Nonlocal Myriad Filters for Cauchy Noise Removal

Abstract

The contribution of this paper is twofold. First, we introduce a generalized myriad filter, which is a method to compute the joint maximum likelihood estimator of the location and the scale parameter of the Cauchy distribution. Estimating only the location parameter is known as myriad filter. We propose an efficient algorithm to compute the generalized myriad filter and prove its convergence. Special cases of this algorithm result in the classical myriad filtering and an algorithm for estimating only the scale parameter. Based on an asymptotic analysis, we develop a second, even faster generalized myriad filtering technique. Second, we use our new approaches within a nonlocal, fully unsupervised method to denoise images corrupted by Cauchy noise. Special attention is paid to the determination of similar patches in noisy images. Numerical examples demonstrate the excellent performance of our algorithms which have moreover the advantage to be robust with respect to the parameter choice.

Appendices

Appendix

This appendix contains the proofs of Sect. 3.

1.1 Proof of Theorem 1:

Proof

• 1. First, we prove that all critical points $({\hat{a}},\hat{\gamma })$ of L are strict minimizers, by showing that $\nabla ^2 L$ is positive definite at all critical points. To simplify the notation, we set $a_i :=x_i - a$ and $\hat{a}_i :=x_i - \hat{a}$, $i=1,\ldots ,n$.

  Estimation of $\frac{\partial ^2 L}{\partial a^2}$: We split the sum in (2) as follows:

  $$\begin{aligned} \sum _{i:a_i^2> \gamma ^2} w_i \frac{a_i^2-\gamma ^2 }{\bigl (a_i^2 + \gamma ^2\bigr )^2}&< \sum _{i:a_i^2> \gamma ^2} w_i \frac{a_i^2-\gamma ^2 }{\bigl ( a_i^2 + \gamma ^2\bigr ) (\gamma ^2 + \gamma ^2 )}\\&= \frac{1}{2\gamma ^2} \sum _{i:a_i^2 > \gamma ^2}w_i \frac{a_i^2-\gamma ^2 }{ a_i^2 + \gamma ^2 } \end{aligned}$$

  and similarly, since in this case the summands are negative,

  $$\begin{aligned} \sum _{i:a_i^2<\gamma ^2} w_i \frac{a_i^2-\gamma ^2 }{\bigl (a_i^2 + \gamma ^2\bigr )^2}< \frac{1}{2\gamma ^2} \sum _{i:a_i^2<\gamma ^2}w_i \frac{a_i^2-\gamma ^2 }{ a_i^2 + \gamma ^2 }. \end{aligned}$$

  Since $n \ge 3$ at least one of these sums exists. Thus, for $\hat{\gamma }$ we have

  $$\begin{aligned} \frac{\partial ^2 L}{\partial a^2}(\hat{a}, \hat{\gamma })&= - 2 \ \sum _{i=1}^n w_i\frac{\hat{a}_i^2 -{\hat{\gamma }}^2}{\bigl (\hat{a}_i^2 + {\hat{\gamma }}^2\bigr )^2}\\&>- \frac{1}{{\hat{\gamma }}^2} \sum _{i=1}^n w_i\frac{\hat{a}_i^2-{\hat{\gamma }}^2 }{ \hat{a}_i^2 + {\hat{\gamma }}^2 }\\&= - \frac{1}{{\hat{\gamma }}^2} \sum _{i=1}^n w_i \frac{\hat{a}_i^2+{\hat{\gamma }}^2-2{\hat{\gamma }}^2 }{ \hat{a}_i^2 + {\hat{\gamma }}^2 }\\&= - \frac{1}{{\hat{\gamma }}^2} + 2 \sum _{i=1}^n w_i \frac{1 }{ \hat{a}_i^2 + {\hat{\gamma }}^2 } = 0, \end{aligned}$$

  where the last equation follows by (5).

  Estimation of $\det \left( \nabla ^2 L({\hat{a}},{\hat{\gamma }}) \right)$: Using (2)–(3) we obtain

  $$\begin{aligned}&\frac{1}{4} \det \left( \nabla ^2 L({\hat{a}},{\hat{\gamma }}) \right) \\&\quad =\left( \sum \limits _{i=1}^n w_i \frac{{\hat{\gamma }}^2 - {\hat{a}}_i^2}{\bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2}\right) \left( \sum \limits _{i=1}^n w_i \frac{{\hat{a}}_i^2-{\hat{\gamma }}^2}{\bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2}+\frac{1}{2{\hat{\gamma }}^2} \right) \\&\qquad -4\left( \sum \limits _{i=1}^n w_i\frac{{\hat{\gamma }} \hat{a}_i}{\bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2}\right) ^2. \end{aligned}$$

  With (5) we rewrite the first term as

  $$\begin{aligned} \sum _{i=1}^n w_i\frac{{\hat{a}}_i^2-{\hat{\gamma }}^2 }{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 }&= \sum _{i=1}^n w_i\frac{{\hat{a}}_i^2+ {\hat{\gamma }}^2 -2{\hat{\gamma }}^2 }{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 }\\&= \frac{1}{2{\hat{\gamma }}^2} - 2{\hat{\gamma }}^2 \sum _{i=1}^n w_i\frac{1}{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 }\\&= -\frac{1}{2{\hat{\gamma }}^2} + 2 \sum _{i=1}^n w_i\frac{1}{{\hat{a}}_i^2 + {\hat{\gamma }}^2}\\&\quad - 2 {\hat{\gamma }}^2\sum _{i=1}^n w_i\frac{1}{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 } \\&= -\frac{1}{2{\hat{\gamma }}^2} \sum _{i=1}^n w_i\frac{ \bigl ({\hat{a}}_i^2- {\hat{\gamma }}^2\bigr )^2 }{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 } . \end{aligned}$$

  With the help of (5), we can simplify the second term as

  $$\begin{aligned} \frac{\partial ^2 L}{\partial \gamma ^2}(a,{\hat{\gamma }}) = 4 \sum _{i=1}^n w_i\frac{a_i^2 }{\bigl (a_i^2 + \hat{\gamma }^2\bigr )^2} \end{aligned}$$

  and the third one using (4) by

  $$\begin{aligned}&\sum _{i=1}^n w_i\frac{{\hat{a}}_i }{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 } \\&\quad = -\frac{1}{2{\hat{\gamma }}^2} \sum _{i=1}^n w_i\frac{{\hat{a}}_i}{{\hat{a}}_i^2 + {\hat{\gamma }}^2} + \sum _{i=1}^n w_i \frac{{\hat{a}}_i }{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 } \\&\quad = -\frac{1}{2{\hat{\gamma }}^2}\sum _{i=1}^n w_i\frac{ {\hat{a}}_i \bigl ({\hat{a}}_i^2 -{\hat{\gamma }}^2 \bigr )}{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 }. \end{aligned}$$

  Therewith we obtain

  $$\begin{aligned}&\frac{1}{4} \det \left( \nabla ^2 L(\hat{a},{\hat{\gamma }}) \right) \\&\quad = \frac{1}{{\hat{\gamma }}^2} \left( \left( \sum _{i=1}^n w_i\frac{ \bigl ({\hat{a}}_i^2- {\hat{\gamma }}^2\bigr )^2 }{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 }\right) \left( \sum _{i=1}^n w_i\frac{{\hat{a}}_i^2 }{\bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2}\right) \right. \\&\left. \qquad - \left( \sum _{i=1}^n w_i\frac{{\hat{a}}_i\bigl ({\hat{a}}_i^2 -{\hat{\gamma }}^2 \bigr )}{ \bigl ({\hat{a}}_i^2 + {\hat{\gamma }}^2\bigr )^2 }\right) ^2 \right) \end{aligned}$$

  and by Cauchy–Schwarz’ inequality finally

  $$\begin{aligned} \det \left( \nabla ^2 L(a,{\hat{\gamma }}) \right) >0. \end{aligned}$$

  Note that we have indeed strict inequality, since otherwise there must exist $\lambda \in {\mathbb {R}}$ such that ${\hat{a}}_i = \lambda ({\hat{a}}_i^2 - {\hat{\gamma }}^2)$ for all $i=1,\ldots ,n$, which is not possible since $n \ge 3$.

