Estimation of term \(T_1\). As was mentioned above, we start with
\(T_1\). It will be more handful to write it in the following form:
$$\begin{aligned} T_1&= \int _{t_{p,d}}^{2t_{p,d}} {\mathbb {P}}\bigl (E_{2,3} E_{1,3} | E_{1,2}(x) \bigr ) f_d(x) dx = \\&= t_{p,d} \int _1^2 {\mathbb {P}}\bigl (E_{2,3} E_{1,3} | E_{1,2}(\alpha t_{p,d}) \bigr ) f_d(\alpha t_{p,d}) d\alpha . \end{aligned}$$
Conditioning on
\(E_{1,2}(\alpha t_{p,d})\), the probability
\({\mathbb {P}}\bigl (E_{2,3} E_{1,3} | E_{1,2}(\alpha t_{p,d})\bigr )\) is just the normalized surface area of the intersection of two caps of angle
\(\arccos (t_{p,d})\) (and
\(\arccos (\alpha t_{p,d})\) is the angle between the axes of these caps).
$$\begin{aligned} {\tilde{p}}(\alpha )&:= P\bigl (E_{2,3} E_{1,3} | E_{1,2}(\alpha t_{p,d})\bigr ) = \nonumber \\&= \frac{A_d(\arccos (t_{p,d}), \arccos (t_{p,d}), \arccos (\alpha t_{p,d}))}{A_d} = \nonumber \\&= \frac{\Gamma \left( \frac{d}{2} \right) }{2\pi ^{d/2}} \left( J_d^{\theta _{min}, \arccos (t_{p,d})} + J_d^{\arccos (\alpha t_{p,d}) - \theta _{min},\arccos (t_{p,d})}\right) {,} \end{aligned}$$
(9)
where
\(J_d^{a,b}\) is defined in Theorem
6, and
$$\begin{aligned} \theta _{min} = \arctan \left( \frac{1}{\sin (\arccos (\alpha t_{p,d}))} - \frac{1}{\tan (\arccos (\alpha t_{p,d}))} \right) = \arctan \left( \sqrt{\frac{1-\alpha t_{p,d}}{1+\alpha t_{p,d}}} \right) . \end{aligned}$$
Because both caps are of the same angle, the parts in the right-hand side of (
9) are equal. Therefore, recalling the definition of
\(J_d^{\theta _{min}, \arccos (t_{p,d})}\),
$$\begin{aligned} {\tilde{p}}(\alpha )&= \frac{\Gamma \left( \frac{d}{2} \right) }{\pi ^{d/2}} J_d^{\theta _{min}, \arccos (t_{p,d})} \\&= \frac{\Gamma \left( \frac{d}{2} \right) \pi ^{(d-1)/2}}{\pi ^{d/2} \Gamma \left( \frac{d-1}{2} \right) } \int _{\theta _{min}}^{\arccos (t_{p,d})} \sin ^{d-2} \phi \, I_{1 - \left( \frac{\tan \theta _{min}}{\tan \phi } \right) ^2} \left( \frac{d-2}{2}, \frac{1}{2} \right) d\phi . \end{aligned}$$
Let us make a change of variables:
\(\sin ^2 \phi = z\). Using this change and the expression for
\(\theta _{min}\), we obtain
$$\begin{aligned} {\tilde{p}}(\alpha ) = \frac{\Gamma \left( \frac{d}{2} \right) \pi ^{(d-1)/2}}{2\pi ^{d/2} \Gamma \left( \frac{d-1}{2} \right) } \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} z^{(d-3)/2} (1-z)^{-1/2} I_{1 - \frac{1-\alpha t_{p,d}}{1+\alpha t_{p,d}}\cdot \frac{1-z}{z}} \left( \frac{d-2}{2}, \frac{1}{2} \right) dz. \end{aligned}$$
Considering the definition of a regularized beta function and the formula
\(\Gamma (d+1) = d\Gamma (d)\), we have
$$\begin{aligned} {\tilde{p}}(\alpha ) = \frac{d}{2\pi } \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} z^{(d-3)/2} (1-z)^{-1/2} \left( \int _0^{1 - \frac{1-\alpha t_{p,d}}{1+\alpha t_{p,d}}\cdot \frac{1-z}{z}} y^{d/2-2} (1-y)^{-1/2} dy \right) dz. \end{aligned}$$
Next, we need a simple double bound on the incomplete beta function
\(I_u(a,1/2)\):
$$\begin{aligned} \frac{u^{a+1}}{a+1}\le \int _0^u t^a (1-t)^{-1/2} dt \le \frac{u^{a+1}}{(a+1)\sqrt{1-u}}{,} \end{aligned}$$
