1
$\begingroup$

I'm reading Achim Klenke's textbook on probability theory, and I've hit a road block in his proof that the strong law of large numbers (SLLN) implies the weak law of large numbers (WLLN).

Let $(X_n)_{n \in \mathbb N}$ be a sequence of real integrable random variables, and let $\widetilde S_n = \sum_{i=1}^n (X_i - \mathbb E[X_i])$, where $\mathbb E[X]$ is the expectation of the random variable $X$. The weak and strong laws of large numbers are:

  • WLLN: $\displaystyle \lim_{n \to \infty} \mathbb P\left[ \left| \frac 1 n \widetilde S_n\right| > \epsilon\right]=0$.
  • SLLN: $\displaystyle \mathbb P\left[ \limsup_{n \to \infty} \left| \frac 1 n \widetilde S_n\right|=0\right] = 1$.

For $\epsilon > 0$, Klenke defines the sets $A_n^\epsilon = \left\{ \left| \frac 1 n \widetilde S_n \right| > \epsilon\right\}$ and $A = \left\{ \limsup_{n \to \infty} \left| \frac 1 n \widetilde S_n \right| > 0 \right\}$, and states that $$A = \mathop\bigcup_{m \in \mathbb N} \limsup_{n \to \infty} A_n^{1/m},$$ and thus $\mathbb P\left[ \limsup_{n \to \infty} A_n^\epsilon\right] = 0$ for $\epsilon>0$. Then, by Fatou's lemma, \begin{align} \limsup_{n \to \infty} \mathbb P\left[A_n^\epsilon\right] &= 1-\liminf_{n \to \infty} \mathbb E\left[ \mathbb 1_{\left(A_n^\epsilon\right)^c}\right] \leq 1 - \mathbb E\left[\liminf_{n \to \infty} \mathbb 1_{\left(A_n^\epsilon\right)^c}\right] = \mathbb E\left[ \limsup_{n \to \infty} \mathbb 1_{A_n^\epsilon}\right] = 0. \end{align} My question: I understand this entire argument, except for the first equality in the above equation. Equivalently, what I'm trying to figure out is, why is $\limsup \mathbb P\left[A_n^\epsilon\right] = 1-\liminf \mathbb P\left[ \left(A_n^\epsilon\right)^c\right]$? Is this some general measure theory result that I'm not seeing?

$\endgroup$
3
$\begingroup$

I think this may just be a real-analysis thing. In particular, I think that the author is using the fact that $$\lim\sup x_n = - (\lim\inf - x_n).$$ This follows from the definition of $\sup$ and $\inf$.

In particular, put $x_n = \mathbb{P}[A^\epsilon_n]$, and note that $\mathbb{P}[A^\epsilon_n] = 1 - \mathbb{P}[(A^\epsilon_n)^c]$.

Then we can rewrite the right-hand side as $$1 - \lim\inf (1 - x_n) = 1 - 1 - \lim\inf - x_n = -\lim\inf - x_n.$$ We can then apply the result above.

$\endgroup$
  • $\begingroup$ Yes, that's exactly what's being used. Took me a second but that's the gist. Thanks! $\endgroup$ – D Ford Nov 26 '17 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.