# strong law of large numbers when mean goes to infinity

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.

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 $$n$$.)
• 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. Apr 1, 2021 at 18:41
• 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 $n\to\infty$? Apr 1, 2021 at 19:05
• This is still not an infinite sequence; what is $Y_{n+1}$? Apr 1, 2021 at 19:16
• $Y_{n+1} = \frac{X_{n+1}}{ \sqrt{n+1} }$. We also have E$[Y_{n+1}]=1$ and Var$[Y_{n+1}]=\sigma$. Apr 1, 2021 at 19:20
• 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. Apr 1, 2021 at 19:22