It’s a Small World for Random Surfers

Abstract

We prove logarithmic upper bounds for the diameters of the random-surfer Webgraph model and the PageRank-based selection Webgraph model, confirming the small world phenomenon holds for them. In the special case when the generated graph is a tree, we provide close lower and upper bounds for the diameters of both models.

Appendix: Omitted Proofs

Proof of Lemma 6

We first prove the upper bound. If $x>1$ then $\exp (-\Upsilon (x) m)=1$, so we may assume that $0 < x \le 1$. We use Chernoff’s technique. Let $\theta = 1 - 1/x$. Then we have

$$\begin{aligned}&\mathbb {P}\left[ {E_1 + E_2 + \dots + E_m \le xm}\right] \\&= \mathbb {P}\left[ {\exp (\theta E_1 + \dots + \theta E_m) \ge \exp (\theta xm)}\right] \\&\le \mathbb {E}\left[ {\exp (\theta E_1 + \dots + \theta E_m)}\right] / \exp (\theta x m) \\&= \mathbb {E}\left[ {\exp (\theta E_1)}\right] \mathbb {E}\left[ {\exp (\theta E_2)}\right] \cdots \mathbb {E}\left[ {\exp (\theta E_m)}\right] / \exp (\theta x m) \\&= (1-\theta )^{-m} \exp (-\theta x m) = \exp (-\Upsilon (x) m) . \end{aligned}$$

We now prove the lower bound. If $x>1$, then the result follows from Markov’s inequality, so we may assume that $0<x\le 1$. Let $\Lambda ^*(x) = \sup \{ \lambda x - \log (\mathbb {E}\left[ {e^{\lambda E_1}}\right] ) : \lambda \le 0\}$. Since $\mathbb {E}\left[ {e^{\lambda E_1}}\right] =1/(1-\lambda )$ for all $\lambda <1$, the supremum here occurs at $\lambda = 1 - 1/x$, which implies $\Lambda ^*(x)=\Upsilon (x)$. Then by Cramér’s Theorem (see, e.g., [10, Theorem 2.2.3, p. 27]) we have

$$\begin{aligned} \mathbb {P}\left[ {E_1 + E_2 + \dots + E_m \le xm}\right] = \exp (-\Lambda ^*(x) m + o(m)) = \exp (-\Upsilon (x) m + o(m)), \end{aligned}$$

as required. $\square$

Proof of Lemma 7

We use Chernoff’s technique. Let $\theta$ satisfy

$$\begin{aligned} e^{\theta } = \frac{\kappa -1}{\kappa (1-p)} . \end{aligned}$$

We have

$$\begin{aligned} \mathbb {E}\left[ {\exp (\theta Z_1)}\right] =\sum _{k=1}^{\infty }p(1-p)^{k-1}e^{\theta k} = \frac{pe^{\theta }}{1-e^{\theta }(1-p)} . \end{aligned}$$

Thus we have

$$\begin{aligned}&\mathbb {P}\left[ {Z_1 + Z_2 + \dots + Z_m \ge \kappa m}\right] \\&= \mathbb {P}\left[ {\exp (\theta Z_1 + \dots + \theta Z_m) \ge \exp (\theta \kappa m)}\right] \\&\le \mathbb {E}\left[ {\exp (\theta Z_1 + \dots + \theta Z_m)}\right] / \exp (\theta \kappa m)\\&= \mathbb {E}\left[ {\exp (\theta Z_1)}\right] \mathbb {E}\left[ {\exp (\theta Z_2)}\right] \cdots \mathbb {E}\left[ {\exp (\theta Z_m)}\right] / \exp (\theta \kappa m) \\&= \left( \frac{pe^{\theta -\theta \kappa }}{1-e^{\theta }(1-p)}\right) ^m = f(2-\kappa )^m . \end{aligned}$$

$\square$

Proof of Lemma 8

We consider two cases.

