to folks?

David
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1 Answers1

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Since $F(x)$ is by definition a non-decreasing function and $F(x)\leq 1$ for all $x$, $\lim_{ x \to \infty} F(x)$ must exist. Therefore, if you showed that $\lim_{n \to \infty} F(x_n) = 1$ for some sequence $x_n$ such that $x_n \to \infty$, it must be the case that $\lim_{x \to \infty} F(x) = 1$.

ilanshom
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