Review of Envelope Statistics Models for Quantitative Ultrasound Imaging and Tissue Characterization

Abstract

The homodyned K-distribution and the K-distribution, viewed as a special case, as well as the Rayleigh and the Rice distributions, viewed as limiting cases, are discussed in the context of quantitative ultrasound (QUS) imaging. The Nakagami distribution is presented as an approximation of the homodyned K-distribution. The main assumptions made are: (1) the absence of log-compression or application of non-linear filtering on the echo envelope of the radiofrequency signal; (2) the randomness and independence of the diffuse scatterers. We explain why other available models are less amenable to a physical interpretation of their parameters. We also present the main methods for the estimation of the statistical parameters of these distributions. We explain why we advocate the methods based on the X-statistics for the Rice and the Nakagami distributions, and the K-distribution. The limitations of the proposed models are presented. Several new results are included in the discussion sections, with proofs in the appendix.

Appendix: Proofs of the New Results

Proof of Theorem 8 From Theorem 6, there is exactly one critical point of $L(\varepsilon ,\sigma ^2)$ for which $\varepsilon >0$, and it is the MLE (here, for $L$ denotes $L_{\text{ Ri }}$). Therefore (using Theorem 5), the function $f(\kappa )$ has exactly one positive root $\kappa _*$ and it corresponds to the MLE. Moreover, one can check that $\kappa =0$ is also a root of the function $f$. Namely, we have $\lim \nolimits_{\kappa \rightarrow 0}f(\kappa )=-1+\frac{1}{N}\sum \nolimits_{i=1}^N y_i^2$, and by construction, $\frac{1}{N}\sum \nolimits_{i=1}^N y_i^2=1$.

We have $\lim \nolimits_{\kappa \rightarrow \infty }f(\kappa )=2(-1+\frac{1}{N}\sum\nolimits_{i=1}^N y_i)=2(-1+\overline{\sqrt{I}}/\sqrt{\overline{I}})$. A direct application of Cauchy-Schwartz’ inequality ensures that $\overline{\sqrt{I}}/\sqrt{\overline{I}}<1$, so that $\lim \nolimits_{\kappa \rightarrow \infty }f(\kappa )<0$. In view of the Intermediate Value Theorem for continuous functions, it follows that $f(\kappa )<0$, if $\kappa >\kappa _*$.

Next, we want to show that $f(\kappa )>0$ for $\kappa \in (0,\kappa _*)$. Since $\kappa _*$ is the only positive root of $f$, and since $f(\kappa )<0$ on $(\kappa _*,\infty )$, it is enough to show that $\frac{\partial f}{\partial \kappa }<0$ at $\kappa _*$; for then, $f(\kappa )>0$ if $\kappa <\kappa _*$ is sufficiently near $\kappa _*$, and hence, $f(\kappa )>0$ on $(0,\kappa _*)$ using the Intermediate Value Theorem.

First of all, we claim that $\frac{\partial f}{\partial \kappa }=\frac{1}{N}\frac{\partial ^2 L}{\partial \kappa ^2}$ at a critical point of $L(\varepsilon ,\sigma ^2)=\sum\nolimits_{i=1}^N \log P_{\text{ Ri }}(A_i\,|\,\varepsilon ,\sigma ^2)$, whenever $\varepsilon >0$ (i.e. $\kappa >0$). Indeed, with the change of variable $\varepsilon = \sqrt{\frac{\mu \kappa }{(\kappa +1)}}$ and $\sigma ^2=\frac{\mu }{2(\kappa +1)}$, we obtain directly from Eq. (10.2)

$$\begin{aligned}\frac{1}{N}L(\mu ,\kappa )= &\frac{1}{N}\sum _{i=1}^N \log A_i -\log \mu +\log 2 +\log (\kappa +1)\nonumber \\& - \frac{(\kappa +1)}{\mu }\frac{1}{N}\sum _{i=1}^N A_i^2 -\kappa +\frac{1}{N}\sum _{i=1}^N \log I_0(\frac{2A_i}{\sqrt{\mu }} \sqrt{\kappa (\kappa +1)}).\nonumber \\ \end{aligned}$$

(10.62)

Next, the derivative of $\frac{1}{N}L(\mu ,\kappa )$ with respect to $\kappa$ is equal to

$$\begin{aligned} \frac{1}{(\kappa +1)} - \frac{1}{\mu }\frac{1}{N}\sum _{i=1}^N A_i^2 -1 +\frac{(2\kappa +1)}{\sqrt{\kappa (\kappa +1)}}\frac{1}{N}\sum _{i=1}^N \frac{A_i}{\sqrt{\mu }}\frac{ I_1(\frac{2A_i}{\sqrt{\mu }} \sqrt{\kappa (\kappa +1)})}{I_0(\frac{2A_i}{\sqrt{\mu }} \sqrt{\kappa (\kappa +1)})}.\end{aligned}$$

(10.63)

But, from Talukdar and Lawing (1991), we have $\mu =\overline{I}=\frac{1}{N}\sum\nolimits_{i=1}^N A_i^2$ at a critical point $(\varepsilon ,a^2)$ of $L_{\text{ Ri }}$. Therefore, we obtain that $\frac{1}{N}\frac{\partial L}{\partial \kappa }=f(\kappa )$ at such a critical point (because $A_i/\sqrt{\mu }$ is then equal to $y_i=A_i/\sqrt{\overline{I}}$). Taking the partial derivative of Eq. (10.63) with respect to $\kappa$, we also see that $\frac{\partial f}{\partial \kappa }=\frac{1}{N}\frac{\partial ^2 L}{\partial \kappa ^2}$ at a critical point $(\varepsilon ,\sigma ^2)$ of $L$.