Case 1: $c\ge 1$. In this case we prove

$$\begin{aligned} -c \Upsilon (1/c) + c \log f ( 2 - \eta / c) < \eta (1-p)\log (1-p^3)-1 . \end{aligned}$$

Notice that we have $1-c\Upsilon (1/c) = c + c \log (1/c)$, so, as $\eta (1-p)\le \eta -c$, the conclusion is implied by

$$\begin{aligned} c + c \log (1/c) + c \log f (2-\eta /c) < (\eta -c) \log (1-p^3) . \end{aligned}$$

Dividing by c and since $c>0$, this statement is equivalent to

$$\begin{aligned} 1 + \log (1/c) + \log f (2-r) < (r-1) \log (1-p^3), \end{aligned}$$

where $r:=\eta /c$. Exponentiating and plugging the definition of f, this statement is equivalent to

$$\begin{aligned} ep (1-p)^{r-1} r^2 (r/(r-1))^{r-1} < \eta (1-p^3)^{r-1}. \end{aligned}$$

Since $(r/(r-1))^{r-1} < e$, and $1-p < (1-p^3)e^{-p}$, for this inequality to hold it suffices to have

$$\begin{aligned} e^{2+p} r^2 p \exp (-pr) \le 4e^p/p \qquad \forall r\in [p^{-1},\infty ) , \end{aligned}$$

which follows from the fact that $x^2 e^{-x} \le 4e^{-2}$ for all $x\ge 1$.

Case 2: $c< 1$. In this case we prove

$$\begin{aligned} -c \Upsilon (1/c) + c \log f ( 2 - \eta / c) < -0.15p\eta -1 . \end{aligned}$$

Since $\Upsilon (1/c)=0$, this is equivalent to

$$\begin{aligned} 1+0.15p \eta + c \log f ( 2 - \eta / c) < 0 . \end{aligned}$$

(36)

Note that

$$\begin{aligned} \left( \frac{\eta /c}{\eta /c - 1}\right) ^{\eta /c - 1} < e, \end{aligned}$$

so we have

$$\begin{aligned} c \log f (2- \eta / c)&= \log \left( (\eta /c)^{\eta }p^c(1-p)^{\eta -c}(\eta /c-1)^{c-\eta } \right) \\&< \log \left( (e \eta p / c)^c (1-p)^{\eta - c} \right) \le c \log (e \eta p / c) + c p - p \eta , \end{aligned}$$

where we have used $\log (1-p)\le -p$ in the last inequality. Hence to prove (36), since $c>0$, it suffices to show that

$$\begin{aligned} \frac{1}{c} + 1 + \log (\eta p / c) + p < 0.85 p \eta / c . \end{aligned}$$

(37)

Since $p\eta \ge 4e^p\ge 4>4c$, we have

$$\begin{aligned}&\qquad \qquad \qquad 1 < 0.25 p \eta / c,\\&\frac{1}{c} +p < \frac{1+p}{c} < \frac{e^p}{c} \le 0.25 p\eta /c ,\\&\qquad \qquad \log (\eta p / c) < 0.35 p \eta / c , \end{aligned}$$

which imply (37). $\square$

Proof of Lemma 14

(a) The conclusion is clear for $a\in \{0,1\}$, so we may assume that $a\in (0,1)$. Since $\Phi (a,a)<0$ and $\Phi (a,1)>0$, there exists at least one $s\in (a,1)$ with $\Phi (a,s)=0$. We now show that there is a unique such s. Fixing a, since $\Phi$ is differentiable with respect to s, it is enough to show that

$$\begin{aligned} \mathrm {if\ } \Phi (a,s)=0\mathrm {,\ then\ } \frac{\partial \Phi }{\partial s} > 0 \end{aligned}$$

(38)

Let $\sigma = p(1-p)$. We have

$$\begin{aligned} \frac{\partial \Phi }{\partial s} = \sigma (2-s)^2 + a - 2 \sigma (2-s)(s-a) . \end{aligned}$$

At a point (a, s) with $\Phi (a,s)=0$, we have

$$\begin{aligned} \sigma (2-s)^2 = \frac{a(1-s)}{s-a}\mathrm {,\ and\ } \sigma (2-s)(s-a) = \frac{a(1-s)}{2-s} , \end{aligned}$$

so at this point,

$$\begin{aligned} \frac{\partial \Phi }{\partial s} = a (1-s) \left( \frac{1}{s-a} + \frac{1}{1-s} - \frac{2}{2-s} \right) , \end{aligned}$$

which is strictly positive because

$$\begin{aligned} \min \left\{ \frac{1}{s-a} , \frac{1}{1-s} \right\} > 1 > \frac{1}{2-s} , \end{aligned}$$

and this proves (38).