which can be established by estimation of
\((1-t)^{-1/2}\) and subsequent explicit integration. That is why
$$\begin{aligned}{}&\frac{1}{\pi } \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} z^{(d-3)/2} (1-z)^{-1/2} \left( 1 - \frac{1-\alpha t_{p,d}}{1+\alpha t_{p,d}}\cdot \frac{1-z}{z} \right) ^{d/2-1} dz \le {\tilde{p}}(\alpha ) \\&\le \frac{2}{\pi } \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} z^{(d-3)/2} (1-z)^{-1/2} \left( \frac{1-\alpha t_{p,d}}{1+\alpha t_{p,d}}\cdot \frac{1-z}{z} \right) ^{-1/2} \left( 1 - \frac{1-\alpha t_{p,d}}{1+\alpha t_{p,d}}\cdot \frac{1-z}{z} \right) ^{d/2-1} dz. \end{aligned}$$
Here we used the fact that
\(1 \le d/(d-2) \le 2\) for
\(d \ge 4\). We can transform:
$$\begin{aligned} 1 - \frac{1-\alpha t_{p,d}}{1+\alpha t_{p,d}}\cdot \frac{1-z}{z} = \frac{2z - 1 + \alpha t_{p,d}}{(1+\alpha t_{p,d})z}, \end{aligned}$$
which gives us the following estimation:
$$\begin{aligned}{}&\frac{1}{2\pi } \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} \frac{1}{\sqrt{z(1-z)}} \left( \frac{2z - 1 + \alpha t_{p,d}}{1+\alpha t_{p,d}}\right) ^{d/2-1} dz \le {\tilde{p}}(\alpha ) \\&\le \frac{1}{\pi } \sqrt{\frac{1+\alpha t_{p,d}}{1-\alpha t_{p,d}}} \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} \frac{1}{1-z} \left( \frac{2z - 1 + \alpha t_{p,d}}{1+\alpha t_{p,d}}\right) ^{d/2-1} dz \end{aligned}$$
It is easy to check that
\(\displaystyle \frac{1}{\sqrt{z(1-z)}}\) has a minimum value 1/2 for
\(z \in [0, 1)\), and
\(\displaystyle \frac{1}{1-z}\) is increasing in
z for
\(z > 0\). Therefore,
$$\begin{aligned}{}&\frac{1}{2\pi } \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} \left( \frac{2z - 1 + \alpha t_{p,d}}{1+\alpha t_{p,d}}\right) ^{d/2-1} dz \le {\tilde{p}}(\alpha ) \\&\le \frac{2}{\pi t^2_{p,d}} \sqrt{\frac{1+\alpha t_{p,d}}{1-\alpha t_{p,d}}} \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} \left( \frac{2z - 1 + \alpha t_{p,d}}{1+\alpha t_{p,d}}\right) ^{d/2-1} dz. \end{aligned}$$
Now we can explicitly compute the integral:
$$\begin{aligned} \int _{\frac{1-\alpha t_{p,d}}{2}}^{1-t^2_{p,d}} \left( \frac{2z - 1 + \alpha t_{p,d}}{1+\alpha t_{p,d}}\right) ^{d/2-1} dz = \frac{1+\alpha t_{p,d}}{d} \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{d/2}, \end{aligned}$$
which implies the final bounds on
\({\tilde{p}}(\alpha )\):
$$\begin{aligned} \frac{1+\alpha t_{p,d}}{2\pi d} \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{d/2} \le {\tilde{p}}(\alpha ) \le \frac{2(1+\alpha t_{p,d})^{3/2}}{\pi d t^2_{p,d} \sqrt{1-\alpha t_{p,d}}} \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{d/2}. \end{aligned}$$
This means that
\(T_1\) can be estimated as follows:
$$\begin{aligned} \frac{t_{p,d}}{2\pi d} \int _1^2 (1+\alpha t_{p,d}) \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{d/2} f_d(\alpha t_{p,d}) d\alpha \le T_1 \nonumber \\ \le \frac{2}{\pi d t_{p,d}} \int _1^2 \frac{(1+\alpha t_{p,d})^{3/2}}{\sqrt{1-\alpha t_{p,d}}} \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{d/2} f_d(\alpha t_{p,d}) d\alpha . \end{aligned}$$
(10)
Let us recall (
8) and rewrite the ‘essential’ part of previous inequalities.