• 2. Next, we show that the first step implies that there is only one critical point. For any fixed $a \in (x_1,x_n)$, let ${\hat{x}} (a) :={\hat{\gamma }} (a)^2$ denote the solution of (5) which is unique due to monotonicity, see Lemma 3. Bringing the summands in (5) to the same nominator, we see that ${\hat{x}} (a)$ is the unique real zero of a polynomial $P(\cdot ,a)$ of degree n whose coefficients are again polynomials in a, say

  $$\begin{aligned} P(x,a)&= x^n + p_{n-1} (a) x^{n-1} + \cdots + p_1(a) x + p_0(a). \end{aligned}$$

  We show that the zero ${\hat{x}}(a)$ of $P(\cdot ,a)$ is differentiable in a. To this end, we consider the smooth function $F:{\mathbb {R}}^n \times {\mathbb {R}} \rightarrow {\mathbb {R}}$ given by

  $$\begin{aligned} F(c,x)&:=x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0,\\ c&:=(c_0,\ldots ,c_{n-1}). \end{aligned}$$

  For an arbitrary fixed $a^* \in (x_1,x_n)$, we define

  $$\begin{aligned} c^* :=(p_0(a^*),\ldots , p_{n-1} (a^*)) \end{aligned}$$

  and $x^* :={\hat{x}} (a^*)$. Then we have $F(c^*, x^*) = 0$ and, since $x^*$ is a simple zero of $P(\cdot ,a^*)$, it holds $\frac{\partial }{\partial x} F(c^*,x^*) = P'(x^*,a^*) \not = 0$. By the implicit function theorem, there exists a continuously differentiable function $\varphi :{\mathbb {R}}^n \rightarrow {\mathbb {R}}$ such that $F(c,\varphi (c)) = 0$ in a neighborhood of $c^*$. Thus, for $c(a) :=(p_0(a), \ldots , p_{n-1}(a))$ with a in a neighborhood of $a^*$ we have ${\hat{x}} (a) = \varphi (c(a))$ and

  $$\begin{aligned} {\hat{x}} '(a) = \varphi (c(a))' = \nabla \varphi (c(a)) \times \left( p_0'(a) , \ldots ,p_{n-1}'(a) \right) , \end{aligned}$$

  which proves the claim.

  Now, the minima of L are given by $({\hat{a}},{\hat{\gamma }}({\hat{a}}))$. Assume that there exist two different minimizers ${\check{a}} < {\tilde{a}}$ of L. Since they are strict, ${\check{a}}$ and ${\tilde{a}}$ are also strict minimizers of the univariate function $g(a) :=L(a,{\hat{\gamma }}(a))$. The function g(a) is continuous, so that there exists a maximizer ${\bar{a}} \in ({\check{a}}, {\tilde{a}})$ of g fulfilling

  $$\begin{aligned} 0 = g'({\bar{a}}) = \nabla L({\bar{a}},{\hat{\gamma }}({\bar{a}})) \times \left( (1,{\hat{\gamma }}'({\bar{a}}) \right) . \end{aligned}$$

  By construction of ${\hat{\gamma }}$, we have $\frac{\partial L}{\partial \gamma } ({\bar{a}},{\hat{\gamma }}({\bar{a}})) = 0$. This implies $\frac{\partial L}{\partial a} ({\bar{a}},{\hat{\gamma }}({\bar{a}})) = 0$. Consequently, $\nabla L({\bar{a}},{\hat{\gamma }}({\bar{a}})) = 0$ so that $({\bar{a}},{\hat{\gamma }}({\bar{a}}))$ is critical point of L which is not a strict minimizer. This yields a contradiction, and thus, we have indeed only one critical point.

• 3. Finally, to see the existence of a critical point, it remains to show that there exists an a such that $S_1(a,{\hat{\gamma }}(a)) = 0$. By part 2 of the proof $S_1(a,{\hat{\gamma }}(a))$ is a continuous function in a. By the proof of the next Lemma 1, we know that $S_1(a,{\hat{\gamma }}(a)) >0$ for $a\le x_1$ and $S_1(a,{\hat{\gamma }}(a)) <0$ for $a\ge x_n$, so that the function has indeed a zero. $\square$

1.2 Proof of Lemma 1:

Proof

By (4), a critical point of $L(\cdot ,\gamma )$ has to fulfill $s_1(a) = 0$, where $s_1 :=\frac{S_1(\cdot ,\gamma )}{\gamma }$. All summands in $s_1(a)$ become positive if $a < x_1$ and negative if $a > x_n$. Since $n \ge 2$ this implies $s_1(a) > 0$ for $a \le x_1$ and $s_1(a) < 0$ for $a \ge x_n$. Hence, the zeros of $s_1$ lie in $(x_1,x_n)$ and there exists at least one zero by continuity of $s_1$. Further,

$$\begin{aligned} s_1(a)&= \frac{P_1(a)}{\prod \nolimits _{i=1}^n \bigl ((x_i-a)^2 + \gamma ^2 \bigr )}, \\ P_1(a)&:=\sum \limits _{i=1}^n w_i (x_i-a) \prod \limits _{j\ne i} \bigl ((x_j-a)^2 + \gamma ^2 \bigr ), \end{aligned}$$

so that the zeros of $s_1$ are the real roots of the nontrivial polynomial $P_1$ of degree $2n-1$, which are at most $2n-1$. $\square$

1.3 Proof of Lemma 2:

Proof

From the previous proof, we know that the zeros of $\frac{\partial L}{\partial a}(\cdot ,\gamma )$ coincide with those of $P_1$. By (2), we have

$$\begin{aligned}&\frac{\partial ^2 L}{\partial a^2}(a,\gamma ) = \frac{-2P_2(a)}{ \prod \nolimits _{i=1}^n \bigl ((x_i-a)^2 + \gamma ^2 \bigr )^2}, \\&P_2(a) :=\sum \limits _{i=1}^n w_i \bigl ((x_i-a)^2 - \gamma ^2\bigr ) \prod \limits _{j\ne i} \bigl ((x_j-a)^2 + \gamma ^2 \bigr )^2, \end{aligned}$$

so that the zeros of $\frac{\partial ^2 L}{\partial a^2}(\cdot ,\gamma )$ are those of the polynomial $P_2$. The coefficients of the polynomials $P_i$, $i=1,2$ are polynomials in $x_1,\ldots ,x_n$. Now, (7) states that $P_1$ and $P_2$ have a common root, which implies that the resultant ${\text {Res}}(P_1,P_2)$ of $P_1$ and $P_2$ equals zero, see, e.g., [19]. The resultant is defined as the determinant of the associated Sylvester matrix, so it is a polynomial expression in the coefficients of $P_1$ and $P_2$ as well, i.e., a polynomial in $x_1,\ldots ,x_n$. Since the set of roots of an n-variate polynomial is a set of measure zero in $\mathbb {R}^n$, this finishes the proof. $\square$

1.4 Proof of Lemma 3:

Proof

• 1. By (5), the critical points of $L(a,\cdot )$ have to fulfill $s_0(\gamma ^2) = \frac{1}{2}$, where $s_0 (\gamma ^2) :=S_0(a,\gamma )$. The continuous function $s_0$ is strictly increasing in $\gamma ^2$. Since $s_0(0) = 0$ and $\lim _{\gamma \rightarrow \infty } s_0(\gamma ) = 1$, we conclude that $s_0(\gamma ^2) = \frac{1}{2}$ has a unique solution ${\hat{\gamma }}^2$. Moreover, in (3), we obtain

  $$\begin{aligned} \frac{\partial ^2 L}{\partial \gamma ^2}(a,{\hat{\gamma }}) = 4 \sum _{i=1}^n w_i\frac{(x_i-a)^2 }{\bigl ((x_i-a)^2 + \hat{\gamma }^2\bigr )^2}>0 \end{aligned}$$

  so that ${\hat{\gamma }}$ is a minimizer.