(b) Plugging the definition of f from (4) and using $\Phi (a,s)=0$ gives this equation.

(c) We first show that $\phi$ is differentiable and increasing on (0, 1). Let $a\in (0,1)$ and let $s=\phi (a)$. We have

$$\begin{aligned} \frac{\partial \Phi }{\partial a} = s-1-p(1-p)(2-s)^2 < 0, \end{aligned}$$

and $\partial \Phi / \partial s$ is positive as proved in part (a). Hence by the implicit function theorem ${\mathrm {d}s}/{\mathrm {d}a}$ exists and is positive, so $\phi$ is differentiable and increasing on (0, 1). Since $\phi (0)=0$ and $\phi (1)=1$, $\phi$ is increasing on [0, 1].

(d) Let $s\in [0,1]$. Then $\Phi (0,s) \Phi (s,s) \le 0$ and so there exists at least one $a_0\in [0,s]$ with $\Phi (a_0,s)=0$. The function $\Phi (a,s)$ is linear in a and the coefficient of a is non-zero, hence this root $a_0$ is unique. The function $\phi ^{-1}$ is increasing since $\phi$ is increasing. The last two statements follow from similar statements proved for $\phi$ in (a). $\square$

Proof of Lemma 15

We have

$$\begin{aligned} \mathbb {P}\left[ {\hat{Y}_1+\dots +\hat{Y}_m \ge am}\right] = \sum _{k=\left\lceil am \right\rceil }^m \left( {\begin{array}{c}m\\ k\end{array}}\right) 2^{-m} \times \mathbb {P}\left[ {Y_1+\dots +Y_k\ge am}\right] , \end{aligned}$$

where k denotes the number of $\hat{Y}_i$’s whose value was determined to be equal to $Y_i$.

If $p> 1/2$ and $0 < a < 1 - \frac{1}{2p}$, then letting $k = \lceil m/2 \rceil$ gives

$$\begin{aligned} \left( {\begin{array}{c}m\\ k\end{array}}\right) 2^{-m} = \Omega \left( \frac{1}{\sqrt{m}} \right) \end{aligned}$$

by Stirling’s approximation, and

$$\begin{aligned} \mathbb {P}\left[ {Y_1+\dots +Y_k\ge am}\right] \ge (1-o(1))^k \end{aligned}$$

by Lemma 13(c). This gives

$$\begin{aligned} \mathbb {P}\left[ {\hat{Y}_1+\dots +\hat{Y}_m \ge am}\right] \ge (1-o(1))^m, \end{aligned}$$

as required.

Otherwise, let $s=\phi (a)$. Then letting $k = \lceil am/s \rceil$ gives

$$\begin{aligned} \left( {\begin{array}{c}m\\ k\end{array}}\right) 2^{-m} = \Omega \left( \left[ \frac{s(s-a)^{a/s}}{2(s-a)a^{a/s}} \right] ^m \Big / \sqrt{m} \right) \end{aligned}$$

(39)

by Stirling’s approximation, and

$$\begin{aligned} \mathbb {P}\left[ {Y_1+\dots +Y_k\ge am}\right] \ge (f(s)-o(1)) ^ k \end{aligned}$$

by Lemma 13(b) and since f is continuous. This implies, in particular, that $f(s)\in [0,1]$. Thus we have

$$\begin{aligned} \mathbb {P}\left[ {Y_1+\dots +Y_k\ge am}\right]&\ge (f(s)-o(1)) ^ {(am/s)+1} = \left( f(s)^{a/s}-o(1) \right) ^ m \\&= \left( \frac{a^{a/s}}{(s-a)^{a/s}}\left( \frac{1-s}{(1-p)(2-s)} \right) ^a -o(1) \right) ^m, \end{aligned}$$