Now, recall that if $u=u(x,y)$ and $v=v(x,y)$ is a change of variable, then

$$\begin{aligned}\frac{\partial ^2 L}{\partial u \partial v}&=\frac{\partial L}{\partial x}\frac{\partial ^2 x}{\partial u \partial v}+\frac{\partial L}{\partial y}\frac{\partial ^2 y}{\partial u \partial v}\nonumber \\&+\frac{\partial ^2 L}{\partial x^2}\frac{\partial x}{\partial u} \frac{\partial x}{\partial v}+ \frac{\partial ^2 L}{\partial y \partial x}\frac{\partial y}{\partial u} \frac{\partial x}{\partial v} + \frac{\partial ^2 L}{\partial x \partial y}\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} +\frac{\partial ^2 L}{\partial y^2}\frac{\partial y}{\partial u} \frac{\partial y}{\partial v};\nonumber \\\frac{\partial ^2 L}{\partial ^2 u}&=\frac{\partial L}{\partial x}\frac{\partial ^2 x}{\partial u^2}+\frac{\partial L}{\partial y}\frac{\partial ^2 y}{\partial u^2}\nonumber \\&+\frac{\partial ^2 L}{\partial x^2}\Bigl (\frac{\partial x}{\partial u}\Bigr )^2+2 \frac{\partial ^2 L}{\partial x \partial y}\frac{\partial x}{\partial u} \frac{\partial y}{\partial u} +\frac{\partial ^2 L}{\partial y^2}\Bigl ( \frac{\partial y}{\partial u}\Bigr )^2.\end{aligned}$$

(10.64)

At this point, we find convenient to use the change of variable $\varepsilon ^2=\mu \kappa /(1+\kappa )$ and $\sigma ^2=\mu /(2(1+\kappa ))$. We develop $\frac{\partial ^2}{\partial \kappa ^2}L=\Bigl ( G_{11} -{G_{12}}+\frac{1}{4}G_{22}\Bigr )\frac{\mu ^2}{(\kappa +1)^4}$ at a critical point of $L$, where $G_{11}=\frac{\partial ^2 L}{\partial \varepsilon ^2 \partial \varepsilon ^2}$, $G_{12}=\frac{\partial ^2 L}{\partial \varepsilon ^2\partial \sigma ^2}$, and $G_{22}=\frac{\partial ^2 L}{\partial \sigma ^2\partial \sigma ^2}$ (we make use of the fact that $\frac{\partial L}{\partial \varepsilon ^2}=0=\frac{\partial L}{\partial \sigma ^2}$ at the critical point). Now, from Carrobi and Cati (2008, Appendix A, p. 686-687), we have $H_{11}H_{22} - H_{12}^2>0$ and $H_{11}<0$ at the critical point of interest, where $H_{11}=\frac{\partial ^2 L}{\partial \varepsilon \partial \varepsilon }$, $H_{12}=\frac{\partial ^2 L}{\partial \varepsilon \partial \sigma ^2}$, and $H_{22}=\frac{\partial ^2 L}{\partial \sigma ^2\partial \sigma ^2}$ ($\sigma ^2$ is viewed as a variable). From there, if one uses the change of variable $\varepsilon =\sqrt{\varepsilon ^2}$ (and $\sigma ^2=\sigma ^2$), one concludes that $G_{11}G_{22} -G_{12}^2=\Bigl ( H_{11}H_{22} - H_{12}^2\Bigr )\frac{1}{4\varepsilon ^2}>0$ and $G_{11}=H_{11}\frac{1}{4\varepsilon ^2}<0$, at that critical point. Thus, we obtain the upper bound $G_{22}<G_{12}^2/G_{11}$ (because $G_{11}<0)$, and therefore $\frac{\partial ^2 L}{\partial \kappa ^2}<\Bigl ( G_{11}-G_{12}+\frac{1}{4}G_{12}^2/G_{11}\Bigr )\frac{\mu ^2}{(\kappa +1)^4}$. But this is equal to $\frac{1}{G_{11}}\frac{\mu ^2}{(\kappa +1)^4} \Bigl ( G_{11}-\frac{1}{2}G_{12} \Bigr )^2$, and it is non-positive since $G_{11}<0$. Therefore, $\frac{\partial f}{\partial \kappa }=\frac{1}{N}\frac{\partial ^2 L}{\partial \kappa ^2}<0$ at the point $\kappa =\kappa _*$ (with $\mu =\overline{I}$). This completes the proof of Theorem 8. $\blacksquare$

Proof of Theorem 9 (a) Setting $\kappa =0$ in Theorem 7, we obtain directly $M_{\text{ Ri }}^{(\nu )}(0)=\varGamma (\nu /2+1)$.

(b) From Luke (1962, pp.7–8), we have the following asymptotic behavior

$$\begin{aligned} \, _1 F_1(a_1;b_1;z)\propto \frac{\varGamma (b_1)}{\varGamma (a_1)} z^\chi e^{z} \Bigl ( 1+\mathcal{O}(1/z)\Bigr ), \end{aligned}$$

(10.65)

where $\chi =a_1-b_1$, valid for $|\arg z|<\pi$ and $|z|\rightarrow \infty$. Therefore, we have

$$\begin{aligned} \, _1 F_1(\nu /2+1;1;\kappa )\propto \frac{1}{\varGamma (\nu /2+1)} \kappa ^{\nu /2} e^{\kappa } \Bigl ( 1+\mathcal{O}(1/\kappa )\Bigr ). \end{aligned}$$

(10.66)

We conclude that $\lim \nolimits_{\kappa \rightarrow \infty } M_{\text{ Ri }}^{(\nu )}(\kappa )=\lim \nolimits_{\kappa \rightarrow \infty } \frac{\varGamma (\nu /2+1)e^{-\kappa }}{(\kappa +1)^{\nu /2}} \times \frac{1}{\varGamma (\nu /2+1)} \kappa ^{\nu /2} e^{\kappa }=1$.

(c) From the definition $M_{\text{ Ri }}^{(\nu )}(\kappa )=\varGamma (\nu /2+1) \frac{\, _1F_1(1+\nu /2,1,\kappa )}{e^{\kappa }(\kappa +1)^{\nu /2}}$, we obtain after algebraic simplifications

$$\begin{aligned} \frac{d}{d\, \kappa } \, M_{\text{ Ri }}^{(\nu )}&= \varGamma (\nu /2+1) \frac{\frac{d}{d\, \kappa }\, _1F_1(1+\nu /2,1,\kappa )- \, _1F_1(1+\nu /2,1,\kappa )\Bigl ( 1+\frac{\nu }{2}(\kappa +1)^{-1}\Bigr )}{e^{\kappa }(\kappa +1)^{\nu /2}}.\nonumber \\ \end{aligned}$$

(10.67)

Now, from Gradshteyn and Ryshik (1994, 9.213, p.1086) and Gradshteyn and Ryshik (1994, 9.212(3), p.1086), we have $\frac{d}{d\, \kappa }\, _1F_1(1+\nu /2,1,\kappa )=(1+\nu /2) \, _1F_1(2+\nu /2,2,\kappa )= \frac{\nu }{2}\, _1F_1(1+\nu /2,2,\kappa )+\, _1F_1(1+\nu /2,1,\kappa )$. So, omitting the positive factor $\varGamma (\nu /2+1)e^{-\kappa }(\kappa +1)^{-\nu /2}$, we obtain

$$\begin{aligned} \frac{\nu }{2}\, _1F_1(1+\nu /2,2,\kappa )- \frac{\nu }{2} \, _1F_1(1+\nu /2,1,\kappa )(\kappa +1)^{-1}. \end{aligned}$$

