How does this limit imply the other

Question

A proof in my lecture notes uses the following

$$ n \log( 1 - P(X \geq n)) \to 0$$

$$\therefore n P(X \geq n) \to 0$$

both limits taken as $n \to \infty$.

I don't think this is a mistake (though if it is please correct me), but I'm hoping someone can help me understand this better.

So as I understand it, as $n \to \infty$ clearly $P(X \geq n) \to 0$ (as otherwise condition of distribution $\to 1$ function is violated). Hence the above step uses the fact that as $x \to 0$ we have

$$x < |\log (1-x)|$$

which is an inequality I found online. Which I think is enough to prove that step.

Any advice is appreciated

Answer 1

Let $p_n = \mathbb{P}(X \geq n) \geq 0$, then you have already known that $$\lim_{n \to \infty} p_n = 0.$$ As a result, $$\lim_{n\to \infty} \frac{p_n}{\log(1 - p_n)} = \lim_{t\to 0^+} \frac{t}{\log(1 - t)} = -1,$$ hence $$ \lim_{n\to \infty} np_n = \lim_{n\to \infty} n\log(1 - p_n) \cdot\lim_{n\to \infty}\frac{p_n}{\log(1 - p_n)} = - \lim_{n\to \infty} n\log(1 - p_n) = 0. $$ Using the inequality $x < |\log (1-x)|$ for $x > 0$ is also good, from which you can get $$ 0 \leq np_n \leq |n\log(1-p_n)|,$$ then the squeeze theorem would lead to the same result.

Answer 2

You may see it as follows.

• Set $x_n = P(X \geq n) \Rightarrow x_n \stackrel{n\to \infty}{\longrightarrow}0$
• If $x_n > 0$ for all $n \in \mathbb{N}$, you can write $$n \log{(1-x_n)} = n\cdot x_n \cdot \frac{\log{(1-x_n)}}{x_n}$$
• Note that $\frac{\log{(1-x_n)}}{x_n} \stackrel{n\to \infty}{\longrightarrow} -1$.
• Since you have $n \log{(1-x_n)} \stackrel{n\to \infty}{\longrightarrow} 0$, it follows $$\boxed{n\cdot x_n \stackrel{n\to \infty}{\longrightarrow} 0}$$

Answer 3

A simple proof can be given using the inequality $1-x \leq e^{-x}$ for $x \geq 0$. Since $n log \, (1-P\{X \geq n\}) \leq n log \, e^{-P\{X \geq n\}} =-n P\{X \geq n\}$ we get $\lim inf (-n P\{X \geq n\}) \geq 0$. It is obvious that $\lim \sup (-n P\{X \geq n\}) \leq 0$ so we are done.