where we have used Lemma 14(b) for the last equality. Using (39) we get

$$\begin{aligned} \mathbb {P}\left[ {\hat{Y}_1+\dots +\hat{Y}_m \ge am}\right]&\ge \left( {\begin{array}{c}m\\ k\end{array}}\right) 2^{-m} \times \mathbb {P}\left[ {Y_1+\dots +Y_k\ge am}\right] \\&\ge \left( \frac{s(1-s)^a}{2(s-a)(1-p)^a(2-s)^a} -o(1) \right) ^ m\\&= \left( 2g_L(a) + o(1) \right) ^{-m} \end{aligned}$$

by the definition of $g_L$ in (19). $\square$

Proof of Lemma 16

(a) The function $r(p)=\log \left( \frac{1-p}{p} \right) - \frac{1-p}{1-2p}$ approaches $+\infty$ when $p\rightarrow 0^+$ and approaches $-\infty$ when $p\rightarrow { \frac{1}{2} }^{-}$. Moreover,

$$\begin{aligned} r'(p)=\frac{-1}{p(1-p)}-\frac{1}{(1-2p)^2} < 0 \end{aligned}$$

for $p\in (0,1/2)$. Hence r(p) has a unique root $p_0$, and $r(p) \ge 0$ if and only if $p\le p_0$.

(b) The function

$$\begin{aligned} \mu (s) = \log (1-p) + \log (2-s) - \log (1-s) - \frac{1}{s} \end{aligned}$$

approaches $-\infty$ as $s \rightarrow 0^+$, and approaches $+\infty$ as $s \rightarrow 1^{-}$, and its derivative is positive in (0, 1), hence it has a unique root $s_0$ in (0, 1). Also we have $\mu (2-p^{-1}) = p/(1-2p)$, which means that if $p>1/2$ then $s_0>2-p^{-1}$. Moreover, if $p_0 < p \le 1/2$, then by part (a),

$$\begin{aligned} \mu \left( \frac{1-2p}{1-p}\right) = \log \left( \frac{1-p}{p}\right) - \frac{1-p}{1-2p}=r(p)<0, \end{aligned}$$

which means $s_0 > \frac{1-2p}{1-p}$. $\square$

Proof of Lemma 19

The conclusion is obvious if $p\ge \frac{1}{2}$ and $a \le 2 - \frac{1}{p}$, or if $a=0$, since in these cases $h(a)=1$. Also, $\mathbb {P}\left[ {X_1+\dots +X_m > m}\right] =0$ so the conclusion is true if $a=1$, so we may assume that $\max \{ 0, 2 - \frac{1}{p} \} < a < 1$.

Observe that if $X_1+\dots +X_m > a m$, there is a subsequence of the form $Y_{m-k+1},\dots ,Y_m$ whose sum is at least am, and this subsequence contains at least am elements since $Y_i \le 1$ for all i. Hence we have

$$\begin{aligned} \mathbb {P}\left[ {X_1+\dots +X_m > am}\right] \le m \max \{ \mathbb {P}\left[ {Y_1+\dots +Y_k \ge am}\right] : k \in [am,m] \cap \mathbb {Z}\} \end{aligned}$$

as the $Y_i$’s are i.i.d.

For any integer $k \in [am, m]$, by Lemma 13(a) we have

$$\begin{aligned} \mathbb {P}\left[ {Y_1+\dots +Y_k \ge am}\right] \le C k (f(am/k))^k \end{aligned}$$

for an absolute constant C, since $am/k \ge a > 2 - \frac{1}{p}$. Let $r = k/m \in [a,1]$. So we find that

$$\begin{aligned} \mathbb {P}\left[ {X_1+\dots +X_m > am}\right] \le C m^2 \big (\sup \{ f(a/r)^r : r\in [a,1]\} \big )^m . \end{aligned}$$

Let us define

$$\begin{aligned} \xi (r) = f(a/r)^r = (2r-a)^{2r-a}p^r(1-p)^{r-a}(r-a)^{a-r}r^{-r} . \end{aligned}$$

So to complete the proof we just need to show that

$$\begin{aligned} \sup \{ \xi (r) : r\in [a,1] \} \le h(a) \quad \forall a \in \left( \max \left\{ 0,2 - \frac{1}{p}\right\} ,1\right) . \end{aligned}$$

(40)

