1
$\begingroup$

Let $X_1,X_2,...,X_n$ be independent and identically distributed (iid) random variables with mean $\mathbb{E}[X_i] = \sqrt{n}$. I am a bit confused about using the strong law of large numbers (SLLN) in this case, because $\mathbb{E}[X_i] = \sqrt{n}$ goes to infinity as $n\to\infty$. I wonder if the assumptions of SLLN holds here and whether we can conclude that $$ \frac{\sum_{i=1}^n X_i}{n} \to \infty $$ almost surely or not? I appreciate any response/comment.

$\endgroup$
2
$\begingroup$

The strong law of large numbers requires that you have an infinite sequence of random variables, from which you can average any number of terms you like. This gives us an infinite sequence $$ X_1, \frac{X_1 + X_2}{2}, \frac{X_1 + X_2 + X_3}{3}, \dots $$ and the strong law of large numbers is about the limit of this sequence.

Here, you have a finite sequence of random variables, and you aren't just extending it as you increase $n$: if you change $n$, the distributions also change, because the expected value changes. So the strong law of large numbers doesn't apply to this situation.

For an example where the conclusion also does not hold, suppose that each of $X_1, \dots, X_n$ is $n^2$ with probability $\frac1{n\sqrt n}$ and $0$ otherwise. Then with probability $(1 - \frac1{n \sqrt n})^n \ge 1 - \frac1{\sqrt n}$, we have $X_1 = X_2 = \dots = X_n = 0$. As a result, $\frac{\sum_{i=1}^n X_i}{n}$ converges in probability to $0$, not $\infty$. (It does not almost surely converge to anything because, in particular, we don't have a common sample space for different values of $n$.)

$\endgroup$
6
  • $\begingroup$ Thank you for your response. I agree, when we change $n$, the distribution changes. However, for any fixed $n$, we have a sequence of iid random variables. Is there any variant of the SLLN or other theorem that can tell something about the limiting distribution of the above sequence? Thanks again. $\endgroup$
    – Arthur
    Apr 1 at 18:41
  • $\begingroup$ For example does the following makes sense: Suppose in addition we have Var$[X_i]= \sigma/n$. Define $Y_i = \frac{X_i}{ \sqrt{n} }$. Then $E [Y_i] = 1$ and Var$[Y_i]=\sigma$. Hence, $Y_1, Y_2, ...$ is an infinite sequence of iid random variables. Using SLLN we conclude $\frac{\sum_{i=1}^n Y_i}{n} \to E[Y_i]=1$. Hence $\frac{\sum_{i=1}^n X_i}{n \sqrt{n} } \to 1$ as $n\to\infty$. From this, can we conclude that $\frac{\sum_{i=1}^n X_i}{n } \to \infty$ as $n\to\infty$? $\endgroup$
    – Arthur
    Apr 1 at 19:05
  • $\begingroup$ This is still not an infinite sequence; what is $Y_{n+1}$? $\endgroup$ Apr 1 at 19:16
  • $\begingroup$ $Y_{n+1} = \frac{X_{n+1}}{ \sqrt{n+1} }$. We also have E$[Y_{n+1}]=1$ and Var$[Y_{n+1}]=\sigma$. $\endgroup$
    – Arthur
    Apr 1 at 19:20
  • 1
    $\begingroup$ There's no such thing as $X_{n+1}$, though! The $X_i$ only go up to $n$. Anyway, I've also added an example where the conclusion you want doesn't hold. $\endgroup$ Apr 1 at 19:22

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.