False Negative probabilities in Tardos codes

Abstract

Forensic watermarking is the application of digital watermarks for the purpose of tracing unauthorized redistribution of content. One of the most powerful types of attack on watermarks is the collusion attack, in which multiple users compare their differently watermarked versions of the same content. Collusion-resistant codes have been developed against these attacks. One of the most famous such codes is the Tardos code. It has the asymptotically optimal property that it can resist $c$ attackers with a code of length proportional to $c^2$. Determining error rates for the Tardos code and its various extensions and generalizations turns out to be a nontrivial problem. In recent work we developed an approach called the convolution and series expansion (CSE) method to accurately compute false positive accusation probabilities. In this paper we extend the CSE method in order to make it possible to compute a bound on the False Negative accusation probabilities.

Appendices

Appendix

Proof of Lemma 1

The proof is similar to the steps taken in Appendix D of [25]. First we split the $q$-dimensional integration $\int \mathrm{d}^q {{\varvec{p}}}\; F({{\varvec{p}}}) r({{\varvec{p}}})$ as follows,

$$\begin{aligned} \mathbb{E }_{{\varvec{p}}}[r({{\varvec{p}}})]= \frac{1}{B(\kappa \mathbf{1}_q)} \int \limits _0^1\! \mathrm{d}p_y\; p_y^{-1+\kappa } \int \limits _0^{1-p_y}\! \mathrm{d}^{q-1}{{\varvec{p}}}_{\setminus y}\delta \left( 1-p_y-{\textstyle \sum \limits _{\beta \in \mathcal{Q }\setminus \{y\}}}\, p_\beta \right) {{\varvec{p}}}_{\setminus y}^{-1+\kappa } r({{\varvec{p}}}).\nonumber \\ \end{aligned}$$

(60)

Then we write ${{\varvec{p}}}_{\setminus y}=(1-p_y){{\varvec{t}}}$. We get $\delta (1-p_y-\sum _{\beta \in \mathcal{Q }\setminus \{y\}}p_\beta )= (1-p_y)^{-1}\delta (1-\sum _{\beta \in \mathcal{Q }\setminus \{y\}}t_\beta )$. Furthermore, $\mathrm{d}^{q-1}{{\varvec{p}}}_{\setminus y}=(1-p_y)^{q-1}\mathrm{d}^{q-1}{{\varvec{t}}}$ and ${{\varvec{p}}}_{\setminus y}^{-1+\kappa }=(1-p_y)^{(q-1)(-1+\kappa )}{{\varvec{t}}}^{-1+\kappa }$. Combined with the fact that $B(\kappa \mathbf{1}_q)=B(\kappa ,\kappa [q-1])B(\kappa \mathbf{1}_{q-1})$, these steps yield the end result. $\square$

Proof of Theorem 1

The guilty user’s symbol is denoted as $X$. The one-segment score is either $g_0(p_y)$ (when $X \ne y$) or $g_1(p_y)$ (when $X=y$). Since no other values are possible, the probability distribution at given ${{\varvec{p}}}$ will consist of delta-function peaks. Each peak is multiplied by the probability that the corresponding event occurs

$$\begin{aligned} \psi _-(u\vert {{\varvec{p}}})&= {\textstyle \sum \limits _{y \in \mathcal{Q }}}\, \delta \left( u-g_0(p_y)\right) \mathrm{Pr}[u=g_0(p_y)\vert {{\varvec{p}}}] \end{aligned}$$

(61)

$$\begin{aligned} \psi _+(u\vert {{\varvec{p}}})&= {\textstyle \sum \limits _{y \in \mathcal{Q }}}\, \delta \left( u-g_1(p_y)\right) \mathrm{Pr}[u=g_1(p_y)\vert {{\varvec{p}}}]. \end{aligned}$$

(62)