$$\begin{aligned}{}&\left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{d/2} f_d(\alpha t_{p,d}) \nonumber \\&= C_f(d) \sqrt{d} \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{3/2} \left( (1-\alpha ^2 t^2_{p,d}) \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) \right)
^{(d-3)/2} \nonumber \\&= C_f(d) \sqrt{d} \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{3/2} \left( 1 - (2 + \alpha ^2)t^2_{p,d} + 2\alpha t^3_{p,d} \right) ^{(d-3)/2} \nonumber \\&= C_f(d) \sqrt{d} \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{3/2} \frac{\left( 1 - (2 + \alpha ^2)t^2_{p,d} + 2\alpha t^3_{p,d} \right) ^{(d-1)/2} }{1 - (2 + \alpha ^2)t^2_{p,d} + 2\alpha t^3_{p,d}} . \end{aligned}$$
(11)
Of course, we are most interested in the term
\(\left( 1 - (2 + \alpha ^2)t^2_{p,d} + 2\alpha t^3_{p,d} \right) ^{(d-1)/2}\), which can be rewritten in the following form:
$$\begin{aligned} \left( 1 - (2 + \alpha ^2))t^2_{p,d} + 2\alpha t^3_{p,d} \right) ^{(d-1)/2} = \exp \left\{ \frac{d-1}{2} \log \left( 1 - (2 + \alpha ^2)t^2_{p,d} + 2\alpha t^3_{p,d} \right) \right\} . \end{aligned}$$
Since
\(t_{p,d} \rightarrow 0\) as
\(n\rightarrow \infty\), one can use Taylor series for logarithm:
$$\begin{aligned}{}&\log \left( 1 - (2 + \alpha ^2)t^2_{p,d} + 2\alpha t^3_{p,d} \right) = -(2 + \alpha ^2)t^2_{p,d} + 2\alpha t^3_{p,d} + O(t^4_{p,d}) \\&= -(3 + (\alpha ^2 - 1))t^2_{p,d} + 2\alpha t^3_{p,d} + O(t^4_{p,d}), \;\; n\rightarrow \infty . \end{aligned}$$
From (
5) one can easily deduce that
$$\begin{aligned} \exp \left( -\frac{d-1}{2} t^2_{p,d}\right) = C_e(p,d) p t_{p,d}\sqrt{d} \exp \left( O(t^4_{p,d}d) \right) , \end{aligned}$$
where
\(1 \le C_e(p,d) \le 12\) is some function that depends only on
p and
d. Therefore,
$$\begin{aligned}{}&\left( 1 - (3 + (\alpha ^2 - 1))t^2_{p,d} + 2\alpha t^3_{p,d} \right) ^{(d-1)/2} \nonumber \\&= \exp \left\{ -3\frac{d-1}{2} t^2_{p,d} \right\} \exp \left\{ -\frac{d-1}{2} \left( (\alpha ^2 - 1)t^2_{p,d} - 2\alpha t^3_{p,d} + O(t^4_{p,d}) \right) \right\} \nonumber \\&= C_e^3(p,d) p^3 t^3_{p,d} d^{3/2} \exp \left\{ -\frac{d-1}{2} \left( (\alpha ^2 - 1)t^2_{p,d} - 2\alpha t^3_{p,d} + O(t^4_{p,d}) \right) \right\} . \end{aligned}$$
(12)
We have transformed the main term of (
11). Let us deal with ‘unimportant’ parts of (
10) and (
11). Denote