• 2. Concerning the range of $\gamma$, it follows from (5) that ${\hat{\gamma }}^2 \in \left( \min _i (x_i-a)^2,\max _i (x_i-a)^2 \right)$ which gives together with the fact $a\in (x_1,x_n)$ the upper bound for ${\hat{\gamma }}$. To see the lower bound, assume that ${\hat{\gamma }}^2 \le d^2 \epsilon ^2$ and distinguish two cases:

  • (i) First, let a be one of the sample points, say $a=x_i$. Then, since $s_0$ is strictly increasing and $(x_j-a)^2 \ge d^2$ for $j\ne i$, it holds

    $$\begin{aligned} S_0(a,\gamma )&< w_i + \sum _{j \not = i} w_j \frac{ d^2 \epsilon ^2}{d^2 + d^2 \epsilon ^2} \\&= w_i + (1-w_i)\frac{\epsilon ^2}{1+\epsilon ^2} \\&\le w_{\text {max}} + (1-w_{\text {max}})\frac{\epsilon ^2}{1+\epsilon ^2}\\&= \frac{1}{1+\epsilon ^2} \frac{1}{2}\le \frac{1}{2}, \end{aligned}$$

    which is in contradiction to (5).

  • (ii) Next, let $a \in (x_i,x_{i+1})$. Similarly we obtain in this case

    $$\begin{aligned}&S_0(a,\gamma )\\&\quad < w_i \frac{d^2 \epsilon ^2}{(a-x_i)^2 + d^2 \epsilon ^2} + w_{i+1} \frac{d^2 \epsilon ^2}{(a-x_{i+1})^2 + d^2 \epsilon ^2}\\&\qquad + \frac{\epsilon ^2}{1+\epsilon ^2} (1-w_i - w_{i+1}). \end{aligned}$$

    If the weights are not equal, say $w_i > w_{i+1}$, then the right-hand side becomes largest for $a = x_i$ and we are in the previous case i). For $w_i = w_{i+1} = w$, we get

    $$\begin{aligned} S_0(a,\gamma )&< \epsilon ^2 w \left( \frac{1}{\left( \frac{a-x_i}{d}\right) ^2 + \epsilon ^2} + \frac{1}{\left( \frac{a-x_{i+1}}{d}\right) ^2 + \epsilon ^2}\right) \\&\quad + \frac{\epsilon ^2}{1+\epsilon ^2}(1-2w) \end{aligned}$$

    and by replacing d by $x_{i+1} -x_i$ and denoting by $z \in \left[ 0,\frac{1}{2}\right]$ the distance of a to the midpoint of the normalized interval,

    $$\begin{aligned} S_0(a,\gamma )&< \epsilon ^2 w \left( \frac{1}{ \left( \frac{1}{2} + z\right) ^2 + \epsilon ^2} + \frac{1}{\left( \frac{1}{2} - z\right) ^2 + \epsilon ^2}\right) \nonumber \\&\quad + \frac{\epsilon ^2}{1+\epsilon ^2}(1-2w)\nonumber \\&= \epsilon ^2 w \left( \frac{2\left( \frac{1}{4} + z^2 + \epsilon ^2\right) }{\left( \frac{1}{4} - z^2\right) ^2 + 2 \epsilon ^2 \left( \frac{1}{4} + z^2\right) + \epsilon ^4} \right) \nonumber \\&\quad + \frac{\epsilon ^2}{1+\epsilon ^2}(1-2w)\nonumber \\&= 2 \epsilon ^2 w \left( \frac{ \frac{1}{4} + z^2 + \epsilon ^2}{\left( \frac{1}{4} + z^2 + \epsilon ^2\right) ^2 - z^2} \right) \nonumber \\&\quad +\, \frac{\epsilon ^2}{1+\epsilon ^2}(1-2w). \end{aligned}$$

    (20)

    Now, $\frac{ \frac{1}{4} + z^2 + \epsilon ^2}{ \left( \frac{1}{4} + z^2 + \epsilon ^2\right) ^2 - z^2}$ becomes largest iff

    $$\begin{aligned}&\frac{\left( \frac{1}{4} + z^2 + \epsilon ^2\right) ^2 - z^2}{ \frac{1}{4} + z^2 + \epsilon ^2}\\&\quad = \left( \frac{1}{4} + z^2 + \epsilon ^2\right) - \frac{z^2}{ \frac{1}{4} + z^2 + \epsilon ^2} \end{aligned}$$

    becomes smallest. Substituting $y :=z^2 + \frac{1}{4} \in \left[ \frac{1}{4},\frac{1}{2}\right]$, we obtain the function

    $$\begin{aligned} f(y)&:=y + \epsilon ^2 - \frac{y - \frac{1}{4}}{y + \epsilon ^2} \\&= - 1 + y + \epsilon ^2 + \frac{\frac{1}{4}+\epsilon ^2}{y + \epsilon ^2}, \end{aligned}$$

    whose derivatives are given by

    $$\begin{aligned} f'(y) = 1-\frac{\epsilon ^2 + \frac{1}{4}}{(y + \epsilon ^2)^2},\qquad f''(y) = 2 \frac{\epsilon ^2 + \frac{1}{4}}{(y + \epsilon ^2)^2}. \end{aligned}$$

    Setting the derivative to zero results in the positive solution $y = -\epsilon ^2 + \sqrt{\epsilon ^2 + \frac{1}{4}}$, which is the global minimum on $\left[ \frac{1}{4},\frac{1}{2}\right]$ since f is convex. Resubstituting and plugging it in (20) yields

    $$\begin{aligned} S_0(a,\gamma )&< 2 \epsilon ^2 w \frac{1}{2\sqrt{\epsilon ^2 + \frac{1}{4}}+1}+ \frac{\epsilon ^2}{1+\epsilon ^2}(1-2w) \\&\le w \epsilon ^2+ \frac{\epsilon ^2}{1+\epsilon ^2}(1-2w)\\&= \underbrace{\epsilon ^2\left( \frac{\epsilon ^2 -1}{\epsilon ^2+1} \right) }_{<0}w + \frac{\epsilon ^2}{1+\epsilon ^2}\\&\le \frac{\epsilon ^2}{1+\epsilon ^2}\le \frac{1}{3}, \end{aligned}$$

    since $\epsilon ^2 \in \left( 0,\frac{1}{2}\right)$.

$\square$

Appendix

This appendix contains the proofs of Sect. 4.