The function $\xi (r)$ is positive and differentiable for each $a\in (0,1)$, hence the supremum here occurs either at a boundary point or at a point with zero derivative. The derivative of $\log (\xi (r))$ equals

$$\begin{aligned} \log \left( \frac{p(1-p)(2r-a)^2}{r(r-a)} \right) . \end{aligned}$$

Thus $\xi '(r)$ has the same sign as $\overline{\xi }(r) = p(1-p)(2r-a)^2 - r(r-a)$ in $r\in [a,1]$. Notice that $\overline{\xi }(r)$ has two roots

$$\begin{aligned} r_1 = \frac{ap}{2p-1} \mathrm {,\ and\ } r_2 = \frac{a(1-p)}{1-2p} . \end{aligned}$$

We may consider several cases.

Case 0: $p=1/2$. The function $\overline{\xi }$ is positive, so $\xi$ is increasing in [a, 1], hence the supremum in (40) happens at $r=1$ and its value is f(a).

Case 1: $p>1/2$. Since $a > 2 - \frac{1}{p}$, we find that $r_1>1$ and $r_2<0$. Moreover, $\overline{\xi }(a)\ge 0$. Thus $\overline{\xi }$ is non-negative in [a, 1], which implies $\xi$ is increasing in [a, 1]. Thus the supremum in (40) happens at $r=1$ and its value is f(a).

Case 2: $p<1/2$ and $a \le \frac{1-2p}{1-p}$. In this case $r_1<0$ and $a \le r_2 \le 1$. Since $\overline{\xi }(a) \ge 0$ and $\overline{\xi }(r_1)=\overline{\xi }(r_2)=0$ and $\overline{\xi }$ is quadratic, the function $\overline{\xi }$ goes from positive to negative at $r_2$. Therefore, the function $\xi$ attains its supremum at $r_2$ and the supremum value in (40) equals

$$\begin{aligned} \xi (r_2) = \left( \frac{p}{1-p}\right) ^a . \end{aligned}$$

Case 3: $p<1/2$ and $a > \frac{1-2p}{1-p}$. We find that $r_1<0$ and $r_2>1$, and $\overline{\xi }(a) \ge 0$, so $\overline{\xi }$ is non-negative in [a, 1], hence $\xi$ is increasing in [a, 1]. Thus the supremum in (40) happens at $r=1$ and its value is f(a). This completes the proof of (40) and the lemma. $\square$

Proof of Lemma 20

(a) First, the case $a=0$ is obvious since $g_U(0) = 1/2$, and the case $a=1$ is easy since $\mathbb {P}\left[ {\hat{X}_1+\dots +\hat{X}_m > m}\right] = 0$. So we may assume that $a\in (0,1)$.

Letting k of the $\hat{X}_i$’s being equal to $X_i$ and the rest equal to zero, we get

$$\begin{aligned}&\mathbb {P}\left[ {\hat{X}_1+\dots +\hat{X}_m > am}\right] \\&= \sum _{k=am}^m \left( {\begin{array}{c}m\\ k\end{array}}\right) 2^{-m} \mathbb {P}\left[ {X_1+\dots +X_k > am}\right] \\&\le m \sup \left\{ \left( {\begin{array}{c}m\\ rm\end{array}}\right) 2^{-m} \mathbb {P}\left[ {X_1+\dots +X_{rm} > am}\right] :r\in [a,1] \right\} . \end{aligned}$$

For a given $r \in [a,1]$, Lemma 19 gives

$$\begin{aligned} \mathbb {P}\left[ {X_1+\dots +X_{rm}> am}\right] \le C(rm)^2 h(a/r)^{rm} \le C m^2 h(a/r)^{rm} . \end{aligned}$$

Moreover, by Stirling’s approximation

$$\begin{aligned} \left( {\begin{array}{c}m\\ rm\end{array}}\right) = O\left( \frac{1}{r^{rm} (1-r)^{(1-r)m}} \right) . \end{aligned}$$

So, we find that

$$\begin{aligned} \mathbb {P}\left[ {\hat{X}_1+\dots +\hat{X}_m > am}\right] \le C' m^3 \left[ \sup \left\{ \frac{h(a/r)^r}{2 r^{r} (1-r)^{1-r}} :r\in [a,1] \right\} \right] ^m . \end{aligned}$$