(10.68)

Multiplying by $(\kappa +1)$ and dividing by $\nu /2$ (both are positive numbers), we obtain

$$\begin{aligned} {}_1F_1(1+\nu /2,2,\kappa )(\kappa +1)- \, _1F_1(1+\nu /2,1,\kappa ). \end{aligned}$$

(10.69)

Using (Gradshteyn and Ryshik 1994, 9.212(2), p.1086), we have $\kappa \, _1F_1(1+\nu /2,2,\kappa )-\, _1F_1(1+\nu /2,1,\kappa )=-\, _1F_1(\nu /2,1,\kappa )$. Therefore, we finally obtain (up to a positive constant)

$$\begin{aligned} \qquad \, _1F_1(1+\nu /2,2,\kappa ) - \, _1F_1(\nu /2,1,\kappa ). \end{aligned}$$

(10.70)

Now, by definition, the hypergeometric function $\, _1F_1(a,b,z)$ is equal to $\sum \nolimits_{n=0}^\infty \frac{(a)_n}{(b)_n}\frac{z^n}{n!}$, where $(a)_n=a(a+1)...(a+n-1)$ is the rising factorial. If $\nu /2<1$, then $\frac{(1+\nu /2)_n}{(2)_n}>\frac{(\nu /2)_n}{(1)_n}$ and hence $\, _1F_1(1+\nu /2,2,\gamma ) - \,_1F_1(\nu /2,1,\gamma )>0$, On the other hand, if $\nu /2>1$, then $\frac{(1+\nu /2)_n}{(2)_n}<\frac{(\nu /2)_n}{(1)_n}$ and hence $\, _1F_1(1+\nu /2,2,\kappa ) - \, _1F_1(\nu /2,1,\kappa )<0$. This completes the proof of the theorem. $\blacksquare$

Proof of Theorem 10 (a) First of all, using the change of variable $I=A^2$, one computes

$$\begin{aligned} \int _0^\infty \log A^2 P_{\text{ Ri }}(A\,|\,\varepsilon ,\sigma ^2)\, d A = \int _0^\infty \log I \frac{1}{2\sigma ^2}I_0\Bigl ( \frac{\varepsilon }{\sigma ^2} \sqrt{I} \Bigr )e^{-\varepsilon ^2/(2\sigma ^2)} e^{-I/(2\sigma ^2)}\, d I, \end{aligned}$$

(10.71)

which is a Laplace transform equal to $\varGamma (0,\frac{\varepsilon ^2}{2\sigma ^2})+\log \varepsilon ^2$, where $\varGamma (0,x)$ is the incomplete gamma function $\int _{x}^\infty \frac{e^{-t}}{t}\, d t$. Then, after subtraction by the term $\log (\varepsilon ^2+2\sigma ^2)$, one obtains $\varGamma (0,\frac{\varepsilon ^2}{2\sigma ^2})+\log (\frac{\varepsilon ^2}{\varepsilon ^2+2\sigma ^2})$, which is equal to $\varGamma (0,\kappa )+\log (\frac{\kappa }{\kappa +1})$ (where $\kappa =\varepsilon ^2/(2\sigma ^2)$).

(b) Again, using the change of variable $I=A^2$, we compute

$$\begin{aligned}&\int _0^\infty A^2 \log A^2 P_{\text{ Ri }}(A\,|\,\varepsilon ,\sigma ^2)\, d A\\&= \int _0^\infty I\, \log I \frac{1}{2\sigma ^2}I_0\Bigl ( \frac{\varepsilon }{\sigma ^2} \sqrt{I} \Bigr )e^{-\varepsilon ^2/(2\sigma ^2)} e^{-I/(2\sigma ^2)}\, d I.\nonumber \end{aligned}$$

(10.72)

This Laplace transform is equal to $4\sigma ^2-2e^{-\frac{\varepsilon ^2}{2\sigma ^2}}\sigma ^2+(\varepsilon ^2+2\sigma ^2)\bigl ( \varGamma (0,\frac{\varepsilon ^2}{2\sigma ^2})+\log \varepsilon ^2 \bigr )$. Dividing by $\varepsilon ^2+2\sigma ^2$ and subtracting $E[\log I]=\varGamma (0,\frac{\varepsilon ^2}{2\sigma ^2})+\log \varepsilon ^2$ (from the proof of part a)), one obtains $(4\sigma ^2-2e^{-\frac{\varepsilon ^2}{2\sigma ^2}}\sigma ^2)\div (\varepsilon ^2+2\sigma ^2)$, which is equal to $\frac{1}{\kappa +1}(2-e^{-\kappa })$, after algebraic simplifications. $\blacksquare$

Proof of Theorem 11 (a) From Abramowitz and Stegun (1972, (6.5.15), p. 262), we have $\varGamma (0,\kappa )=E_1(\kappa )$ (the exponential integral). Moreover, from Abramowitz and Stegun (1972, (5.1.11), p. 229), $E_1(\kappa )=-\gamma _E-\log \kappa +\sum \nolimits_{n=1}^\infty \frac{(-1)^n\kappa ^n}{nn!}$. We conclude that $U_{\text{ Ri }}(\kappa )=-\gamma _E-\log (1+\kappa )+\sum \nolimits_{n=1}^\infty \frac{(-1)^n\kappa ^n}{nn!}$. Henceforth, $\lim \nolimits_{\kappa \rightarrow 0} U_{\text{ Ri }}(\kappa )=-\gamma _E$.

(b) Since $\varGamma (0,\kappa )=\int _\kappa ^\infty \frac{e^{-t}}{t}\, d t$, it follows that $\lim \nolimits_{\kappa \rightarrow \infty }\varGamma (0,\kappa )=0$. Moreover, $\lim \nolimits_{\kappa \rightarrow \infty }\log \frac{\kappa }{\kappa +1}=0$.

(c) We compute $\frac{d}{d\, \kappa }U_{\text{ Ri }}(\kappa )=-\frac{e^{-\kappa }}{\kappa }+\frac{1}{\kappa }-\frac{1}{\kappa +1}$. This is positive because $e^\kappa >1+\kappa$, for $\kappa >0$. $\blacksquare$

Proof of Theorem 12 Parts (a) and (b) follow from basic Calculus.