Notice that

$$\begin{aligned} \mathrm{Pr}[u=g_0(p_y)\vert {{\varvec{p}}}] =\mathrm{Pr}[X\ne y \wedge Y=y \vert {{\varvec{p}}}]&\quad ;\quad&\mathrm{Pr}[u\!=\!g_1(p_y)\vert {{\varvec{p}}}] \!=\!\mathrm{Pr}[X = y \wedge Y=y \vert {{\varvec{p}}}] \nonumber \\ \end{aligned}$$

(63)

and that

$$\begin{aligned} \mathrm{Pr}[X\ne y \wedge Y=y \vert {{\varvec{p}}}]+\mathrm{Pr}[X = y \wedge Y=y \vert {{\varvec{p}}}] = \mathrm{Pr}[Y=y \vert {{\varvec{p}}}] = \tau _{y|{{\varvec{p}}}}. \end{aligned}$$

(64)

Next step is to compute $\mathrm{Pr}[u=g_1(p_y)\vert {{\varvec{p}}}]$ in (62). Let be ${{\varvec{e}}}_y$ a $q$-ary vector entirely set to $0$ except for the $y$-th element that is instead equal to 1.

$$\begin{aligned} \mathrm{Pr}[u=g_1(p_y)\vert {{\varvec{p}}}] = \mathrm{Pr}[X_{ji}=y]\mathrm{Pr}[Y=y\vert X_{ji}=y, {{\varvec{p}}}] = p_y \sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{qc}} \left( {\begin{array}{c}c-1\\ {\varvec{\sigma }}- {{\varvec{e}}}_y\end{array}}\right) {{\varvec{p}}}^{{\varvec{\sigma }}-{{\varvec{e}}}_y} \theta _{y \vert {\varvec{\sigma }}}. \nonumber \\ \end{aligned}$$

(65)

The last equation is obtained as follows: $\mathrm{Pr}[X_{ji}=y]=p_y$; $\mathrm{Pr}[Y=y\vert X_{ji}=y, {{\varvec{p}}}]$ is equal to the sum over all the possible ${\varvec{\sigma }}$ vectors that have at least one occurrence of $y$ (expressed with the condition $\sigma _y>0$). Knowing that $X_{ji}=y$, the multinomial factor is needed to count the remaining $c-1$ pirate symbols in ${\varvec{\sigma }}$, subtracting 1 from $\sigma _y$ (using the ${{\varvec{e}}}_y$ vector).

$$\begin{aligned} \mathrm{Pr}[u=g_1(p_y)\vert {{\varvec{p}}}]=\sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{qc}} \frac{\sigma _y}{c}\left( {\begin{array}{c}c\\ {\varvec{\sigma }}\end{array}}\right) {{\varvec{p}}}^{{\varvec{\sigma }}} \theta _{y \vert {\varvec{\sigma }}}. \end{aligned}$$

(66)

In the last equation we used ${{\varvec{p}}}^{\varvec{\sigma }}=p_y {{\varvec{p}}}^{{\varvec{\sigma }}-{{\varvec{e}}}_y}$ and $\left( {\begin{array}{c}c-1\\ {\varvec{\sigma }}-{{\varvec{e}}}_y\end{array}}\right) =\frac{\sigma _y}{c}\left( {\begin{array}{c}c\\ {\varvec{\sigma }}\end{array}}\right)$. Then the condition $\sigma _y>0$ becomes superfluous and (27) trivially follows. Notice that

$$\begin{aligned} p_y \frac{\partial T_{y|{{\varvec{p}}}}}{\partial p_y} = p_y \frac{\partial }{\partial p_y} \sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{qc}} {c\atopwithdelims (){\varvec{\sigma }}}{{\varvec{p}}}^{\varvec{\sigma }}\theta _{y|{\varvec{\sigma }}} = \sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{qc}} {c\atopwithdelims (){\varvec{\sigma }}} \theta _{y|{\varvec{\sigma }}} p_y \frac{\partial {{\varvec{p}}}^{\varvec{\sigma }}}{\partial p_y} = \sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{qc}} {c\atopwithdelims (){\varvec{\sigma }}} \theta _{y|{\varvec{\sigma }}} \sigma _y {{\varvec{p}}}^{\varvec{\sigma }}\nonumber \\ \end{aligned}$$

