Let $\{X_i\}$ be i.i.d random variables uniform on a measurable, symmetric set $A$ contained in $[-1,1]$. Let $g_{n}$ be density of $X_1+\ldots + X_n$.

Question (general): Is there any non-trivial bound from below on $g_n(0)$ in terms of the Lebesgue measure of $A$.

Let us consider an example, say sets with $|A|=1/4$. It is easy to find an example of $A$ such that $g_1(0)=g_3(0)=0$. One checks also that $g_2(0)=1/|A|$.

Question (specific): What about $g_4(0)$?

  • $\begingroup$ local limit theorems don't care much about the details of your distribution, but they say for large n the one that maximizes the variance minimizes the density. $\endgroup$
    – user83457
    Jun 16 '16 at 9:38

I'll only discuss even $n$, I don't have any ideas for odd $n$, plus, as your comments indicate, there may not be any lower bounds for odd $n$.

Let's write $f(t)=|A|^{-1}\widehat{\chi}_A(t)=(1/|A|)\int_A e^{2\pi itx}\, dx$. Then $$ g_n(0)=\int \widehat{g_n}(x)\, dx = \int f^n(x)\, dx . $$

First attempt: To obtain a lower bound, I'll just focus on the positive contribution coming from $x$ close to zero. Notice that $f(0)=1$, $f'(0)=0$ (because $A$ is symmetric), $|f'(x)|, |f''(x)|\le C$ (the constants could be worked out here). So $h=f^n$ satisfies $h(0)=1$, $h'(0)=0$, $|h''(x)|\lesssim n^2$, thus $$ |f^n(x)|\ge 1-cn^2x^2 . $$ It follows that $$ g_n(0)\ge C/n $$ for even $n$.

This was clearly not a very sophisticated estimate and there's probably room for improvement; for example, for $n=2$, we're off by a factor of $|A|$ for small $|A|$.

Second attempt: A perhaps more useful bound that tries more seriously to keep track of the dependence on $A$ is obtained if we use Hölder as follows: $$ \frac{1}{|A|}=\int f^2 = \int f^{a+2-a} \le \left(\int f^n\right)^{a/n}\left( \int |f|^{(2-a)n/(n-a)}\right)^{(n-a)/n} $$ If I could take $a=n/(n-1)$ here, this would (formally) give that $$ g_n(0)\ge |A|^{-1/(n-1)} \left( \int |f|\right)^{-(n-2)/(n-1)^2} . $$ That was not correct because $f\notin L^1$, but if $f\in L^{1+\epsilon}$ (this is an assumption on $A$, it's not true automatically), I can take a slightly smaller $a$ to produce a bound of the form $g_n(0)\gtrsim |A|^{-1/(n-1)+\delta}$.

  • $\begingroup$ Thanks! These are nice arguments. I see some hopes for the second argument. Note that in order to get a good bound one needs to take into account that the A is in the unit interval (I can construct bad examples when this is not valid). $\endgroup$ Jun 23 '16 at 15:12

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