Thus to complete the proof of part (a) we just need to show

$$\begin{aligned} g_U(a) = \inf \left\{ \frac{\zeta -a}{\zeta } \left( \frac{a}{(\zeta -a)h(\zeta )}\right) ^{a/\zeta } : \zeta \in [a,1] \right\} , \end{aligned}$$

(41)

where we have used the change of variable $\zeta = a/r$. For analysing this infimum we define the two variable function

$$\begin{aligned} \psi (a,\zeta ) = \frac{\zeta -a}{\zeta } \left( \frac{a}{(\zeta -a)h(\zeta )}\right) ^{a/\zeta } \end{aligned}$$

with domain $\{(a,\zeta ) : 0 < a < 1, a \le \zeta \le 1 \}$, and consider two cases depending on the value of p.

Case 1: $p \ge 1/2$. By the definition of h in (22) we have

$$\begin{aligned} \psi (a,\zeta ) = {\left\{ \begin{array}{ll} \frac{\zeta -a}{\zeta } \left( \frac{a}{\zeta -a}\right) ^{a/\zeta } &{} \mathrm {\ if\ }a \le \zeta \le 2 - p^{-1} \\ \frac{\zeta -a}{\zeta } \left( \frac{a}{(\zeta -a)f(\zeta )}\right) ^{a/\zeta } &{} \mathrm {\ otherwise}, \end{array}\right. } \end{aligned}$$

where f is defined in (4). Since $f(2-p^{-1}) = 1$, $\psi$ is continuous here. Let us define $\psi _1(\zeta ) = \frac{\zeta -a}{\zeta } \left( \frac{a}{\zeta -a}\right) ^{a/\zeta }$ and $\psi _2(\zeta )= \frac{\zeta -a}{\zeta } \left( \frac{a}{(\zeta -a)f(\zeta )}\right) ^{a/\zeta }$.

The derivative of $\log \psi _1(\zeta )$ is

$$\begin{aligned} a \log \left( \frac{\zeta -a}{a} \right) / \zeta ^2 , \end{aligned}$$

which is negative for $\zeta <2a$ and positive for $\zeta >2a$. This implies $\psi _1(\zeta )$ is decreasing when $\zeta \le 2a$ and increasing when $\zeta \ge 2a$. So $\psi _1$ achieves its minimum at $\zeta = 2a$, and its minimum value is 1 / 2.

The derivative of $\log \psi _2(\zeta )$ is

$$\begin{aligned} \frac{a}{\zeta ^2} \left[ \log \left( p(1-p)(2-\zeta )^2 (\zeta -a) \right) - \log \left( a(1-\zeta ) \right) \right] . \end{aligned}$$

Comparing with (17) we find that this derivative has the same sign as $\Phi (a,\zeta )$. So by Lemma 14(a) it vanishes at a unique point $\zeta = \phi (a)$. Also at $\zeta =\phi (a)$ we have $\partial \Phi / \partial \zeta > 0$ (see (38)), which implies $\Phi (a,\zeta )$ is non-positive when $\zeta \le \phi (a)$ and non-negative when $\zeta \ge \phi (a)$. Thus $\psi _2$ achieves its minimum at $\phi (a)$, and its minimum value is

$$\begin{aligned} \psi _2(\phi (a))= & {} \frac{\phi (a)-a}{\phi (a)} \left( \frac{a}{(\phi (a)-a)f(\phi (a))}\right) ^{a/\phi (a)}\\= & {} \frac{\phi (a)-a}{\phi (a)} \left( \frac{(1-p) (2-\phi (a))}{1-\phi (a)} \right) ^a \end{aligned}$$

by Lemma 14(b).

We conclude that:

(i) If $2a \le 2 - 1/p$, then the infimum of $\psi$ occurs at $\zeta =2a$ and its value is $\psi (a,2a) = \psi _1(2a)=1/2$. The reason is that on $[a,2-1/p]$, $\psi = \psi _1$ achieves its minimum at 2a, and on $[2-1/p,1]$, $\psi = \psi _2$ is increasing since $\Phi (a,2-1/p) \ge 0$.