(c) We compute $\frac{d}{d\, \kappa } X_{\text{ Ri }}(\kappa )=\frac{e^{-\kappa }(\kappa +1)-(2-e^{-\kappa })}{(\kappa +1)^2}$. Ignoring the positive factor $1/(\kappa +1)^2$, we obtain $e^{-\kappa }(2+\kappa )-2$. This is negative since $e^{\kappa }>1+\kappa /2$, for $\kappa >0$. $\blacksquare$

Lemma 5

Let $\alpha >0$ be fixed. Denote any root of $\frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha )$ by $\sigma ^2(\alpha ,\tilde{A})$, where $\tilde{A}=\{A_1,A_2,...,A_N\}$.

• (a) If $0<\alpha \le 1/2$, then $\sigma ^2(\alpha ,\tilde{A})\ge \frac{\alpha +\alpha ^2+\sqrt{2\alpha ^3+\alpha ^4}}{\alpha ^2}\bigl ( \overline{A}\bigr )^2$.

• (b) If $1/2 < \alpha \le 3$, then $\sigma ^2(\alpha ,\tilde{A})\ge \frac{1}{2\alpha ^2}\bigl ( \overline{A}\bigr )^2$.

• (c) If $\alpha >3$, then $\sigma ^2(\alpha ,\tilde{A})\ge \frac{2\alpha -3+\sqrt{4\alpha -7}}{4(\alpha -2)^2}\bigl ( \overline{A}\bigr )^2$.

• (d) If $0<\alpha \le 1/2$, then $\sigma ^2(\alpha ,\tilde{A})\le \frac{1}{2\alpha ^2}\bigl (\overline{A}\bigr )^2$.

• (e) If $1/2<\alpha \le 3/2$, then $\sigma ^2(\alpha ,\tilde{A})\le \frac{1}{2(\alpha /2+1/4)^2} \bigl (\overline{A}\bigr )^2$.

• (f) If $3/2<\alpha \le 3$, then $\sigma ^2(\alpha ,\tilde{A})\le \frac{1}{2} \bigl (\overline{A}\bigr )^2$.

• (g) If $\alpha >3$, then $\sigma ^2(\alpha ,\tilde{A})\le \frac{1}{2(\alpha -2)}\overline{A^2}$.

• (h) The function $\frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha )$ is positive at the lower bounds mentioned in parts (a) to (c), whereas it is negative at the upper bounds of parts (d) to (g).

Proof

We compute

$$\begin{aligned} \sigma ^2 \frac{\frac{\partial }{\partial \sigma ^2}P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )}{P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )} = -\alpha + \Bigl ( \frac{1}{\sqrt{2\sigma ^2}}A \Bigr ) \frac{K_{\alpha }\bigl ( \sqrt{\frac{2}{\sigma ^2}}A \bigr )}{K_{\alpha -1}\bigl ( \sqrt{\frac{2}{\sigma ^2}}A \bigr )}.\end{aligned}$$

(10.73)

Part a). If $0<\alpha \le 1/2$, then $K_{\alpha -1}(x)=K_{1-\alpha }(x)<K_{1}(x)$ and $K_\alpha (x)>K_0(x)$ for any $x> 0$. Also, the inequality $K_0(x)/K_1(x)>1-\frac{1}{(x+1)}$ holds for any $x>0$. Therefore, from Eq. (10.73), we obtain $\sigma ^2 \frac{\frac{\partial }{\partial \sigma ^2}P_{\text{ HK }}(A\,|\,\sigma ^2,\alpha )}{P_{\text{ HK }}(A\,|\,\sigma ^2,\alpha )} > -\alpha + \frac{1}{2}f(\sqrt{\frac{2}{\sigma ^2}}A)$, where $f(x)=x(1-\frac{1}{(x+1)})$. Thus, we obtain that

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) > -N\alpha + \frac{1}{2} \sum _{i=1}^N f(\sqrt{\frac{2}{\sigma ^2}}A_i).\end{aligned}$$

(10.74)

Here, $L$ denotes $L_{\text{ K }}$. Now, the function $f(x)$ is convex. Therefore, from Jensen’s inequality (Jensen 1906), we conclude that

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) > -N\alpha + \frac{N}{2} f(\sqrt{\frac{2}{\sigma ^2}}\overline{A}).\end{aligned}$$

(10.75)

But the right-hand side of Eq. (10.75) is positive if $\sigma ^2<\frac{\alpha +\alpha ^2+\sqrt{2\alpha ^3+\alpha ^4}}{\alpha ^2}\bigl ( \overline{A}\bigr )^2$. This proves part a).

Part b). If $\alpha >1/2$, then $K_{\alpha }(x)>K_{\alpha -1}(x)$ for any $x> 0$. Therefore, from Eq. (10.73), we obtain $\sigma ^2 \frac{\frac{\partial }{\partial \sigma ^2}P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )}{P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )} > -\alpha + \frac{1}{\sqrt{2\sigma ^2}}A$. Thus, we conclude that

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) > -N\alpha + \frac{1}{\sqrt{2\sigma ^2}}\sum _{i=1}^NA_i.\end{aligned}$$

(10.76)

But the right-hand side of Eq. (10.76) is positive if $\sigma ^2<\frac{1}{2\alpha ^2}\bigl (\overline{A}\bigr )^2$. This proves part b).

Part c). If $\alpha >3$, then $\frac{x}{2}\frac{K_\alpha (x)}{K_{\alpha -1}(x)}>\frac{x}{2}\times (\frac{2(\alpha -1)}{x}+\frac{1}{\frac{2(\alpha -2)}{x}+1})$. Thus, $\sigma ^2 \frac{\frac{\partial }{\partial \sigma ^2}P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )}{P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )}$ has lower bound $-1+f(\frac{A}{\sqrt{2\sigma ^2}})$, where $f(x)=(\frac{(\alpha -2)}{x^2}+\frac{1}{x})^{-1}$. Thus, we conclude that

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) > -N+ \sum _{i=1}^N\frac{1}{\frac{2\sigma ^2(\alpha -2)}{A_i^2}+\frac{\sqrt{2\sigma ^2}}{A_i}}.\end{aligned}$$

(10.77)

From Jensen’s inequality, we then obtain

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) > -N+N \frac{1}{\frac{2\sigma ^2(\alpha -2)}{\overline{A}^2}+\frac{\sqrt{2\sigma ^2}}{\overline{A}}},\end{aligned}$$

(10.78)

because the function $f(x)$ above is convex. But the right-hand side of Eq. (10.78) is positive if $\sigma ^2<\frac{2\alpha -3+\sqrt{4\alpha -7}}{4(\alpha -2)^2}\bigl ( \overline{A}\bigr )^2$. This proves part c).

