Analytic formula and an exponential bound
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For $r \in [0, 1]$, define $\tau_m(r) \in [0, 1]$ by
$$
\tau_m(r) := \frac{1}{\lambda(B_m(r))}\int_{B_m(r)} \frac{\lambda(B_m(1) \cap (B_m(1) + x))}{\lambda(B_m(1))}dx.
$$

It has been proven by other users that

$$
\tau_m(r) = \frac{1}{V_m^{cap}(r, 0)}\int_{0}^r ms^{m-1}V_m^{cap}(1, s/2)ds = \frac{1}{r^mV_m^{cap}(1, 0)}\int_{0}^r ms^{m-1}V_m^{cap}(1, s/2)ds,
$$

where $V_m(r; h)$ is the volume (i.e Lebesgue measure) of the half-lens $\{x \in B_m(r) \mid x_1 \ge h\}$. I propose to do the actual calculations and get an analytic formula in terms of special functions (beta, gamma, etc.).

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The end result will be the following result on the exponential decay of $\tau_m(r)$ as a function on the dimension $m$.

>**Theorem.** *For every $r \in [0, 1]$ and large integer $m$, we have the bound*
$$
\tau_m(r) \le e^{-\frac{r^4}{16}} + r^{-m}e^{-m\frac{(1-r^2/2)^2}{2}}.
$$
*In particuar, when $r=1$ it holds that*
$$
\tau_m(1) \le 2e^{-\frac{m}{16}}.
$$

[![enter image description here][1]][1]
*Proof.*
Now, it is a classical computaiton that $V_m^{cap}(1, h) = \beta_{1-h^2}\left(\frac{m+1}{2},\frac{1}{2}\right)$, where $\beta_x(a, b) := \int_0^xt^{a-1}(1-t)^{b-1}dt$ defines the incomplete beta function. In particular, $V_m^{cap}(0) = \beta_{1}\left(\frac{m+1}{2},\frac{1}{2}\right)$.

Thus, integrating by parts, we get

$$
\begin{split}
r^mV_m^{cap}(1,0)\tau_m(1) &= \int_{0}^r s^{m-1}V_m^{cap}(s/2)ds = \int_{0}^r \beta_{1-s^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)ds^m\\
&= \left[s^m\beta_{1-s^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)\right]_0^r + \int_0^r s^m(1-s^2/4)^{\frac{m-1}{2}}ds \\
&= r^m \beta_{1-r^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right) + I_m,
\end{split}
\tag{1}
$$
where $I_m := \int_0^r s^m(1-s^2/4)^{\frac{m-1}{2}}ds$. Now, consider the change of variable $s = 2\sqrt{t}$. This gives $s^m(1-s^2/4)^{(m-1)/2}ds=2^mt^{n/2}\cdot(1-t)^{(m-1)/2}\cdot s^{-1/2}ds = 2^mt^{(m-1)/2}(1-t)^{(m-1)/2}$. Thus,

$$
\begin{split}
I_m &= \int_0^r s^m(1-s^2/4)^{\frac{m-1}{2}}ds = 2^m\int_0^{r^2/4}t^{(m-1)/2}(1-r)^{(m-1)/2}dt\\
&=: 2^m\beta_{1/4}\left(\frac{m+1}{2},\frac{1}{2}\right).
\end{split}
\tag{2}
$$

Combining (1) and (2) then gives the analytic formula,

>**Analytic formula.**
$$
\tau_m(r) = \frac{r^m\beta_{1-r^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right) + 2^m\beta_{r^2/4}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{m\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)}
$$

Now, by [well-known sub-Gaussian concentration inequalities for the beta distribution][2], one has

$$
\begin{split}
\frac{\beta_{1-r^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)}{\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)} &= \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{1}{2}} \le 1-r^2/4\right) = \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{1}{2}} - 1 \le -r^2/4\right)\\
&\le e^{-(r^2/4)^2\cdot \frac{1}{2\cdot 1/(2m)}} = e^{-mr^4/16}.
\end{split}
$$

Likewise,

$$
\begin{split}
\frac{\beta_{r^2/4}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{\beta_{1}\left(\frac{m+1}{2},\frac{m+1}{2}\right)} &= \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{m+1}{2}} \le r^2/4\right) = \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{m+1}{2}} - \frac{1}{2} \le -\frac{1-r^2/2}{2}\right)\\
&\le e^{-((1-r^2/2)/2)^2/2\cdot 4(m+2)} = e^{-(m+2)(1-r^2/2)^2/2} \le e^{-\frac{m(1-r^2/2)^2}{2}}.
\end{split}
$$

On the other hand, setting $\alpha_m = (m+1)/2$ for large $m$, Stirling's formula gives

$$
\beta_{1}\left(\alpha_m,\alpha_m\right) \approx \sqrt{2\pi}\frac{\alpha_m^{2\alpha_m-1}}{(2\alpha_m)^{2\alpha_m-1/2}} = \frac{\sqrt{\pi}}{2^{2\alpha_m-1}} = \frac{1}{2^m} \sqrt{\frac{2\pi}{m+1}}\approx \frac{1}{2^m}\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)
$$

Putting things together gives

$$
\tau_m(r) \le e^{-\frac{mr^2}{16}}+r^{-m}e^{-\frac{m(1-r^2/2)^2}{2}}.
$$


  [1]: https://i.sstatic.net/q6Ckb.png
  [2]: https://projecteuclid.org/download/pdfview_1/euclid.ecp/1507860211