(67)

proving that (28)=(27) and (26)=(25). Finally, from (64) combined with (63) we have

$$\begin{aligned} \psi _-(u\vert {{\varvec{p}}}) = {\textstyle \sum \limits _{y \in \mathcal{Q }}}\, \delta \left( u-g_0(p_y)\right) \left( \tau _{y|{{\varvec{p}}}} - \mathrm{Pr}[X = y \wedge Y =y \vert {{\varvec{p}}}]\right) . \end{aligned}$$

(68)

This, together with (66), completes the proof. $\square$

Proof of Theorem 2

The full $\psi (u)$, without conditioning, is obtained by taking the expectation over ${{\varvec{p}}}$ of (25)+(27).

$$\begin{aligned} \psi (u)=\mathbb{E }_{{\varvec{p}}}[\psi (u \vert {{\varvec{p}}})]=\varTheta (-u)\mathbb{E }_{{\varvec{p}}}[\psi _-(u \vert {{\varvec{p}}})]+\varTheta (u)\mathbb{E }_{{\varvec{p}}}[\psi _+(u \vert {{\varvec{p}}})]. \end{aligned}$$

(69)

We first prove (29) starting from $\mathbb{E }_{{\varvec{p}}}[\psi _-(u \vert {{\varvec{p}}})]$ with $\psi _-(u \vert {{\varvec{p}}})$ as given in (25).

$$\begin{aligned} \mathbb{E }_{{\varvec{p}}}[\psi _-(u \vert {{\varvec{p}}})]&= \mathbb{E }_{{\varvec{p}}}\left[ \sum \limits _{y \in \mathcal{Q }}\delta \left( u-g_0(p_y)\right) \sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{qc}} \left( {\begin{array}{c}c\\ {\varvec{\sigma }}\end{array}}\right) \left( 1-\frac{\sigma _y}{c}\right) {{\varvec{p}}}^{\varvec{\sigma }}\theta _{y\vert {\varvec{\sigma }}} \right] \end{aligned}$$

(70)

$$\begin{aligned}&= \sum \limits _{y \in \mathcal{Q }}\sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{qc}} \left( {\begin{array}{c}c\\ {\varvec{\sigma }}\end{array}}\right) \left( 1-\frac{\sigma _y}{c}\right) \theta _{y\vert {\varvec{\sigma }}} \mathbb{E }_{{\varvec{p}}}\left[ \delta \left( u-g_0(p_y) \right) {{\varvec{p}}}^{\varvec{\sigma }}\right] . \end{aligned}$$

(71)

From Lemma 1 and ${{\varvec{p}}}_{\setminus y}^{{\varvec{\sigma }}_{\setminus y}}=\left( 1-p_y\right) ^{c-\sigma _y}\prod _{\alpha \in \mathcal{Q }\setminus \{y\}} t_\alpha ^{\sigma _\alpha }$ we have that

$$\begin{aligned} \mathbb{E }_{{\varvec{p}}}\left[ \delta \left( u-g_0(p_y)\right) {{\varvec{p}}}^{\varvec{\sigma }}\right]&= \frac{1}{B(\kappa \mathbf{1}_q)} \int \limits _0^1 \mathrm{d}p_y \; \delta \left( u-g_0(p_y)\right) p_y^{\sigma _y +\kappa -1} (1-p_y)^{c-\sigma _y+\kappa [q-1]-1}\nonumber \\&\quad \quad \int \limits _0^1 \mathrm{d}^{q-1} {{\varvec{t}}}\; \delta (1-\sum \limits _{\beta \in \mathcal{Q }\setminus \{y\}} t_\beta ) \prod _{\alpha \in \mathcal{Q }\setminus \{y\}} t_\alpha ^{\sigma _\alpha + \kappa -1}. \end{aligned}$$

(72)