Part d). If $0<\alpha \le 1/2$, then $K_{\alpha }(x)<K_{\alpha -1}(x)$ for any $x> 0$. Therefore, from Eq. (10.73), we obtain $\sigma ^2 \frac{\frac{\partial }{\partial \sigma ^2}P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )}{P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )} < -\alpha + \frac{1}{\sqrt{2\sigma ^2}}A$. This yields the inequality

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) < -N\alpha + \frac{1}{\sqrt{2\sigma ^2}}\sum _{i=1}^N A_i.\end{aligned}$$

(10.79)

But the right-hand side of Eq. (10.79) is negative if $\sigma ^2>\frac{1}{2\alpha ^2}\bigl ( \overline{A}\bigr )^2$. This proves part d).

Part e). If $1/2<\alpha \le 3/2$, then $\frac{K_{\alpha }(x)}{K_{\alpha -1}(x)} <1+\frac{(\alpha -1/2)}{x}$ for any $x>0$. Therefore, we have $\sigma ^2 \frac{\frac{\partial }{\partial \sigma ^2}P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )}{P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )} < -\frac{\alpha }{2}- \frac{1}{4} +\frac{1}{\sqrt{2\sigma ^2}}A$. It follows that

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) < -N(\alpha /2+1/4) +\frac{1}{\sqrt{2\sigma ^2}}\sum _{i=1}^N A_i.\end{aligned}$$

(10.80)

But the right-hand side of Eq. (10.80) is negative if $\sigma ^2>\frac{1}{2(\alpha /2+1/4)^2}\bigl ( \overline{A}\bigr )^2$. This proves part e).

Part f). If $3/2<\alpha \le 3$, then $\frac{K_{\alpha }(x)}{K_{\alpha -1}(x)}< 1+\frac{2(\alpha -1)}{x}$ for any $x>0$. Thus, we obtain $\sigma ^2 \frac{\frac{\partial }{\partial \sigma ^2}P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )}{P_{\text{ K }}(A\,|\,\sigma ^2,\alpha )} < -\alpha + (\alpha -1) +\frac{1}{\sqrt{2\sigma ^2}}A$. From there, we conclude that

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) < -N+ \frac{1}{\sqrt{2\sigma ^2}}\sum _{i=1}^N A_i.\end{aligned}$$

(10.81)

But the right-hand side of Eq. (10.81) is negative if $\sigma ^2>\frac{1}{2} \bigl ( \overline{A} \bigr )^2$. Hence, part f) of the Theorem.

Part g). If $3<\alpha$, then $\frac{x}{2}\frac{K_{\alpha }(x)}{K_{\alpha -1}(x)}< (\alpha -1)+\frac{x^2}{4(\alpha -2)}$ for any $x>0$. Therefore, we obtain

$$\begin{aligned} \sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha ) < -N+ \frac{1}{2\sigma ^2(\alpha -2)}\sum _{i=1}^N A_i^2.\end{aligned}$$

(10.82)

But the right-hand side of Eq. (10.82) is negative if $\sigma ^2>\frac{1}{2(\alpha -2)} \overline{A^2}$. Hence, part g) of the Theorem.

Finally, part h) follows from the proof of parts a) to g). $\blacksquare$

Proof of Theorem 18 From Lemma 5, for any $\alpha >0$, there exist two values $0<\sigma _1^2<\sigma _2^2$ for which $\frac{\partial }{\partial \sigma ^2}L(\sigma _1^2,\alpha )>0$ and $\frac{\partial }{\partial \sigma ^2}L(\sigma _2^2,\alpha )<0$, where $L$ denotes $L_{\text{ K }}$. Thus, by the Intermediate Value Theorem, there exists $\sigma ^2=\sigma ^2(\alpha ,\tilde{A})$ such that $\frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha )=0$. $\blacksquare$

Proof of Theorem 19 Part a). Let $0<\alpha <1/2$. In Eq. (10.32), the term $-\psi (\alpha ) +\log \bigl ( \frac{x}{2} \bigr ) - \frac{\frac{\partial }{\partial \alpha }K_{1-\alpha }(x)}{K_{1-\alpha }(x)}$ is an increasing function of $x>0$. Also, from Lemma 5 part d), we have $\sigma ^2(\alpha ,\tilde{A})\le \frac{1}{2\alpha ^2} \bigl ( \overline{A} \bigr )^2$. Therefore, we obtain $\sqrt{\frac{2}{\sigma ^2}}A_i\ge 2 \alpha \frac{A_i}{\overline{A}}$. It follows that $LB(\alpha )=-\psi (\alpha ) +\log \bigl ( \alpha \frac{A_i}{\overline{A}} \bigr ) - \frac{\frac{\partial }{\partial \alpha }K_{1-\alpha }\bigl (2 \alpha \frac{A_i}{\overline{A}}\bigr )}{K_{1-\alpha }\bigl (2 \alpha \frac{A_i}{\overline{A}}\bigr )}$ is a lower bound for that term. Now, from Abramowitz and Stegun (1972, Eq. (9.6.45), p. 377), we have $\frac{\frac{\partial }{\partial \alpha }K_{1-\alpha }(x)}{K_{1-\alpha }(x)}\approx \frac{\frac{\partial }{\partial \alpha }K_{1}(x)}{K_{1}(x)}=\frac{K_0(x)}{x\, K_1(x)}$ as $\alpha \rightarrow 0$. Moreover, from Abramowitz and Stegun (1972, Eqs. (9.6.8) and (9.6.9), p. 375), we have $\frac{K_0(x)}{x\, K_1(x)}\sim -\log x$ for small values of $x>0$. But $x=2 \alpha \frac{A_i}{\overline{A}}$ has small values for $\alpha \rightarrow 0$. Thus, we obtain $\lim \nolimits_{\alpha \rightarrow 0} \alpha LB(\alpha )=\lim \nolimits_{\alpha \rightarrow 0} \alpha \Bigl \{ -\psi (\alpha )+\log \bigl ( \alpha \frac{A_i}{\overline{A}} \bigr )+ \log \bigl ( 2 \alpha \frac{A_i}{\overline{A}} \bigr )\Bigr \}=1$. This proves part a).

Part b). First of all, we observe that there exist constants $0<C_1<C_2$, such that $\frac{1}{C_2} \le \liminf \nolimits_{\alpha \rightarrow \infty } \frac{\sigma ^2(\alpha ,\tilde{A})}{1/\alpha }\le \limsup \nolimits_{\alpha \rightarrow \infty } \frac{\sigma ^2(\alpha ,\tilde{A})}{1/\alpha } \le \frac{1}{C_1}$. The first inequality follows from Lemma 5 part c), whereas the third inequality follows from Lemma 5 part g).

