I'm practicing probability and came across this question which I think could be solved by appealing to the Central Limit Theorem.


Let $X_i, i\geq 1$ be i.i.d. uniform random variables on $[0,1]$. Let $S_n=X_1 + \dots + X_n$. Find constants $\alpha$ and $\beta$ such that the limit $\lim_{n\to\infty} \mathbb{P}(S_n<\alpha n + n^\beta)$ exists and is strictly between $\frac{1}{2}$ and $1$.

My attempt:

I tried re-arranging to get it in a form where I could use CLT, using $\alpha=\beta=\frac{1}{2}$ to get

$\mathbb{P}(S_n<\alpha n + n^\beta) = \mathbb{P}(\frac{S_n - n\cdot \frac{1}{2}}{\frac{1}{\sqrt{12}} \cdot \sqrt{n}}<\sqrt{12})$

And then taking limits via CLT to get $\Phi(\sqrt{12})$.

Is this right? Is there a simpler way? Also if anyone could provide insight into why this question is being asked/is interesting I'd appreciate that.


1 Answer 1


Yes, everything looks alright.

The CLT yields that $$\sqrt{12n}\left(\frac{S_n}{n} - \frac{1}{2}\right) \to N(0, 1),$$ so we have that $$\lim_{n \to \infty} P\left(\sqrt{12n}\left(\frac{S_n}{n} - \frac{1}{2}\right)< z\right) = \Phi(z).$$ Rearranging, $$\lim_{n \to \infty} P\left(S_n < z\sqrt{\frac{n}{12}}+\frac{n}{2}\right) = \Phi(z),$$ so if you take $z = \sqrt{12}$ we get that $1/2 < \lim_{n \to \infty} P(S_n < \sqrt{n} + n/2) = \Phi(\sqrt{12}) < 1$. So the choice $\alpha = \beta = 1/2$ is a correct choice.

But moreover, it is the only actual choice. We have that $$P\left(S_n < \alpha n + n^\beta\right) = P\left(\sqrt{12n}\left(\frac{S_n}{n} - \frac{1}{2}\right) < \sqrt{12n}\left(\frac{\alpha n + n^\beta}{n} - \frac{1}{2}\right)\right) = P\left(\sqrt{12n}\left(\frac{S_n}{n} - \frac{1}{2}\right) < \sqrt{12n}\left(\alpha - \frac{1}{2}\right) + \sqrt{12}n^{\beta-1/2}\right).$$

Since the LHS of the the inequality goes to a $N(0, 1)$ random variable, if we want the probability to be in $[1/2, 1]$, we need the RHS to be nonnegative and finite as $n \to \infty$. But then clearly we need $\alpha = \beta = 1/2$.

The interpretation is that we are trying to match how $S_n$ "grows". Certainly we expect $S_n$ to on average increase by $1/2$ every increment of $n$ as $n \to \infty$ (that is, $P(S_n > n/2) = 1/2$ and $P(S_n > \alpha n)$ for $\alpha \neq 1/2$ is either 0 or 1); the issue is capturing how the distribution changes as $n \to \infty$ so, and the term $n^{1/2}$ captures the fact that the CLT tells you that $S_n$ will have second moment that is asymptotically on the order of $n$ (since $S_n/n$ has second moment that vanishes on the order of $n$).


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