The second integral in (72) evaluates to $B({\varvec{\sigma }}_{\setminus y}+\kappa \mathbf{1}_{q-1})$, having the structure shown in Def. 1. In order to evaluate the $p_y$-integral we have to rewrite the delta function into the form $\delta \left( p_y-\cdots \right)$. We use the rule

$$\begin{aligned} \delta \left( u-w(p)\right) =\frac{\delta \left( p-w^\mathrm{inv}(u)\right) }{\vert \mathrm{d}w/ \mathrm{d}p \vert } \end{aligned}$$

(73)

for any monotonic function $w(p)$. This gives

$$\begin{aligned} \delta \left( u-g_0(p)\right) =\varTheta (-u) \frac{2 \vert u\vert }{\left( 1+u^2\right) ^2} \delta \left( p-\frac{u^2}{1+u^2}\right) . \end{aligned}$$

(74)

We substitute (74) into (72) and solve the integral

$$\begin{aligned} \mathbb{E }_{{\varvec{p}}}\left[ \delta \left( u\!-\!g_0(p_y)\right) {{\varvec{p}}}^{\varvec{\sigma }}\right]&= 2 \vert u\vert \varTheta (-u)\left( \frac{1}{1\!+\!u^2}\right) ^2 \frac{B({\varvec{\sigma }}_{\setminus y}\!+\!\kappa \mathbf{1}_{q-1})}{B(\kappa \mathbf{1}_q)} \int \limits _0^1 \mathrm{d}p_y\, \delta \left( p_y\!-\!\frac{u^2}{1+u^2}\right) \nonumber \\&\quad \quad p_y^{\sigma _y +\kappa -1} (1-p_y)^{c-\sigma _y+\kappa [q-1]-1} \nonumber \\&= 2 \vert u\vert \varTheta (-u)\left( \frac{1}{1+u^2}\right) ^2 \frac{B({\varvec{\sigma }}_{\setminus y}+\kappa \mathbf{1}_{q-1})}{B(\kappa \mathbf{1}_q)}\nonumber \\&\quad \left( \frac{u^2}{1+u^2}\right) ^{\sigma _y+\kappa -1} \left( \frac{1}{1+u^2}\right) ^{c-\sigma _y+\kappa [q-1]-1} \nonumber \\&= 2\varTheta (-u)\frac{B({\varvec{\sigma }}_{\setminus y}+\kappa \mathbf{1}_{q-1})}{B(\kappa \mathbf{1}_q)} \frac{(u^2)^{\sigma _y+\kappa -1/2}}{(1+u^2)^{c+\kappa q}}. \end{aligned}$$

(75)

Substituting (75) into (71) we have

$$\begin{aligned} \mathbb{E }_{{\varvec{p}}}[\psi _-(u \vert {{\varvec{p}}})]= 2 \sum \limits _{y \in \mathcal{Q }} \sum \limits _{{\varvec{\sigma }}\in \mathcal{S }_{q c}} \left( {\begin{array}{c}c\\ {\varvec{\sigma }}\end{array}}\right) \left( 1-\frac{\sigma _y}{c}\right) \frac{B({\varvec{\sigma }}_{\setminus y}+\kappa \mathbf{1}_{q-1})}{B(\kappa \mathbf{1}_q)}\frac{(u^2)^{\sigma _y+\kappa -1/2}}{(1+u^2)^{c+\kappa q}}\theta _{y\vert {\varvec{\sigma }}}. \qquad \quad \end{aligned}$$

(76)

Now we change the summations as follows: the $\sum _{\varvec{\sigma }}$ can be written as $\sum _b \sum _{{\varvec{x}}}$ with $b=\sigma _y$ and ${{\varvec{x}}}={\varvec{\sigma }}_{\setminus y}$, so $\theta _{y|{\varvec{\sigma }}}=\Psi _b({{\varvec{x}}})$. Then the summand is a function of only $b$ and ${{\varvec{x}}}$, which allows us to write

$$\begin{aligned} \sum \limits _y \sum _{\varvec{\sigma }}\left( {\begin{array}{c}c\\ {\varvec{\sigma }}\end{array}}\right) \rightarrow q \sum \limits _{b=0}^c \sum \limits _{{\varvec{x}}}\left( {\begin{array}{c}c\\ b\end{array}}\right) \left( {\begin{array}{c}c-b\\ {{\varvec{x}}}\end{array}}\right) . \end{aligned}$$