1.1 Proof of Theorem 2:

Proof

• 1. We show that the objective function $L(a_r,\gamma _r)$ decreases for increasing r. By concavity of the logarithm, we have

  $$\begin{aligned}&L(a_{r+1},\gamma _{r+1}) - L(a_r,\gamma _r) \\&\quad = \sum _{i=1}^n w_i \log \left( \frac{(x_i-a_{r+1})^2 + \gamma _{r+1}^2}{(x_i-a_{r})^2 + \gamma _{r}^2} \frac{\gamma _{r}}{\gamma _{r+1}}\right) \\&\quad \le \log \Biggl ( \underbrace{\sum _{i=1}^n w_i\frac{(x_i-a_{r+1})^2 + \gamma _{r+1}^2}{(x_i-a_{r})^2 + \gamma _{r}^2} \frac{\gamma _{r}}{\gamma _{r+1}} }_{\varUpsilon }\Biggr ), \end{aligned}$$

  so that it suffices to show that $\varUpsilon \le 1$. Setting $S_{0r} :=S_0(a_r,\gamma _r)$ and $S_{1r} :=S_1(a_r,\gamma _r)$, we obtain with Algorithm 1

  $$\begin{aligned} \varUpsilon&= \sqrt{\frac{S_{0r}}{1-S_{0r}}} \sum _{i=1}^n w_i\frac{(x_i-a_r + a_r- a_{r+1})^2 + \frac{1-S_{0r}}{S_{0r}} \gamma _{r}^2}{(x_i-a_{r})^2 + \gamma _{r}^2} \\&= \sqrt{\frac{S_{0r}}{1-S_{0r}}} \Bigg ( \underbrace{\sum _{i=1}^n w_i\frac{(x_i-a_r )^2 }{(x_i-a_{r})^2 + \gamma _{r}^2}}_{1-S_{0r}}\\&\quad +2 \underbrace{(a_r - a_{r+1} )}_{-\gamma _r \tfrac{S_{1r}}{S_{0r}}} \underbrace{\sum _{i=1}^n w_i\frac{x_i-a_r }{(x_i-a_{r})^2 + \gamma _{r}^2}}_{\tfrac{S_{1r}}{\gamma _r}}\\&\quad + \sum _{i=1}^n w_i\frac{\overbrace{(a_r-a_{r+1})^2}^{\gamma _r^2\tfrac{S_{1r}^2}{S_{0r}^2}} }{(x_i-a_{r})^2 + \gamma _{r}^2}\\&\quad + \frac{1-S_{0r}}{S_{0r}} \underbrace{\sum _{i=1}^n w_i \frac{\gamma _r^2 }{(x_i-a_{r})^2 + \gamma _{r}^2}}_{S_{0r}} \Bigg )\\&= 2 \sqrt{S_{0r} (1-S_{0r})} - 2 \frac{S_{1r}^2}{\sqrt{S_{0r} (1-S_{0r}) }}\\&\quad +\sqrt{\frac{S_{0r}}{1-S_{0r}}} \frac{S_{1r}^2}{S_{0r}^2} S_{0r}\\&= 2 \sqrt{S_{0r} (1-S_{0r})} - \frac{S_{1r}^2}{\sqrt{S_{0r} (1-S_{0r})}}. \end{aligned}$$

  The function

  $$\begin{aligned} f:(0,1) \rightarrow {\mathbb {R}},\quad f(z) :=2\sqrt{z(1-z)} - \frac{\alpha ^2}{\sqrt{z(1-z)}} \end{aligned}$$

  attains its global maximum in $z = \frac{1}{2}$, where $f(\frac{1}{2}) = 1 - 2\alpha ^2 \le 1$. Consequently, $\varUpsilon \le 1$ with equality if and only if $S_{1r} = 0$ and $S_{0r} = \frac{1}{2}$, that is, $(a_{r+1},\gamma _{r+1}) = (a_r,\gamma _r)$.

• 2. By (5) and (4), we know that $(a_{r+1},\gamma _{r+1}) = (a_r,\gamma _r)$ is a fixed point of $(a_{r+1},\gamma _{r+1}) :=T(a_r,\gamma _r)$ in Algorithm 1 if and only if it is the minimizer of L. Let $(a_{r+1},\gamma _{r+1}) \not = (a_r,\gamma _r)$ for all $r \in {\mathbb {N}}_0$. The sequence $\{(a_r,\gamma _r)\}_{r\in \mathbb {N}}$ is bounded: for $a_r$ we have by (8) that $a_r$ is always a convex combination of the $x_i$ so that $a_r \in (x_1,x_n)$; for $\gamma _r$ this follows from Lemma 3 and Theorem 5, that is shown later on. Together with part 1 of the proof, we see that $L_r :=L(a_r,\gamma _r)$ is a strictly decreasing, bounded sequence of numbers which must converge to some number ${\hat{L}}$. Further, $\{ (a_r,\gamma _r)\}_{r\in \mathbb {N}}$ contains a convergent subsequence $\{ (a_{r_j},\gamma _{r_j})\}_{j\in \mathbb {N}}$ which converges to some $({\hat{a}},{\hat{\gamma }})$. By the continuity of L and T, we obtain

  $$\begin{aligned} L ({\hat{a}},{\hat{\gamma }})&= \lim _{j\rightarrow \infty } L(a_{r_j}, \gamma _{r_j}) = \lim _{j\rightarrow \infty } L_{r_j}=\lim _{j\rightarrow \infty } L_{r_j+1} \\&= \lim _{j\rightarrow \infty } L(a_{r_j+1}, \gamma _{r_j+1}) \\&= \lim _{j\rightarrow \infty } L\left( T(a_{r_j}, \gamma _{r_j}) \right) = L\left( T({\hat{a}},{\hat{\gamma }}) \right) . \end{aligned}$$

  However, this implies $({\hat{a}},{\hat{\gamma }}) = T({\hat{a}},{\hat{\gamma }})$ so that $({\hat{a}},{\hat{\gamma }})$ is a fixed point of T and consequently the minimizer. Since the minimizer is unique, the whole sequence $\{(a_r,\gamma _r)\}_{r\in \mathbb {N}}$ converges to $({\hat{a}},{\hat{\gamma }})$ and we are done.

$\square$

1.2 Proof of Theorem 3:

Proof

We follow the lines of the proof of Theorem 2. Recall that

$$\begin{aligned} Q(a) = L(a,\gamma ) + \log (\gamma ) = \sum _{i=1}^n w_i \log \left( (x_i-a)^2 + \gamma ^2 \right) . \end{aligned}$$

• 1. First, we show that the objective function $Q(a_r)$ decreases for increasing r. By concavity of the logarithm, we have

  $$\begin{aligned} Q(a_{r+1})-Q(a_r)&= L(a_{r+1},\gamma ) - L(a_r,\gamma ) \\&= \sum _{i=1}^n w_i \log \left( \frac{(x_i-a_{r+1})^2 + \gamma ^2}{(x_i-a_{r})^2 + \gamma ^2} \right) \\&\le \log \Biggl ( \underbrace{\sum _{i=1}^n w_i\frac{(x_i-a_{r+1})^2 + \gamma ^2}{(x_i-a_{r})^2 + \gamma ^2} }_{\varUpsilon }\Biggr ), \end{aligned}$$

  and it suffices to show that $\varUpsilon \le 1$. Setting $S_{0r} :=S_0(a_r,\gamma )$ and $S_{1r} :=S_1(a_r,\gamma )$ (note that $\gamma$ is fixed here) we obtain with Algorithm 2

  $$\begin{aligned} \varUpsilon&= \sum _{i=1}^n w_i\frac{(x_i-a_{r+1})^2 + \gamma ^2}{(x_i-a_{r})^2 + \gamma ^2}\\&= 1-S_{0r} - \frac{S_{1r}^2}{S_{0r}} + S_{0r}= 1 - \frac{S_{1r}^2}{S_{0r} }\le 1 \end{aligned}$$

  with equality if and only if $S_{1r} = 0$, i.e., $a_{r+1}= a_r$, in which case ${\hat{a}} :=a_r$ is a critical point of Q.