Let $L$ denote $L_{\text{ K }}$. Since by definition $\frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha )\vert _{\sigma ^2(\alpha ,\tilde{A})}=0$, we might as well consider the expression $\alpha \frac{\partial }{\partial \alpha }L(\sigma ^2,\alpha )-\sigma ^2 \frac{\partial }{\partial \sigma ^2}L(\sigma ^2,\alpha )$. From Eqs. (10.32) and (10.33), each term of that expression is equal to

$$\begin{aligned}&-\alpha \psi (\alpha )+\alpha \log \Bigl ( \frac{1}{\sqrt{2\sigma ^2}}A_i \Bigr ) +\alpha \frac{\frac{\partial }{\partial \alpha }K_{\alpha -1}\Bigl (\sqrt{\frac{2}{\sigma ^2}}A_i\Bigr )}{K_{\alpha -1}\Bigl (\sqrt{\frac{2}{\sigma ^2}}A_i\Bigr )}\\&+\alpha -\Bigl ( \frac{1}{\sqrt{2\sigma ^2}}A_i \Bigr ) \frac{K_{\alpha }\bigl ( \sqrt{\frac{2}{\sigma ^2}}A_i \bigr )}{K_{\alpha -1}\bigl ( \sqrt{\frac{2}{\sigma ^2}}A_i \bigr )}.\nonumber \end{aligned}$$

(10.83)

From Abraham and Lyons (2002, Eq. (46)), we have $\frac{\frac{\partial }{\partial \alpha }K_{\alpha -1}(x)}{K_{\alpha -1}(x)} \sim \psi (\alpha -1)-\log (x/2)+\frac{x^2}{4\alpha ^2}$ for large values of $\alpha$. Also, from Abraham and Lyons (2002, Eq. (45)), we have $K_{\alpha }(x) \sim \frac{\varGamma (\alpha )}{2(x/2)^\alpha } \Bigl ( 1- \frac{(x/2)^2}{(\alpha -1)(\alpha -2)}\Bigr )^{\alpha -2}$ for large values of $\alpha$. Therefore, taking $\sigma ^2=1/(C\alpha )$, we obtain the asymptotic expression

$$\begin{aligned} -\alpha \psi (\alpha )+\alpha \psi (\alpha -1) +\frac{CA_i^2}{2} +\alpha -(\alpha -1) \frac{\Bigl ( 1- \frac{C\alpha A_i^2}{2(\alpha -1)(\alpha -2)}\Bigr )^{\alpha -2}}{\Bigl ( 1- \frac{C\alpha A_i^2}{2(\alpha -2)(\alpha -3)}\Bigr )^{\alpha -3}}. \end{aligned}$$

(10.84)

Finally, Eq. (10.84) tends to $0$ as $\alpha$ tends to infinity. This proves part b). $\blacksquare$

Proof of Theorem 20 Part a). Let $0<\alpha <1$. We consider again Eq. (10.83). Using the asymptotic forms (for small values of $x$ and of $\alpha$) $\frac{\frac{\partial }{\partial \alpha }K_{1-\alpha }(x)}{K_{1-\alpha }(x)}\approx \frac{K_0(x)}{x\, K_1(x)}$, $\frac{K_0(x)}{x\, K_1(x)}\sim -\log x$ and $(\frac{x}{2})\frac{K_0(x)}{K_1(x)} \sim - \frac{x^2}{2}\log x$, and setting $x=\sqrt{\frac{2}{\sigma ^2}}A_i$ with $\sigma ^2=1/(C\alpha )$ and $C=2/\overline{I}$, we obtain the asymptotic expression

$$\begin{aligned} -\alpha \psi (\alpha )+ \alpha \log \bigl ( \frac{1}{2}\sqrt{\alpha }\sqrt{C} A_i \bigr )+ \alpha \log \bigl ( \sqrt{\alpha }\sqrt{C} A_i \bigr )+\alpha + \alpha ^2 C A_i^2 \log \bigl ( \sqrt{\alpha }\sqrt{C} A_i \bigr ).\end{aligned}$$

(10.85)

Part a) then follows by taking the limit of Eq. (10.85) as $\alpha \rightarrow 0$.

Part b). Taking $\sigma ^2=1/(C\alpha )$, where $C=2/\overline{I}$, into Eq. (10.84), we obtain the limit $0$ as $\rightarrow \infty$. This proves part b). $\blacksquare$

Proof of Theorem 21 a) At $\alpha =0$, we have $\varGamma (\alpha +\nu /2)=\varGamma (\nu /2)$. Also, $\varGamma (\alpha )$ has a simple pole with residue $1$ at $\alpha =0$. Therefore, $\frac{\varGamma (\alpha +1/2)}{\alpha ^{\nu /2}\varGamma (\alpha )} \sim \varGamma (\nu /2) \alpha ^{1-\nu /2}$ at $\alpha \approx 0$, which shows part a).

b) Using Sterling’s formula, we have $\frac{\varGamma (\alpha +\nu /2)}{\alpha ^{\nu /2}\varGamma (\alpha )} \sim \frac{e^{-\alpha -\nu /2}(\alpha +\nu /2)^{\alpha +\nu /2-1/2}}{\alpha ^{\nu /2}e^{-\alpha }\alpha ^{\alpha -1/2}}$, which is equal to $e^{-\nu /2} \Bigl ( 1+\frac{\nu /2}{\alpha }\Bigr )^\alpha \Bigl ( 1+\frac{\nu /2}{\alpha }\Bigr )^{\nu /2-1/2}$. Therefore, $\lim \nolimits_{\alpha \rightarrow \infty } \frac{\varGamma (\alpha +\nu /2)}{\alpha ^{\nu /2}\varGamma (\alpha )}=1$, and we are done.

c) Using the logarithmic derivative, we have $\frac{d}{d\, \alpha }M_{\text{ K }}^{(\nu )}(\alpha )=M_{\text{ K }}^{(\nu )}(\alpha ) \Bigl ( \psi (\alpha +\nu /2) -\psi (\alpha ) - \frac{\nu }{2\alpha } \Bigr )$. Now, we have $M_{\text{ K }}^{(\nu )}(\alpha )>0$. Also, $\frac{1}{\alpha }=\psi (\alpha +1)-\psi (\alpha )$, and hence $\psi (\alpha +\nu /2) -\psi (\alpha ) - \frac{\nu }{2\alpha }=\psi (\alpha +\nu /2)-\Bigl (\frac{\nu }{2} \psi (\alpha +1)+(1-\frac{\nu }{2})\psi (\alpha )\Bigr )$. Since the function $\psi$ is convex, we conclude that $\psi (\alpha +\nu /2)-\Bigl (\frac{\nu }{2} \psi (\alpha +1)+(1-\frac{\nu }{2})\psi (\alpha )\Bigr )>0$, if $\nu /2<1$, whereas it is negative if $\nu /2>1$. $\blacksquare$