(77)

Now we have

$$\begin{aligned} \mathbb{E }_{{\varvec{p}}}[\psi _-(u \vert {{\varvec{p}}})]=2 q \sum \limits _{b=0}^{c}\sum \limits _{{{\varvec{x}}}}\left( {\begin{array}{c}c\\ b\end{array}}\right) \left( {\begin{array}{c}c-b\\ {{\varvec{x}}}\end{array}}\right) \frac{c-b}{c} \frac{B({{\varvec{x}}}+\kappa \mathbf{1}_{q-1})}{B(\kappa \mathbf{1}_q)}\frac{(u^2)^{b+\kappa -1/2}}{(1+u^2)^{c+\kappa q}}\Psi _b({{\varvec{x}}}) \qquad \end{aligned}$$

(78)

where

$$\begin{aligned} \sum \limits _{{\varvec{x}}}\left( {\begin{array}{c}c\\ b\end{array}}\right) \left( {\begin{array}{c}c-b\\ {{\varvec{x}}}\end{array}}\right) \frac{B({{\varvec{x}}}+\kappa \mathbf{1}_{q-1})}{B(\kappa \mathbf{1}_{q})}\Psi _b({{\varvec{x}}})&= \left( {\begin{array}{c}c\\ b\end{array}}\right) \sum \limits _{{\varvec{x}}}\left( {\begin{array}{c}c-b\\ {{\varvec{x}}}\end{array}}\right) \frac{B({{\varvec{x}}}\!+\!\kappa \mathbf{1}_{q-1})}{B(\kappa \mathbf{1}_{q-1})B(\kappa ,\kappa [q\!-\!1])}\Psi _b({{\varvec{x}}})\nonumber \\ \end{aligned}$$

(79)

$$\begin{aligned}&= \left( {\begin{array}{c}c\\ b\end{array}}\right) \frac{1}{B(\kappa ,\kappa [q-1])}\sum \limits _{{\varvec{x}}}\mathbb{P }_{q-1}({{\varvec{x}}}\vert b)\Psi _b({{\varvec{x}}})\nonumber \\&= \left( {\begin{array}{c}c\\ b\end{array}}\right) \frac{K_b}{B(\kappa ,\kappa [q-1])}. \end{aligned}$$

(80)

In the last line we used Definition 3. Substituting (80) into (78) and removing $0$ and $c$ from the $b$-range, we have (29).

We can use exactly the same steps to obtain (30) from (27). The only significant difference is the delta function which in this case will be

$$\begin{aligned} \delta \left( u-g_1(p)\right) =\varTheta (u) \frac{2 u}{\left( 1+u^2\right) ^2} \delta \left( p-\frac{1}{1+u^2}\right) . \end{aligned}$$

(81)

$\square$

Proof of consistency check 1

Integration of (29) and (30) gives

$$\begin{aligned} \int \limits _{-\infty }^{\infty }\mathrm{d}u \, \psi (u)&= \frac{2q}{B(\kappa ,\kappa [q-1])} \sum \limits _{b=1}^c \left( {\begin{array}{c}c\\ b\end{array}}\right) K_b \left[ \left( 1-\frac{b}{c}\right) \int \limits _{-\infty }^{0}\mathrm{d}u \, \frac{\left( u^2\right) ^{b+\kappa -\frac{1}{2}}}{\left( 1+u^2\right) ^{c+\kappa q}} \nonumber \right. \\&\left. + \frac{b}{c} \int \limits _{0}^{\infty }\mathrm{d}u \, \frac{\left( u^2\right) ^{c-b+\kappa [q-1]-\frac{1}{2}}}{\left( 1+u^2\right) ^{c+\kappa q}}\right] . \end{aligned}$$

