2
$\begingroup$

Let $X_1, X_2, \ldots $ be an infinite sequence of i.i.d. random variables with $E(X_i)=\mu$ and $\mbox{Var}(X_i) < \infty$.

The law of large numbers states $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{X_i}{n} = \mu$.

Wikipedia mentions that "other formulas that look similar are not verified".

In other words $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} X_i = n\mu$ is not correct?

Is there a counter-example or an obvious reason why this is not the case?

$\endgroup$
8
  • 1
    $\begingroup$ Note also that the Law of Large Numbers does not require the existence of the variance of the $X_i$'s. $\endgroup$
    – Xi'an
    Oct 13, 2022 at 6:49
  • 5
    $\begingroup$ A warning that arises every time in the classroom:$$\lim_{n \rightarrow \infty} \sum_{i=1}^{n} X_i = n\mu$$does not make sense (mathematically) since the lhs does not depend on $n$ as a limit, while the rhs does. $\endgroup$
    – Xi'an
    Oct 13, 2022 at 6:50
  • 3
    $\begingroup$ No it does not make sense either: $\mathbb E(X_i)=\mu/n$ has the lhs depending on $i$ and the rhs depending on $n$. $\endgroup$
    – Xi'an
    Oct 13, 2022 at 7:23
  • 2
    $\begingroup$ Sorry, but the fundamental reason is that $\mathbb E(X_i)=\mu/n$ does not make sense mathematically. The random variable $X_i$ for one given $i\in\mathbb N^*$, does not depend on $n$. $\endgroup$
    – Xi'an
    Oct 13, 2022 at 10:12
  • 1
    $\begingroup$ In that case, you need to rephrase the question differently. It sounds like you are dealing with the law of large numbers for triangular arrays. $\endgroup$
    – Xi'an
    Oct 14, 2022 at 14:10

1 Answer 1

2
$\begingroup$

What do LLNs say?

At the risk of getting too oversimplistic, for independent sequence $\langle X_i\rangle_{i\in\mathbb N},$ they say the partial sum (or average, precisely)

$$\frac{1}{n}(X_1+X_2+\ldots+ X_n) \tag 1$$

converges to common mean $\mu$ in certain sense provided second moment or forth moment be finite and uniformly bounded depending on what convergence one is bothered with.

When $\rm i.i.d.$ is imposed, only the first moment needs to be finite.

What is Wikipedia saying?

It is only saying form like

$$\sum_{i=1}^n X_i -n\times \bar X$$

doesn't make sense for

not only does it not converge toward zero as $n$ increases, but it tends to increase in absolute value as $n$ increases.

Finally as Xi'an noted in the comments, be careful with indexing set and limiting variable.


Further Read:

$\rm [I]$ A First Look at Rigorous Probability Theory, Jeffrey S. Rosenthal, World Scientific Publishing, $2006, $ section $5.3,~5.4,$ pp. $60-64.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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