• 2. If $a_{r+1}\not = a_r$ for all $r \in {\mathbb {N}}_0$, the sequence $Q_r :=Q( a_r)$ is strictly decreasing and bounded below by $\log (\gamma ^2)$, so that $Q_r \rightarrow {\hat{Q}}$ as $r \rightarrow \infty$. Further, since Q is continuous and coercive, the sequence $\{a_r\}_{r\in \mathbb {N}}$ is bounded. Consequently, it contains a convergent subsequence $\{a_{r_j} \}_{j\in \mathbb {N}}$ which converges to some ${\hat{a}}$ and by continuity of Q we have $Q({\hat{a}}) = {\hat{Q}}$. By continuity of Q and the operator $T_1$ given by $a_{r+1} :=T_1(a_r)$ in Algorithm 2, it follows

  $$\begin{aligned} Q({\hat{a}})&= \lim _{j \rightarrow \infty } Q(a_{r_j}) = \lim _{j \rightarrow \infty } Q_{r_j} = \lim _{j \rightarrow \infty } Q_{r_j+1}\\&= \lim _{j \rightarrow \infty } Q(a_{r_j+1}) = \lim _{j \rightarrow \infty } Q(T_1(a_{r_j}) = Q(T_1({\hat{a}})). \end{aligned}$$

  By the first part of the proof, this implies that ${\hat{a}}$ is a fixed point of $T_1$ and thus a critical point of Q.

• 3. Observing that $\frac{S_{1r}^2}{S_{0r}} = \frac{S_0r}{\gamma ^2}(a_{r+1}-a_r)^2$ and $- \log (1-y) \ge y$, $y \in (0,1)$, we have

  $$\begin{aligned} Q(a_r) - Q(a_{r+1})&\ge - \log \left( 1- \frac{S_{0r}}{\gamma ^2}(a_{r+1}-a_r)^2\right) \\&\ge \frac{S_{0r}}{\gamma ^2} (a_r-a_{r+1})^2. \end{aligned}$$

  Since $1+|x-y|^2 <2 (1+|x|^2)(1+|y|^2)$ and $a_r\in [x_1,x_n]$, we estimate

  $$\begin{aligned} \frac{S_0r}{\gamma ^2}&= \sum _{i=1}^n w_i \frac{1 }{(x_i-a_r)+\gamma ^2} \\&\ge \frac{\gamma ^2}{2\bigl ( a_r^2+\gamma ^2 \bigr )} \sum _{i=1}^n w_i \frac{1 }{x_i^2+ \gamma ^2} \\&\ge \frac{\gamma ^2}{2\bigl (\max \{ x_1^2,x_n^2\} +\gamma ^2 \bigr )} \sum _{i=1}^n w_i\frac{1 }{x_i^2+\gamma ^2} \\&=:\tau _0 > 0, \end{aligned}$$

  which results in

  $$\begin{aligned} Q(a_r) - Q(a_{r+1}) \ge \tau _0 (a_r-a_{r+1})^2. \end{aligned}$$

  Since by the second part of the proof $\lim _{r\rightarrow \infty } Q(a_r) - Q(a_{r+1}) = 0$, we also have $\lim _{r\rightarrow \infty } |a_r-a_{r+1}| = 0$.

• 4. Assume now that there exists a subsequence $\{a_{r_l} \}_{l\in \mathbb {N}}$ which converges to some $a^*\ne {\hat{a}}$. Since the set of critical points is finite, there exists $\varepsilon > 0$ such that $|{\hat{a}} - a^*| \ge \varepsilon$. On the other hand, we have by the third part of the proof for l, j large enough that $\varepsilon > |a_{r_l} - a_{r_j}|$. For $l,j \rightarrow \infty$, this leads to a contradiction. $\square$

1.3 Proof of Theorem 4:

Proof

By Theorem 3, we know that ${\hat{a}} = \lim _{r\rightarrow \infty } a_r$ exists and ${\hat{a}} = T({\hat{a}})$ is a stationary points of Q fulfilling $Q'({\hat{a}}) = 0$. We distinguish the following cases:

$$\begin{aligned} \text{ Case } \text{ I } \text{: } \; \exists \, r_0\in \mathbb {N}:&a_{r_0 + 1} = a_{r_0},\\ \text{ Case } \text{ II: } \; \forall \, r \in \mathbb {N}:&a_{r+1}\ne a_r. \end{aligned}$$

We show that Case I occurs with probability zero and that in Case II, the probability of ${\hat{a}}$ being a local minimum is one. By (8), we get

$$\begin{aligned} a_{r+1}&= T(a_r) = \frac{ \sum \nolimits _{i=1}^n w_i x_i \prod \nolimits _{j\ne i}\left( (x_j-a_r)^2 + \gamma ^2\right) }{ \sum \nolimits _{i=1}^n \prod \nolimits _{j\ne i}\left( (x_j-a_r)^2 + \gamma ^2\right) }. \end{aligned}$$

Rearranging yields

$$\begin{aligned} 0&= a_{r+1} \underbrace{ \sum \limits _{i=1}^n \prod \limits _{j\ne i} \left( (x_j-a_r)^2 + \gamma ^2\right) }_{p_1(a_r)} \\&\quad -\, \underbrace{\sum \limits _{i=1}^n w_i x_i \prod \limits _{j\ne i}\left( (x_j-a_r)^2 + \gamma ^2\right) }_{p_2(a_r)}\\&= a_{r+1} p_1(a_r) - p_2(a_r) , \end{aligned}$$

where $p_1$ and $p_2$ are polynomials. This polynomial equation in $a_r$ has only finitely many solutions (up to $2(n-1)$). Recursively, backtracking each of the possible values for $a_r$ in a similar way, we end up with at most $2^{r+1} (n-1)^{r+1}$ starting points $a_0 \in (x_1,x_n)$ that can lead to the point $a_{r+1}$ after exactly $r+1$ iterations.

Case I As seen above, there are only finitely many starting points $a_0$ for which the sequence $\{ a_r\}_{r\in \mathbb {N}}$ reaches a fixed point after exactly $r_0$ steps. Since the set of natural numbers $\mathbb {N}$ is countable and countable unions of finite sets are countable, the set of starting points leading to Case I is countable and has consequently Lebesgue measure zero.

Case II Since Q is smooth, there might occur the following cases for the critical point ${\hat{a}}$: a) $Q''({\hat{a}}) <0$ (local maximum), b) $Q''({\hat{a}}) =0$ (saddle point), c) $Q''({\hat{a}}) >0$ (local minimum). Indeed, case a) cannot happen since we have seen in the proof of Theorem 3 that $\{Q_r\}_{r\in \mathbb {N}}$ is decreasing. Addressing case b), according to Lemma 2 the function Q has with probability one only minima and maxima, but no saddle points. Since cases a) and b) occur each with probability zero, case c) occurs with probability one. This finishes the proof. $\square$