(82)

Let be $\lambda {:=} b+\kappa$ and $w {:=} c-b+\kappa [q-1]$. Applying Lemma 2 we have

$$\begin{aligned}&\frac{2q}{B(\kappa ,\kappa [q-1])} \sum \limits _{b=1}^c \left( {\begin{array}{c}c\\ b\end{array}}\right) K_b \left[ \left( 1-\frac{b}{c}\right) \frac{1}{2}B(\lambda , w) + \frac{b}{c}\frac{1}{2}B(w, \lambda )\right] \nonumber \\&\quad =\frac{q}{B(\kappa ,\kappa [q-1])} \sum \limits _{b=1}^c \left( {\begin{array}{c}c\\ b\end{array}}\right) K_b B(\lambda ,w). \end{aligned}$$

(83)

The result follows applying Lemma 3 followed by Lemma 4. $\square$

Proof of consistency check 2

Taking (29) and (30), the integral $\int \limits _{-\infty }^\infty \!\mathrm{d}u\; u\psi (u)$ can be written as

$$\begin{aligned}&\frac{2q}{B(\kappa ,\kappa [q-1])} \sum \limits _{b=1}^c \left( {\begin{array}{c}c\\ b\end{array}}\right) K_b \left[ \left( 1-\frac{b}{c}\right) \int \limits _{-\infty }^{0}\mathrm{d}u \, \frac{u\left( u^2\right) ^{b+\kappa -\frac{1}{2}}}{\left( 1+u^2\right) ^{c+\kappa q}}\nonumber \right. \\&\quad \left. + \frac{b}{c} \int \limits _{0}^{\infty }\mathrm{d}u \, \frac{u\left( u^2\right) ^{c-b+\kappa [q-1]-\frac{1}{2}}}{\left( 1+u^2\right) ^{c+\kappa q}}\right] . \end{aligned}$$

(84)

Let $\lambda {:=} b+\kappa -\frac{1}{2}$ and $w {:=} c-b+\kappa [q-1]-\frac{1}{2}$. Applying Lemma 2 and the property $\varGamma (x+1)=x\varGamma (x)$ we have

$$\begin{aligned} \int \limits _{-\infty }^{\infty }\mathrm{d}u \, u\psi (u) \!=\!\frac{2q}{B(\kappa ,\kappa [q\!-\!1])} \sum \limits _{b=1}^c \left( {\begin{array}{c}c\\ b\end{array}}\right) K_b \left[ \left( \frac{b}{c}\!-\!1\right) \frac{\varGamma (\lambda ) \varGamma (w)\lambda }{2\varGamma (c\!+\!\kappa q)} \!+\! \frac{b}{c} \frac{\varGamma (\lambda ) \varGamma (w)w}{2\varGamma (c+\kappa q)} \right] .\nonumber \\ \end{aligned}$$

(85)

To obtain $\tilde{\mu }$ as in (24) we use Lemma 3 to substitute $\left( {\begin{array}{c}c\\ b\end{array}}\right) \frac{1}{B(\kappa ,\kappa [q-1])}$ with $\frac{\mathbb{P }_1(b)}{B(\lambda +1/2,w+1/2)}$. After some simplifications, the result follows. $\square$

Proof of Lemma 6

The integral $\int \limits _{-\infty }^{\infty }\mathrm{d}u\; u^2\psi (u)$ can be written as

$$\begin{aligned}&\frac{2q}{B(\kappa ,\kappa [q-1])} \sum \limits _{b=1}^c \left( {\begin{array}{c}c\\ b\end{array}}\right) K_b \left[ \left( 1-\frac{b}{c}\right) \int \limits _{-\infty }^{0}\mathrm{d}u \, \frac{u^2\left( u^2\right) ^{b+\kappa -\frac{1}{2}}}{\left( 1+u^2\right) ^{c+\kappa q}}\nonumber \right. \\&\quad \left. + \frac{b}{c} \int \limits _{0}^{\infty }\mathrm{d}u \, \frac{u^2\left( u^2\right) ^{c-b+\kappa [q-1]-\frac{1}{2}}}{\left( 1+u^2\right) ^{c+\kappa q}}\right] . \end{aligned}$$