Proof of Theorem 22 a) We consider the function $f(\alpha )=\frac{\varGamma (\nu +1)\varGamma (\nu +\alpha )\varGamma (\alpha )}{\varGamma ^2(\nu /2+1)\varGamma ^2(\nu /2+\alpha )}$, noting that $R_{\text{ K }}^{(\nu )}=\bigl (f(\alpha )-1\bigr )^{-1/2}$. Now, as $\alpha \rightarrow 0$, we have $\varGamma (\alpha )\rightarrow \infty$, whereas $\frac{\varGamma (\nu +1)\varGamma (\nu +\alpha )}{\varGamma ^2(\nu /2+1)\varGamma ^2(\nu /2+\alpha )}\rightarrow \frac{\varGamma (\nu +1)\varGamma (\nu )}{\varGamma ^2(\nu /2+1)\varGamma ^2(\nu /2)}>0$. This proves part a).

b) Next, using directly Sterling’s formula for $\varGamma (\nu +\alpha )$, $\varGamma (\alpha )$ and $\varGamma (\nu /2+\alpha )$, one finds that $\lim \nolimits_{\alpha \rightarrow \infty }f(\alpha )=\frac{\varGamma (\nu +1)}{\varGamma ^2(\nu /2+1)}$, which proves part b).

c) Finally, taking the logarithmic derivative of $f(\alpha )$ yields $\frac{d\, f}{d\, \alpha }=f(\alpha )\bigl ( \psi (\nu +\alpha ) + \psi (\alpha ) - 2 \psi (\nu /2+\alpha )\bigr )$. This is negative, since $f(\alpha )>0$ and $\psi (\nu /2+\alpha )>\frac{1}{2}(\psi (\nu +\alpha ) + \psi (\alpha ))$ (because $\psi$ is a convex function). It follows that $f(\alpha )> \lim \nolimits_{\alpha \rightarrow \infty }f(\alpha )=\frac{\varGamma (\nu +1)}{\varGamma ^2(\nu /2+1)}$. We claim that $g(\nu )=\frac{\varGamma (\nu +1)}{\varGamma ^2(\nu /2+1)}>1$, for any $\nu >0$. In fact, the function $g(\nu )$ is increasing (its derivative is equal to $g(\nu )(\psi (\nu +1)-\psi (\nu /2+1))$) and $g(0)=1$. Therefore, $f(\alpha )>1$, and it follows that $\bigl (f(\alpha )-1\bigr )^{-1/2}$ is an increasing function. This completes the proof of part c). $\blacksquare$

Proof of Theorem 23 This is an immediate consequence of Theorem 28. $\blacksquare$

Proof of Theorem 24 Starting with Eq. (10.44), we compute

$$\begin{aligned}M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )&= \frac{\varGamma (\nu /2+1)}{(\gamma +\alpha )^{\nu /2}} \int _0^\infty w^{\nu /2} \, _1F_1(-\nu /2,1,-\frac{\gamma }{w}) \mathcal{G}(w\,|\,\alpha ,1)\, dw\\ & =\frac{\varGamma (\nu /2+1)}{(\gamma +\alpha )^{\nu /2}} \int _0^\infty w^{\nu /2} e^{-\gamma /w}\, _1F_1(1+\nu /2,1,\frac{\gamma }{w}) \mathcal{G}(w\,|\,\alpha ,1)\, dw\nonumber \\& =\frac{1}{(\gamma +\alpha )^{\nu /2}} \frac{\varGamma (\nu /2+1)}{\varGamma (\alpha )} \sum _{n=0}^\infty \frac{(1+\nu /2)_n}{n!n!}\gamma ^n \int _0^\infty w^{\nu /2+\alpha -n-1}e^{-\gamma /w}e^{-w}\, d w.\nonumber \end{aligned}$$

(10.86)

Using (Erdélyi 1954, I, p. 146, (29)), this is equal to

$$\begin{aligned}&\frac{1}{(\gamma +\alpha )^{\nu /2}} \frac{\varGamma (\nu /2+1)}{\varGamma (\alpha )} \sum _{n=0}^\infty \frac{(1+\nu /2)_n}{n!n!}\gamma ^n 2 (\sqrt{\gamma })^{\nu /2+\alpha -n} K_{\nu /2+\alpha -n}(2\sqrt{\gamma })\nonumber \\&=\frac{2}{(\gamma +\alpha )^{\nu /2}} \frac{\varGamma (\nu /2+1)}{\varGamma (\alpha )} \sum _{n=0}^\infty \frac{(1+\nu /2)_n}{n!n!}(\sqrt{\gamma })^{\nu /2+\alpha +n} K_{\nu /2+\alpha -n}(2\sqrt{\gamma }).\nonumber \\ \end{aligned}$$

(10.87)

This completes the proof of Theorem 25. $\blacksquare$

Proof of Theorem 26 a) This follows from Theorem 24, part b), and Theorem 13.