1.4 Proof of Theorem 5:

Proof

• 1. First, we show property (11). From (5), we see immediately

  $$\begin{aligned} S_0(a,\gamma ) {\left\{ \begin{array}{ll}< \frac{1}{2}&{}\quad \text {if } \gamma < {\hat{\gamma }},\\ = \frac{1}{2}&{}\quad \text {if } \gamma = {\hat{\gamma }},\\> \frac{1}{2}&{}\quad \text {if } \gamma > {\hat{\gamma }}, \end{array}\right. } \end{aligned}$$

  so that $\gamma _r < {\hat{\gamma }}$ implies $\gamma _{r+1} > \gamma _r$ and $\gamma _r > {\hat{\gamma }}$ results in $\gamma _{r+1} < \gamma _r$. To see that the iterates cannot skip ${\hat{\gamma }}$ we consider the quotient

  $$\begin{aligned} \frac{\gamma _{r+1}^2}{{\hat{\gamma }}^2} = \frac{\gamma _r^2}{{\hat{\gamma }}^2}\frac{1-S_0(a,\gamma _r)}{S_0(a,\gamma _r)} = \frac{\sum \nolimits _{i=1}^n w_i \frac{a_i^2}{\alpha a_i^2+ {\hat{\gamma }}^2} }{\sum \nolimits _{i=1}^n w_i \frac{{\hat{\gamma }}^2}{\alpha a_i^2 + {\hat{\gamma }}^2 }}, \end{aligned}$$

  where $\alpha :=\left( \frac{{\hat{\gamma }} }{\gamma _r} \right) ^2$. We have to show that $\alpha < 1$ implies $\frac{\gamma _{r+1}^2}{{\hat{\gamma }}^2} > 1$ and conversely, $\alpha > 1$ implies $\frac{\gamma _{r+1}^2}{{\hat{\gamma }}^2} < 1$. Alternatively, we can prove that the function

  $$\begin{aligned} f(\alpha ) = \sum _{i=1}^n w_i \frac{\gamma ^2-a_i^2 }{\alpha a_i^2 + \gamma ^2}, \ \end{aligned}$$

  fulfills

  $$\begin{aligned} f(\alpha ) \left\{ \begin{array}{ll} < 0 &{}\quad \mathrm {if} \; \alpha \in (0,1), \\ > 0 &{}\quad \mathrm {if} \; \alpha \in (1,+\infty ). \end{array} \right. \end{aligned}$$

  We have $f(1) = 0$, and the derivatives of f are given by

  $$\begin{aligned} f'(\alpha )&= \sum _{i=1}^nw_i \frac{a_i^2(a_i^2 -\gamma ^2)}{(\alpha a_i^2 + \gamma ^2)^2},\\ f''(\alpha )&=2 \sum _{i=1}^n w_i \frac{a_i^4(\gamma ^2-a_i^2)}{(\alpha a_i^2 + \gamma ^2)^3}. \end{aligned}$$

  For $f'$ we estimate similarly as in the proof of Theorem 1,

  $$\begin{aligned}&\sum _{i:a_i^2> \gamma ^2} w_i \frac{a_i^2(a_i^2 -\gamma ^2)}{(\alpha a_i^2 + \gamma ^2)^2}\\&\quad> \sum _{i:a_i^2> \gamma ^2}^n w_i \frac{a_i^2(a_i^2 -\gamma ^2)}{(\alpha a_i^2 + a_i^2)(\alpha a_i^2 + \gamma ^2)} \\&\quad =\frac{1}{\alpha +1}\sum _{i:a_i^2 > \gamma ^2}^n w_i \frac{a_i^2 - \gamma ^2 }{\alpha a_i^2 + \gamma ^2} \end{aligned}$$

  and analogously for the negative summands, so that in summary

  $$\begin{aligned} f'(\alpha )&> \frac{1}{\alpha +1}\sum _{i=1}^n w_i \frac{a_i^2 -\gamma ^2}{\alpha a_i^2 + \gamma ^2} = -\frac{1}{\alpha +1}f(\alpha ). \end{aligned}$$

  (21)

  Analogously, we obtain for $f''$

  $$\begin{aligned} f''(\alpha )&< \frac{2}{\alpha +1}\sum _{i=1}^n w_i \frac{a_i^2(\gamma ^2-a_i^2)}{(\alpha a_i^2 + \gamma ^2)^2} = -\frac{2}{\alpha +1}f'(\alpha ).\nonumber \\ \end{aligned}$$

  (22)

  From (21), it follows $f'(1) > \frac{1}{2} f(1) = 0$ and therewith further $f(\alpha ) <0$ for $\alpha \in (0,1)$. Consider the case $\alpha > 1$. By continuity of f, we have $f(\alpha )>0$ for $\alpha$ sufficiently close to 1. Since $\lim _{\alpha \rightarrow \infty } f(\alpha ) = 0$, we conclude that $f'$ has at least one root for $\alpha > 1$. On the other hand, $f'$ has at most one root, since according to (22) any root of $f'$ is a local maximum of f. Thus, $f'$ has exactly one root, so that f has exactly one critical point (a local maximum). Since $\lim _{\alpha \rightarrow \infty } f(\alpha ) = 0$, this implies $f(\alpha )>0$ for all $\alpha >1$.

• 2. Since $L(a,\cdot )$ is continuous and has only one critical point, it follows immediately from (11) that $L(a,\gamma _r) \ge L(a,\gamma _{r+1}) = L(a,T_2 (\gamma _r))$ with equality if and only if $\gamma _r = \gamma _{r+1} = {\hat{\gamma }}$. In the latter case, we are done, so assume that $\gamma _r \not = \gamma _{r+1}$ for all $r \in {\mathbb {N}}_0$. By part 1 of the proof, the sequence $\{\gamma _r\}_r$ is monotone and bounded, so it converges to some $\gamma ^*$. By continuity of $L(a,\cdot )$ and $T_2$, we get

  $$\begin{aligned} L(a,\gamma ^*)&= \lim _{r \rightarrow \infty } L(a,\gamma _r) = \lim _{r \rightarrow \infty } L(a,\gamma _{r+1})\\&= \lim _{r \rightarrow \infty } L(a,T_2(\gamma _r)) = L(a,T_2(\gamma ^*)), \end{aligned}$$

  which is only possible it $\gamma ^* = T_2(\gamma ^*)$, i.e., if $\gamma ^* = {\hat{\gamma }}$.

$\square$

Appendix

This appendix contains the proofs of Sect. 5.

1.1 Proof of Lemma 4:

Proof

Consider the functions $g(x) :=\frac{1}{1 + x^2}$ and $h(x) :=\frac{x}{1 + x^2}$. Both functions are measurable and since

$$\begin{aligned}&\int _{-\infty }^\infty g(x) p(x)\,\mathrm {d}x\le \int _{-\infty }^\infty p(x) \,\mathrm {d}x=1 ,\\&\int _{-\infty }^\infty |h(x) p(x)|\,\mathrm {d}x\le \int _{-\infty }^\infty \frac{1}{2} p(x) \,\mathrm {d}x=\frac{1}{2}, \end{aligned}$$

where p denotes the density function of $C(a,\gamma )$; the expected values $\mathbb {E}(Y),\mathbb {E}(Z)$ exist.

For g and $a\ne 0$, we compute

$$\begin{aligned}&\int _{-\infty }^\infty g(x) p(x)\,\mathrm {d}x= \int _{-\infty }^\infty \frac{1}{1+x^2} \frac{1}{\pi \gamma }\frac{\gamma ^2}{(x-a)^2 + \gamma ^2} \,\mathrm {d}x\\&\quad \scriptscriptstyle {= \left[ \frac{ a\gamma \log \left( \frac{(x-a)^2 + \gamma ^2}{x^2 + 1}\right) + \gamma (a^2 + \gamma ^2-1)\arctan (x) + (a^2 -\gamma ^2 + 1)\arctan \left( \tfrac{x-a}{\gamma }\right) }{\pi \bigl (a^4 + 2a^2(\gamma ^2 + 1) + (\gamma ^2-1)^2\bigr ) } \right] _{-\infty }^{\infty }}. \end{aligned}$$