$$\begin{aligned} h_l(\alpha , p, d)&= \frac{C_f(d) C_e^3(p,d)}{2\pi } \cdot \frac{(1+\alpha t_{p,d}) \left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{3/2}}{1 - (3 + (\alpha ^2 - 1))t^2_{p,d} + 2\alpha t^3_{p,d}};\\ h_u(\alpha , p, d)&= \frac{2C_f(d) C_e^3(p,d)}{\pi } \cdot \frac{(1+\alpha t_{p,d})^{3/2}}{\sqrt{1-\alpha t_{p,d}}} \cdot \frac{\left( 1 - \frac{2t^2_{p,d}}{1+\alpha t_{p,d}} \right) ^{3/2}}{1 - (3 + (\alpha ^2 - 1))t^2_{p,d} + 2\alpha t^3_{p,d}}. \end{aligned}$$
Since
\(1 \le \alpha \le 2\) and
\(t_{p,d} \rightarrow 0\), for sufficiently large
n and some constants
\(C_l > 0\) and
\(C_u > 0\),
$$\begin{aligned} h_l(\alpha , p, d) \ge \frac{C_f(d) C_e^3(p,d)}{4\pi } \cdot \frac{(1+t_{p,d}) \left( 1 - \frac{2t^2_{p,d}}{1+t_{p,d}} \right) ^{5/2}}{1 - 3t^2_{p,d} + 4t^3_{p,d}} \ge 6\sqrt{2\pi } C_l. \end{aligned}$$
(13)
and
$$\begin{aligned} h_u(\alpha , p, d) \le \frac{C_f(d) C_e^3(p,d) (1+2t_{p,d})^{3/2} \left( 1 - \frac{2t^2_{p,d}}{1+2t_{p,d}} \right) ^{3/2}}{\pi \sqrt{1-t_{p,d}} (1 - 6t^2_{p,d} + 2t^3_{p,d})} \le 6\sqrt{2\pi } C_u. \end{aligned}$$
(14)
Then, plugging (
11), (
12), (
13) and (
14) into (
10), we obtain the following final bounds on
\(T_1\) at this step:
$$\begin{aligned} 6\sqrt{2\pi }C_l dt^4_{p,d} p^3 \int _1^2 \exp \left\{ -\frac{d-1}{2} \left( (\alpha ^2 - 1) t^2_{p,d} - 2\alpha t^3_{p,d} + O(t^4_{p,d}) \right) \right\} d\alpha \le T_1 \le \nonumber \\ \le 6\sqrt{2\pi }C_l dt^2_{p,d} p^3 \int _1^2 \exp \left\{ -\frac{d-1}{2} \left( (\alpha ^2 - 1)t^2_{p,d} - 2\alpha t^3_{p,d} + O(t^4_{p,d}) \right) \right\} d\alpha . \end{aligned}$$
(15)
Expression of bounds on \(T_1\) with the CDF of the standard normal distribution One can easily get that
$$\begin{aligned} (\alpha ^2 - 1)t^2_{p,d} - 2\alpha t^3_{p,d} + O(t^4_{p,d}) = t^2_{p,d} \left( (\alpha - t_{p,d})^2 - 1 + O(t^2_{p,d}) \right) {,} \end{aligned}$$
which implies
$$\begin{aligned}{}&\int _1^2 \exp \left\{ -\frac{d-1}{2} \left( (\alpha ^2 - 1)t^2_{p,d} - 2\alpha t^3_{p,d} + O(t^4_{p,d}) \right) \right\} d\alpha \\&= \exp \left\{ \frac{d-1}{2} t^2_{p,d} (1+O(t^2_{p,d}))\right\} \int _1^2 \exp \left\{ -\frac{d-1}{2} t^2_{p,d} (\alpha - t_{p,d})^2 \right\} d\alpha . \end{aligned}$$
Let us treat the integral in the right-hand side of the last equation. Changing the variable
\(\beta = (\alpha - t_{p,d})t_{p,d} \sqrt{d-1}\), we obtain that