(ii) If $a \le 2 - 1/p$ and $2a > 2 - 1/p$, then the infimum occurs at $\zeta =\phi (a)$ and its value is $\frac{\phi (a)-a}{\phi (a)} \left( (1-p) (2-\phi (a)) / (1-\phi (a)) \right) ^a$. The reason is that on $[a,2-1/p]$, $\psi = \psi _1$ is decreasing, and on $[2-1/p,1]$, $\psi = \psi _2$ achieves its minimum at $\phi (a)$ since $\Phi (a,2-1/p)\le 0$ and $\Phi (a,1)\ge 0$.

(iii) If $a > 2 - 1/p$, then the infimum occurs at $\zeta =\phi (a)$ and its value is equal to $\frac{\phi (a)-a}{\phi (a)} \left( (1-p) (2-\phi (a)) / (1-\phi (a)) \right) ^a$. The reason is that on [a, 1], $\psi =\psi _2$ achieves its minimum at $\phi (a)$ since $\Phi (a,a)\le 0$ and $\Phi (a,1)\ge 0$.

Case 2: $p < 1/2$. By the definition of h in (22) we have

$$\begin{aligned} \psi (a,\zeta ) = {\left\{ \begin{array}{ll} \left( \frac{1-p}{p}\right) ^{a} \frac{\zeta -a}{\zeta } \left( \frac{a}{\zeta -a}\right) ^{a/\zeta } &{} \mathrm {\ if\ } a \le \zeta \le \frac{1-2p}{1-p} \\ \frac{\zeta -a}{\zeta } \left( \frac{a}{(\zeta -a)f(\zeta )}\right) ^{a/\zeta } &{} \mathrm {\ otherwise.} \end{array}\right. } \end{aligned}$$

The function $\psi$ is continuous here since

$$\begin{aligned} f\left( \frac{1-2p}{1-p}\right) =\left( \frac{p}{1-p}\right) ^{\frac{1-2p}{1-p}} . \end{aligned}$$

Let us define $\psi _3(\zeta ) = \left( \frac{1-p}{p}\right) ^{a} \frac{\zeta -a}{\zeta } \left( \frac{a}{\zeta -a}\right) ^{a/\zeta }$. Since $\psi _3(\zeta ) = \left( \frac{1-p}{p}\right) ^{a} \psi _1(\zeta )$, the function $\psi _3(\zeta )$ is decreasing when $\zeta \le 2a$ and increasing when $\zeta \ge 2a$. So $\psi _3$ achieves its minimum at $\zeta = 2a$ and its minimum value is $\left( \frac{1-p}{p}\right) ^{a}/2$. We conclude that

(iv) If $a \le 1 - p/(1-p)$ and $2a \le 1 - p/(1-p)$, then the infimum in (41) occurs at $\zeta =2a$ and at this point we have $\psi (a,\zeta ) = \left( \frac{1-p}{p}\right) ^a/2$. The reason is that on $[a, 1 - p/(1-p)]$, $\psi = \psi _3$ achieves its minimum at 2a, and on $[1-p/(1-p),1]$, $\psi = \psi _2$ is increasing since $\Phi (a,1 - p/(1-p))\ge 0$.

(v) If $a \le 1 - p/(1-p)$ and $2a > 1 - p/(1-p)$, then the infimum in (41) occurs at $\zeta =\phi (a)$ and its value is equal to $\frac{\phi (a)-a}{\phi (a)} \left( (1-p) (2-\phi (a)) / (1-\phi (a)) \right) ^a$. The reason is that on $[a, 1 - p/(1-p)]$, $\psi = \psi _3$ is decreasing, and on $[1-p/(1-p),1]$, $\psi = \psi _2$ achieves its minimum at $\phi (a)$ since $\Phi (a,1 - p/(1-p))\le 0$ and $\Phi (a,1)\ge 0$.

(vi) If $a > 1 - p/(1-p)$, then the infimum in (41) occurs at $\zeta =\phi (a)$ and its value is equal to $\frac{\phi (a)-a}{\phi (a)} \left( (1-p) (2-\phi (a)) / (1-\phi (a)) \right) ^a$. The reason is that on [a, 1], $\psi = \psi _2$ achieves its minimum at $\phi (a)$ since $\Phi (a,a)\le 0$ and $\Phi (a,1) \ge 0$.