Since $\lim _{x\rightarrow \pm \infty }\log \left( \frac{(x-a)^2 + \gamma ^2}{x^2 + 1}\right) = 0$ and $\lim _{x\rightarrow \pm \infty } \arctan (x) = \pm \frac{\pi }{2}$, we obtain the first equation in (13). For $a=0$, we have

$$\begin{aligned} \int _{-\infty }^\infty g(x) p(x)\,\mathrm {d}x&= \int _{-\infty }^\infty \frac{1}{1+x^2} \frac{1}{\pi \gamma }\frac{\gamma ^2}{x^2 + \gamma ^2} \,\mathrm {d}x\\&=\left[ \frac{\gamma ^2 \arctan (x) - \gamma \arctan \left( \tfrac{x}{\gamma }\right) }{\pi \gamma (\gamma ^2-1)}\right] _{-\infty }^\infty , \end{aligned}$$

which results in the second equation in (13). Similarly, for h and $(a,\gamma ) \not = (0,1)$,

$$\begin{aligned}&\int _{-\infty }^\infty h(x) p(x) \,\mathrm {d}x= \int _{-\infty }^\infty \frac{x}{1+x^2} \frac{1}{\pi \gamma }\frac{\gamma ^2}{(x-a)^2 + \gamma ^2} \,\mathrm {d}x\\&\quad \scriptscriptstyle =\left[ \frac{\gamma (a^2 + \gamma ^2 - 1) \log \left( \frac{x^2 + 1}{(x-a)^2 + \gamma ^2}\right) + 2a(a^2 + \gamma ^2 + 1)\arctan \left( \tfrac{x-a}{\gamma }\right) -4a\gamma \arctan (x)}{2\pi \bigl (a^4 + 2a^2(\gamma ^2 + 1) + (\gamma ^2-1)^2\bigr ) }\right] _{-\infty }^\infty , \end{aligned}$$

so that the first equation in (14) follows. Finally, for $(a,\gamma ) = (0,1)$ it holds

$$\begin{aligned} \int _{-\infty }^\infty h(x) p(x)\,\mathrm {d}x&= \int _{-\infty }^\infty \frac{x}{1+x^2} \frac{1}{\pi }\frac{1}{x^2 + 1} \,\mathrm {d}x\\&= \left[ -\frac{1}{2\pi (1 + x^2)}\right] _{-\infty }^\infty , \end{aligned}$$

and consequently $E(Z) = 0$. $\square$

1.2 Proof of Corollary 1:

Proof

By Proposition 1, we have $X_r\sim C\left( \tfrac{a-a_r}{\gamma _r},\tfrac{\gamma }{\gamma _r}\right)$. Setting $\tilde{a} = \frac{a-a_r}{\gamma _r}$ and $\tilde{\gamma } = \frac{\gamma }{\gamma _r}$, and applying the results of Lemma 4, we obtain

$$\begin{aligned} \mathbb {E}\bigl (g(X_r)\bigr )&= \frac{\tilde{\gamma }(\tilde{a}^2 + \tilde{\gamma }^2-1) + \tilde{a}^2-\tilde{\gamma }^2 + 1}{(\tilde{a}^2 + \tilde{\gamma }^2 + 1)^2 - 4\tilde{\gamma }^2} \\&\textstyle = \frac{\frac{\gamma }{\gamma _r}\left[ \left( \frac{a-a_r}{\gamma _r}\right) ^2 + \left( \frac{\gamma }{\gamma _r}\right) ^2-1\right] +\left( \frac{a-a_r}{\gamma _r}\right) ^2 - \left( \frac{\gamma }{\gamma _r}\right) ^2+1}{\left[ \left( \frac{a-a_r}{\gamma _r}\right) ^2 + \left( \frac{\gamma }{\gamma _r}\right) ^2 + 1\right] ^2 - 4\left( \frac{\gamma }{\gamma _r}\right) ^2}\\&= \frac{\gamma _r(\gamma + \gamma _r) }{(a-a_r)^2 + (\gamma + \gamma _r)^2}. \end{aligned}$$

Similarly, we compute

$$\begin{aligned} \mathbb {E}\bigl (h(X_r)\bigr )&= \frac{\tilde{a}(\tilde{a}^2 + \tilde{\gamma }^2+1-2\tilde{\gamma })}{(\tilde{a}^2 + \tilde{\gamma }^2 + 1)^2 - 4\tilde{\gamma }^2} \\&= \frac{\frac{a-a_r}{\gamma _r}\left[ \left( \frac{a-a_r}{\gamma _r}\right) ^2 + \left( \frac{\gamma }{\gamma _r}\right) ^2+1-2\frac{\gamma }{\gamma _r}\right] }{\left[ \left( \frac{a-a_r}{\gamma _r}\right) ^2 + \left( \frac{\gamma }{\gamma _r}\right) ^2 + 1\right] ^2 - 4\left( \frac{\gamma }{\gamma _r}\right) ^2}\\&= \frac{\gamma _r(a-a_r) }{(a-a_r)^2 + (\gamma + \gamma _r)^2}. \end{aligned}$$

$\square$

1.3 Proof of Theorem 6:

Proof

(i) Since $\gamma ,\tilde{\gamma }_0>0$, it follows inductively from  (18) that $\tilde{\gamma }_r>0$. Further, if $\tilde{\gamma }_r<\gamma$, then

$$\begin{aligned} \tilde{\gamma }^2_{r+1}-\tilde{\gamma }^2_r = \tilde{\gamma }_r(\gamma -\tilde{\gamma }_r) + \tilde{\gamma }_r \frac{(a-\tilde{a}_r)^2}{\gamma + \tilde{\gamma }_r}>0. \end{aligned}$$

On the other hand, if $\tilde{\gamma }_r>\gamma$, then it holds

$$\begin{aligned} \tilde{\gamma }^2_{r+1}-\gamma ^2 = \gamma (\tilde{\gamma }_r - \gamma ) + \tilde{\gamma }_r \frac{(a-\tilde{a}_r)^2}{\gamma + \tilde{\gamma }_r}>0. \end{aligned}$$

Thus, to summarize it holds $\tilde{\gamma }^2_{r+1}\ge \min \{\tilde{\gamma }^2_r,\gamma ^2\}$ and inductively we have $\tilde{\gamma }^2_{r+1}\ge \min \{\tilde{\gamma }^2_0,\gamma ^2\}$.

(ii) Since $\gamma ,\tilde{\gamma }_r>0$, this is a direct consequence of (16).

(iii) Let $q = \max \left\{ \frac{1}{2},\frac{\gamma }{\gamma + \tilde{\gamma }_0}\right\}$, so that $\frac{1}{2} \le q < 1$. For the sequence $\{\tilde{a}_r\}_{r\in \mathbb {N}}$ we estimate

$$\begin{aligned} |\tilde{a}_{r+1} - a|&= \left| \tilde{a}_r + \tilde{\gamma }_r \frac{a-\tilde{a}_r}{\gamma + \tilde{\gamma }_r} - a \right| \\&= \left| \left( 1-\frac{\tilde{\gamma }_r}{\gamma + \tilde{\gamma }_r}\right) (\tilde{a}_r - a) \right| = \frac{\gamma }{\gamma + \tilde{\gamma }_r} |\tilde{a}_r - a|\\&\le q |\tilde{a}_r - a| \le \cdots \le q^{r+1}|\tilde{a}_0 - a| \overset{r\rightarrow \infty }{\rightarrow } 0. \end{aligned}$$