$$\begin{aligned}{}&\int _1^2 \exp \left\{ -\frac{d-1}{2} t^2_{p,d} (\alpha - t_{p,d})^2 \right\} d\alpha = \frac{1}{t_{p,d}\sqrt{d-1}} \int _{(1-t_{p,d})t_{p,d}\sqrt{d-1}}^{(2-t_{p,d})t_{p,d}\sqrt{d-1}} e^{-\beta ^2/2} d\beta \\&= \frac{1}{t_{p,d}\sqrt{d-1}} \left( \Phi ((2-t_{p,d})t_{p,d}\sqrt{d-1}) - \Phi ((1-t_{p,d})t_{p,d}\sqrt{d-1}) \right) . \end{aligned}$$
Since
\(t_{p,d}\sqrt{d-1} \rightarrow \infty\) as
\(n\rightarrow \infty\),
$$\begin{aligned} \Phi ((2-t_{p,d})t_{p,d}\sqrt{d-1}) - \Phi ((1-t_{p,d})t_{p,d}\sqrt{d-1}) \\ = \frac{e^{-(1-t_{p,d})^2 t^2_{p,d}(d-1)/2}}{(1-t_{p,d}) t_{p,d}\sqrt{2\pi (d-1)}}(1+o(1)) - \frac{e^{-(2-t_{p,d})^2 t^2_{p,d}(d-1)/2}}{(2-t_{p,d}) t_{p,d}\sqrt{2\pi (d-1)}}(1+o(1)). \end{aligned}$$
But the ratio of the second and the first terms in the right-hand side converges to 0 as
\(n\rightarrow \infty\). Indeed,
$$\begin{aligned}{}&\frac{e^{-(2-t_{p,d})^2 t^2_{p,d}(d-1)/2}}{e^{-(1-t_{p,d})^2 t^2_{p,d}(d-1)/2}} \cdot \frac{2-t_{p,d}}{1-t_{p,d}} \le 3 \exp \left\{ {\frac{d-1}{2} t^2_{p,d} ((1-t_{p,d})^2-(2-t_{p,d})^2)} \right\} \nonumber \\&= 3 \exp \left\{ {\frac{d-1}{2} t^2_{p,d} (-3+2t_{p,d})} \right\} \le 3\exp \left\{ -2\cdot \frac{d-1}{2} t^2_{p,d} \right\} \rightarrow 0, \;\; n\rightarrow \infty . \end{aligned}$$
(16)
The condition
\(t_{p,d} \rightarrow 0\) implies that
$$\begin{aligned} \int _1^2 \exp \left\{ -\frac{d-1}{2} t_{p,d} (\alpha - t_{p,d})^2 \right\} d\alpha = \frac{e^{-(1-t_{p,d})^2 t^2_{p,d}(d-1)/2}}{t^2_{p,d} d \sqrt{2\pi }} (1 + o(1)), \end{aligned}$$
and, therefore,
$$\begin{aligned}{}&\exp \left\{ \frac{d-1}{2} t^2_{p,d} (1+O(t^2_{p,d}))\right\} \int _1^2 \exp \left\{ -\frac{d-1}{2} t^2_{p,d} (\alpha - t_{p,d})^2 \right\} d\alpha \nonumber \\&= \frac{e^{d(t^3_{p,d} + O(t^4_{p,d}))}}{t^2_{p,d} d \sqrt{2\pi }} (1+o(1)) = \frac{e^{dt^3_{p,d}}}{t^2_{p,d} d \sqrt{2\pi }} (1+o(1)). \end{aligned}$$
(17)
The last equality holds because under the condition
\(d \gg \log ^2 n\), it is true that
\(dt^4_{p,d} \rightarrow 0\) as
\(n\rightarrow \infty\), and
\(e^{dt^4_{p,d}} \rightarrow 1\). Putting (
17) in (
15) gives the final bounds on
\(T_1\):
$$\begin{aligned} 6C_l p^3 t^2_{p,d} e^{t^3_{p,d} d} (1+o(1)) \le T_1 \le 6C_u p^3 e^{t^3_{p,d} d} (1+o(1)). \end{aligned}$$
This concludes the first part the proof.