In all cases we proved that $g_U(a)$ actually gives the value of the infimum in (41), and this concludes the proof of (41) and of part (a).

(b) Consider the definition of $g_U$ in (23). The formulae in (25) for the cases ‘$p\ge 1/2$ and $0 < a \le 1 - 1/2p$’ and ‘$p < 1/2$ and $0 < a \le \frac{1-2p}{2-2p}\,$’ are clearly true, so we assume that a is in the ‘otherwise’ case. We use the equality (41). Note that as proved in part (a), the infimum in (41) occurs at the point $\zeta =\phi (a)$ that has $\left. \frac{\partial \psi }{\partial \zeta }\right| _{(a,\phi (a))} = 0$. This implies for every $a_0$,

$$\begin{aligned} \frac{\mathrm {d} g_U}{\mathrm {d} a}(a_0)&= \frac{\partial \psi }{\partial a} {(a_0,\phi (a_0))} + \frac{\partial \psi }{\partial \zeta } {(a_0,\phi (a_0))} \times \frac{\mathrm {d} \phi }{\mathrm {d} a}({a_0}) \\&=\frac{\partial \psi }{\partial a} {(a_0,\phi (a_0))} =\left. \frac{\partial }{\partial a} \left[ \frac{\zeta -a}{\zeta } \left( \frac{a}{(\zeta -a)f(\zeta )}\right) ^{a/\zeta }\right] \right| _{(a_0,\phi (a_0))} , \end{aligned}$$

and (25) follows from computing this partial derivative and putting $\zeta = \phi (a_0)$.

We next prove the continuity of $g_U$ and its derivative. Note that by Lemma 14(a), if $a\in (0,1)$ then $\phi (a)\in (0,1)$. First, $g_U$ is continuous at $a=1$ since

$$\begin{aligned} \lim _{a\rightarrow 1} \frac{\phi (a)-a}{\phi (a)} \left( \frac{(1-p) (2-\phi (a))}{1-\phi (a)} \right) ^a&= \lim _{a\rightarrow 1} \frac{\phi (a)-a}{\phi (a)} \left( \frac{a}{p(2-\phi (a))(\phi (a)-a)} \right) ^a \\&= \lim _{a\rightarrow 1} \frac{(\phi (a)-a)^{1-a}}{\phi (a)} \left( \frac{a}{p(2-\phi (a))} \right) ^a = \frac{1}{p} . \end{aligned}$$

For $p\ge 1/2$, the only discontinuity for $g_U$ can possibly occur at $b = 1 - 1/2p$. However at this point we have $\phi (b) = 2b = 2 - p^{-1}$ so that $(1-p)(2-\phi (b)) = 1-\phi (b)$. Hence the left and right limits of $g_U$ equal 1 / 2, and the left and right limits of $g'_U$ equal 0. Therefore, both $g_U$ and $g'_U$ are continuous at b.

For $p < 1/2$, the only discontinuity for $g_U$ can possibly occur at $c = (1-2p)/(2-2p)$. However at this point $\phi (c) = 2c = (1-2p)/(1-p)$ so that $\frac{(1-p)(2-\phi (c))}{1-\phi (c)} = \frac{1-p}{p}$. Hence the left and right limits of $g_U$ equal $\left( p^{-1}-1\right) ^{c}/2$, and the left and right limits of $g'_U$ equal $\displaystyle \left( \log \left( p^{-1}-1\right) \right) \left( p^{-1}-1\right) ^{c}/2$. Therefore, both $g_U$ and $g'_U$ are continuous at c.

(c) Note that $g_U$ is positive everywhere, so $\log (g_U)$ is (strictly) increasing if and only if $g_U$ is (strictly) increasing. By the formulae for $g'_U$ in part (b), it is easy to see that $g'_U$ is always non-negative, and is positive when $g_U(a) > 1/2$. To show $\log (g_U)$ is convex, we need to show its derivative, i.e., $g'_U/g_U$ is increasing. This also follows from part (b), noting that $\phi$ is increasing by Lemma 14(c). $\square$