(86)

Let $\lambda {:=} c-b+\kappa [q-1]$ and $w {:=} b+\kappa$. Applying Lemma 2 with (2) and the property $\varGamma (x+1)=x\varGamma (x)$, we get

$$\begin{aligned}&\frac{2q}{B(\kappa ,\kappa [q-1])} \sum \limits _{b=1}^c \left( {\begin{array}{c}c\\ b\end{array}}\right) K_b \left[ \left( 1-\frac{b}{c}\right) \frac{\varGamma (\lambda -1) \varGamma (w-1)w(w-1)}{2\varGamma (c+\kappa q)}\right. \nonumber \\&\quad \left. + \frac{b}{c} \frac{\varGamma (\lambda -1) \varGamma (w-1) \lambda (\lambda -1)}{2\varGamma (c+\kappa q)} \right] . \end{aligned}$$

(87)

Then using (18) we have

$$\begin{aligned} \int \limits _{-\infty }^{\infty }\mathrm{d}u \, u^2\psi (u)=q \sum \limits _{b=1}^c K_b \mathbb{P }_1(b)\left[ \left( 1-\frac{b}{c}\right) \frac{w}{\lambda -1} + \frac{b}{c} \frac{\lambda }{w-1} \right] \end{aligned}$$

(88)

and (33) follows after some rewriting. $\square$

Proof of Theorem 5

We start from Corollary 3 and write a general power series expansion,

$$\begin{aligned} \tilde{\chi }(k) = 1-(V/2) k^2 +{\textstyle \sum \limits _{t=0}^\infty }\, \gamma _t |k|^{r_t}, \end{aligned}$$

(89)

where the $r_t\ge 2+2\kappa$ are powers and the $\gamma _t\in \mathbb{C }$ are coefficients of the form $i^{\beta _t\,\mathrm{sgn}\,k}$ times a real factor. In this expression the desired relation $\tilde{\chi }(-k)=[\tilde{\chi }(k)]^*$ evidently holds, and the properties $\tilde{\chi }(0)=1$, $\tilde{\chi }^{\prime }(0)=0$, $\tilde{\chi }^{\prime \prime }(0)=-V$ are clearly present. Then we write

$$\begin{aligned}{}[\tilde{\chi }(k/{\sqrt{m}})]^m =\exp [ m\ln \tilde{\chi }(k/\sqrt{m})] =e^{-\frac{V}{2} k^2}\exp \left[ m{\textstyle \sum \limits _{t=0}^\infty }\, (\frac{|k|}{\sqrt{m}})^{r_t^{\prime }}\delta _t\right] , \end{aligned}$$

(90)

where the powers $r_t^{\prime }\ge 2+2\kappa$ and coefficients $\delta _t\propto i^{\beta _t^{\prime }\mathrm{sgn}\,k}$ are obtained (laboriously) by substituting (89) into the Taylor series for the logarithm, $\ln (1+\varepsilon )=\varepsilon -\varepsilon ^2/2+\varepsilon ^3/3-\varepsilon ^4/4+\cdots$. It is worth noting that $m$ disappears from the $k^2$ term, but not from the others. Equation (56) is obtained from (90) by using the Taylor series for the $\exp$ function,

$$\begin{aligned} \exp \varepsilon =1+\varepsilon +\varepsilon ^2/2!+\varepsilon ^3/3!+\cdots \end{aligned}$$

(91)

and (again laboriously) collecting terms with equal powers of $k$. Since we started out with powers $r_t\ge 2+2\kappa$, we end up with powers $\nu _t\ge 2+2\kappa$. Finally, (57) follows by applying Lemma 10 and Lemma 11 to evaluate the integrals that arise when (56) is substituted into Theorem 4. $\square$