Average deviation from mean of Irwin-Hall distribution

Question

Consider the Irwin-Hall distribution, which for me means $X_n$ is the sum of $n$ uniform iid variables in $[-1/2, 1/2]$. I would like an absolute lower bound for $E[|X_n|]$.

By general theory we know that it approaches an explicit constant times $\sqrt{n}$, but I would like a statement of the form: for all $n > N$, we have $E[|X_n|] > c\sqrt{n}$ for explicit $c$ and $N$. I am aware that one way to achieve such a statement is from Holder's inequality, applied to say the first and fourth moments and comparing with the second moment.

However, this may be a bit weak, and the $N$ we get may be quite large. Ideally I would want a statement that is true for $N$ relatively small (e.g. 1-digit would be great), and maybe a better bound if possible. Of course I can explicitly compute using a program what it is for small $N$. So one strategy that might work for what I need is a statement that the actual $c$ one gets for $n$ gets closer to the limiting value of $c$ after a certain point, and check the values by hand for all $n$ up to that value.

Edit: to clarify, and in light of Henry's answer, it seems that these $c$-values decrease with $N$, so that the limiting value is indeed an absolute lower bound. So I am really looking for a proof that these $c$-values do indeed decrease.

In some other contexts it seems that some results have been proven regarding monotonic convergence towards the central limit theorem, which is the flavor of result I need, but I haven't found anything that directly implies it.

Answer 1

I am not quite clear what you are asking for.

For $n=1$ you get $\mathbb E[|X_1|]=\frac14=0.25\times \sqrt{1}$.

For $n=2$ you get $\mathbb E[|X_2|]=\frac13\approx 0.2357\times \sqrt{2}$.

The expectation of $X_n$ is $0$ and the variance of $X_n$ is $\frac{n}{12}$ with the central limit theorem applying to $X_n$, while the expectation of a half-normal distribution is $\sqrt{\frac{2 }{\pi}}\sigma$, so I would expect your limiting value $c$ to be $\sqrt{\frac{1}{6\pi}}$ and for this to be the lower bound, i.e. for all $n$

$$\mathbb E[|X_n|] \ge \sqrt{\frac{n}{6\pi}} \approx 0.2303 \times \sqrt{n}.$$