Similarly, we obtain for the sequence $\{\tilde{\gamma }_r\}_{r\in \mathbb {N}}$ ,

$$\begin{aligned}&|\tilde{\gamma }^2_{r+1} - \gamma ^2|\\&\quad = \left| \tilde{\gamma }_r \left( \gamma + \frac{(a-\tilde{a}_r)^2}{\gamma + \tilde{\gamma }_r}\right) - \gamma ^2 \right| \\&\quad = \left| \gamma (\tilde{\gamma }_r-\gamma ) +\frac{\tilde{\gamma }_r}{\gamma + \tilde{\gamma }_r}(a-\tilde{a}_r)^2 \right| \\&\quad = \left| \frac{\gamma }{\gamma + \tilde{\gamma }_r}(\tilde{\gamma }_r^2 - \gamma ^2)+\frac{\tilde{\gamma }_r}{\gamma + \tilde{\gamma }_r}(a-\tilde{a}_r)^2 \right| \\&\quad \le \frac{\gamma }{\gamma + \tilde{\gamma }_r} \left| \tilde{\gamma }_r^2 - \gamma ^2\right| + \frac{\tilde{\gamma }_r}{\gamma + \tilde{\gamma }_r}(a-\tilde{a}_r)^2 \\&\quad \le q \left| \tilde{\gamma }_r^2 - \gamma ^2\right| + (a-\tilde{a}_r)^2\\&\quad \le q \left| \tilde{\gamma }_r^2 - \gamma ^2\right| + q^{2r}(a-\tilde{a}_0)^2\\&\quad \le q\bigl (q \left| \tilde{\gamma }_r^2 - \gamma ^2\right| + q^{2(r-1)}(a-\tilde{a}_0)^2 \bigr )+ q^{2r}(a-\tilde{a}_0)^2\\&\quad \;\;\vdots \\&\quad \le q^{r+1} \left| \tilde{\gamma }_0^2 - \gamma ^2\right| + \left( \sum \limits _{i=0}^r q^{r+i}\right) (a-\tilde{a}_0)^2\\&\quad = q^{r+1} \left| \tilde{\gamma }_0^2 - \gamma ^2\right| + q^r \frac{1-q^{r+1}}{1-q}(\tilde{a}_0 - a)^2\overset{r\rightarrow \infty }{\rightarrow } 0. \end{aligned}$$

$\square$

1.4 Proof of Theorem 7:

Proof

By strict concavity of the logarithm function and since $w_i > 0$, we have

$$\begin{aligned}&L(a_{r+1},\gamma _{r+1}) - L(a_{r},\gamma _{r})\\&\quad \le \log \Biggl (\underbrace{\sum _{i=1}^n w_i\frac{(x_i-a_{r+1})^2 + \gamma _{r+1}^2}{(x_i-a_{r})^2 + \gamma _{r}^2} \frac{\gamma _{r}}{\gamma _{r+1}}}_{\varUpsilon }\Biggr ), \end{aligned}$$

with equality if and only if $(a_r,\gamma _r) = (a_{r+1},\gamma _{r+1})$. From Algorithm 4, we obtain similarly as in the proof of 2,

$$\begin{aligned} \varUpsilon&= \frac{S_{0r}^2+ S_{1r}^2}{S_{0r} (1-S_{0r}) - S_{1r}^2} \left( (1-S_{0r}) - 2 \gamma _r \frac{S_{1r}}{S_{0r}^2+ S_{1r}^2} \frac{S_{1r}}{\gamma _r} \right. \\&\quad \left. + \frac{S_{0r}S_{1r}^2}{(S_{0r}^2+ S_{1r}^2)^2} + \frac{S_{0r} \left( S_{0r} (1-S_{0r}) - S_{1r}^2 \right) ^2}{(S_{0r}^2+ S_{1r}^2)^2}\right) \\&\scriptstyle {=\frac{\left( S_{0r}^2+ S_{1r}^2\right) ^2 \left( 1-S_{0r}\right) - 2S_{1r}^2\left( S_{0r}^2+ S_{1r}^2\right) + S_{0r} S_{1r}^2 + S_{0r}\left( S_{0r} \left( 1-S_{0r}\right) - S_{1r}^2 \right) ^2 }{\left( S_{0r} \left( 1-S_{0r}\right) - S_{1r}^2 \right) \left( S_{0r}^2+ S_{1r}^2\right) }}\\&=1. \end{aligned}$$

Thus, $L(a_{r+1},\gamma _{r+1}) \le L(a_{r},\gamma _{r})$ with equality if and only if $(a_r,\gamma _r) = (a_{r+1},\gamma _{r+1})$. The convergence result follows as in part 2 of the proof of Theorem 2. $\square$

Appendix

This appendix contains the proof of Sect. 6.

1.1 Proof of Lemma 5:

Proof

Under $\mathcal {H}_0$ (i.e., $n=2$), the ML estimate is not unique, but one easily verifies using (4) that

$$\begin{aligned} \hat{\theta }&=\frac{1}{2}(x_1 + y_1)\in {{\mathrm{\mathop {\hbox {arg max}}\limits }}}_{\theta \in \mathcal {H}_0}\bigl \{\mathcal {L}(\theta |x_1,y_1)\bigr \}\\&={{\mathrm{\mathop {\hbox {arg max}}\limits }}}_{\theta \in \varTheta } \bigl \{\mathcal {L}(\theta |x_1)\mathcal {L}(\theta |y_1)\bigr \}, \end{aligned}$$

and therewith

$$\begin{aligned}&\sup _{\theta \in \mathcal {H}_0} \mathcal {L}(\theta |x_1,y_1) = \mathcal {L}(\hat{\theta }_0|x_1)\mathcal {L}(\hat{\theta }_0|y_1)\\&\quad = \frac{1}{\pi \gamma } \frac{\gamma ^2 }{\left( x_1-\frac{x_1+ y_1}{2}\right) ^2 + \gamma ^2} \frac{1}{\pi \gamma } \frac{\gamma ^2 }{\left( y_1-\frac{x_1+ y_1}{2}\right) ^2 + \gamma ^2} \\&\quad = \frac{1}{\pi ^2 \gamma ^2 }\frac{1 }{\left( \left( \frac{x_1- y_1}{2\gamma }\right) ^2 + 1\right) ^2}. \end{aligned}$$

Note that although the ML estimate $\hat{\theta }$ is not unique, the value of the log-likelihood function does not change when using another solution.

Under $\mathcal {H}_1$ (i.e., $n=1$), the ML estimate simply reads as

$$\begin{aligned} \hat{\theta }_i = \mathop {{{\mathrm{\mathop {\hbox {arg max}}\limits }}}}\limits _{\theta \in \varTheta }\mathcal {L}(\theta |x_i)=x_i,\qquad i=1,2, \end{aligned}$$

resulting in

$$\begin{aligned} \sup _{\theta \in \varTheta } \mathcal {L}(\theta |x_i)= \frac{1}{\pi \gamma } \frac{\gamma ^2 }{\left( x_i-x_i\right) ^2 + \gamma ^2}=\frac{1}{\pi \gamma },\qquad i=1,2. \end{aligned}$$

Therewith, the LR statistic becomes

$$\begin{aligned} \lambda (x_1,y_1)&= \frac{\sup \nolimits _{\theta \in \varTheta }\bigl \{\mathcal {L}(\theta |x_1)\mathcal {L}(\theta |y_1)\bigr \}}{\sup \nolimits _{\theta \in \varTheta }\bigl \{\mathcal {L}(\theta |x_1)\bigr \}\sup \nolimits _{\theta \in \varTheta }\bigl \{\mathcal {L}(\theta |y_1)\bigr \}}\\&=\frac{\frac{1}{\pi ^2 \gamma ^2 }\frac{1 }{\left( \left( \frac{x_1- y_1}{2\gamma }\right) ^2 + 1\right) ^2}}{\frac{1}{\pi \gamma } \frac{1}{\pi \gamma } }\\&=\left( \left( \frac{x_1- y_1}{2\gamma }\right) ^2 + 1\right) ^{-2}. \end{aligned}$$

$\square$