b) From Theorem 24, part a), we know that $M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )$ is equal to $\frac{\varGamma (\nu /2+1)}{(\gamma +\alpha )^{\nu /2}} \int _0^\infty w^{\nu /2}\, _1F_1(-\nu /2,1,-\gamma /w)\mathcal{G}(w\,|\,\alpha ,1)\, d w$. From Luke (1962, pp. 7–8), we have $\varGamma (\nu /2+1)\, _1F_1(-\nu /2,1,-z)=\varGamma (\nu /2+1)e^{-z}\, _1F_1(1+\nu /2,1,z) =z^{\nu /2}(1+O(1/z))$, for large values of $z$. Let $\eta >0$ be a real number (arbitrarily small). Take $z_0$ sufficiently large so that $(1-\eta ) z^{\nu /2}\le \varGamma (\nu /2+1)\, _1F_1(-\nu /2,1,-z)\le (1+\eta ) z^{\nu /2}$, for any $z\ge z_0$. Then, if $\gamma /w\ge z_0$, i.e. $w\le \gamma /z_0$, we have $(1-\eta ) \gamma ^{\nu /2}\le \varGamma (\nu /2+1)w^{\nu /2}\, _1F_1(-\nu /2,1,-\gamma /w)\le (1+\eta ) \gamma ^{\nu /2}$. Therefore, the integral $\frac{\varGamma (\nu /2+1)}{(\gamma +\alpha )^{\nu /2}} \int _0^{\gamma /z_0} w^{\nu /2}\, _1F_1(-\nu /2,1,-\gamma /w)\mathcal{G}(w\,|\,\alpha ,1)\, d w$ has lower bound $(1-\eta )\frac{\gamma ^{\nu /2}}{(\gamma +\alpha )^{\nu /2}}Pr(w\le \gamma /z_0)$ and upper bound $(1+\eta )\frac{\gamma ^{\nu /2}}{(\gamma +\alpha )^{\nu /2}}Pr(w\le \gamma /z_0)$. On the other hand, the function $\, _1F_1(-\nu /2,1,-z)$ equals $1$ at $z=0$, and hence there is a real number $C>0$ such that $0<\, _1F_1(-\nu /2,1,-\gamma /w)\le C$ for any $w>\gamma /z_0$. Thus, the integral $\frac{\varGamma (\nu /2+1)}{(\gamma +\alpha )^{\nu /2}} \int _{\gamma /z_0}^\infty w^{\nu /2}\, _1F_1(-\nu /2,1,-\gamma /w)\mathcal{G}(w\,|\,\alpha ,1)\, d w$ has lower bound $0$ and upper bound $\frac{\varGamma (\nu /2+1)}{(\gamma +\alpha )^{\nu /2}} C \frac{\varGamma (\nu /2+\alpha )}{\varGamma (\alpha )}$. But now, $\lim \nolimits_{\gamma \rightarrow \infty } \frac{\gamma ^{\nu /2}}{(\gamma +\alpha )^{\nu /2}}=1$, $\lim \nolimits_{\gamma \rightarrow \infty } Pr(w\le \gamma /z_0)=1$, and $\lim \nolimits_{\gamma \rightarrow \infty } \frac{1}{(\gamma +\alpha )^{\nu /2}}=0$. Therefore, we obtain $\liminf \nolimits_{\gamma \rightarrow \infty }M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )\ge 1-\eta$ and $\limsup \nolimits_{\gamma \rightarrow \infty }M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )\le 1+\eta$. Since $\eta$ is arbitrarily small, we conclude that $\lim \nolimits_{\gamma \rightarrow \infty }M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )=1$.

c) We consider the function $f(\gamma ,w)=\frac{\, _1F_1(1+\nu /2,1,\gamma /w)}{e^{\gamma /w}(\gamma +\alpha )^{\nu /2}}$. From Theorem 24, part a), we have $M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )=\varGamma (\nu /2+1) \int _0^\infty w^{\nu /2}f(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w$. Thus, we obtain $\frac{\partial }{\partial \, \gamma } M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )= \varGamma (\nu /2+1) \int _0^\infty w^{\nu /2}\frac{\partial }{\partial \, \gamma } f(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w$.

We compute the value of $\frac{\partial }{\partial \, \gamma } \, f(\gamma ,w)$ as

$$\begin{aligned} \frac{\frac{d}{d\, z}\, _1F_1(1+\nu /2,1,\gamma /w)w^{-1}- \, _1F_1(1+\nu /2,1,\gamma /w)\Bigl ( w^{-1}+\frac{\nu }{2}(\gamma +\alpha )^{-1}\Bigr )}{e^{\gamma /w}(\gamma +\alpha )^{\nu /2}}.\end{aligned}$$

(10.88)

Using (Gradshteyn and Ryshik 1994, 9.213, p. 1086) and (Gradshteyn and Ryshik 1994, 9.212(3), p. 1086), we have $\frac{d}{d\, z}\, _1F_1(1+\nu /2,1,\gamma /w)=(1+\nu /2) \, _1F_1(2+\nu /2,2,\gamma /w)= \frac{\nu }{2}\, _1F_1(1+\nu /2,2,\gamma /w)+\, _1F_1(1+\nu /2,1,\gamma /w)$. So, we obtain after algebraic simplifications

$$\begin{aligned} \frac{\nu /2}{e^{\gamma /w}(\gamma +\alpha )^{\nu /2+1}} \Bigl \{ \, _1F_1(1+\nu /2,2,\gamma /w)(\frac{\gamma }{w}+\frac{\alpha }{w})- \, _1F_1(1+\nu /2,1,\gamma /w) \Bigr \}.\end{aligned}$$

(10.89)

Using (Gradshteyn and Ryshik (1994), 9.212(2), p. 1086), we have $\frac{\gamma }{w}\, _1F_1(1+\nu /2,2,\gamma /w)-\, _1F_1(1+\nu /2,1,\gamma /w)=-\, _1F_1(\nu /2,1,\gamma /w)$. Therefore, we finally obtain

$$\begin{aligned} \frac{\alpha \nu /2}{e^{\gamma /w}(\gamma +\alpha )^{\nu /2+1}w} \Bigl \{ \, _1F_1(1+\nu /2,2,\gamma /w)- \frac{w}{\alpha }\, _1F_1(\nu /2,1,\gamma /w) \Bigr \}.\end{aligned}$$

(10.90)

Now, let $\nu /2<1$. Then, we obtain the strict lower bound for $\frac{\partial }{\partial \, \gamma } \, f(\gamma ,w)$

$$\begin{aligned} \frac{\alpha \nu /2}{e^{\gamma /w}(\gamma +\alpha )^{\nu /2+1}w} \Bigl \{ \, _1F_1(1+\nu /2,2,\gamma /w)- \frac{w}{\alpha }\, _1F_1(1+\nu /2,2,\gamma /w) \Bigr \}. \end{aligned}$$

(10.91)

Consider the function $g(\gamma ,w)=\varGamma (\nu /2+1)w^{\nu /2}\frac{\alpha \nu /2}{e^{\gamma /w}(\gamma +\alpha )^{\nu /2+1}w} \, _1F_1(1+\nu /2,2,\gamma /w)$. We have shown that

$$\begin{aligned} \frac{\partial }{\partial \, \gamma } M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )> \int _0^\infty g(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty \frac{w}{\alpha }g(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w. \end{aligned}$$

(10.92)

But, $\frac{w}{\alpha }\mathcal{G}(w\,|\,\alpha ,1)=\mathcal{G}(w\,|\,\alpha +1,1)$. So, we obtain

$$\begin{aligned} \frac{\partial }{\partial \, \gamma } M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )> \int _0^\infty g(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty g(\gamma ,w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w. \end{aligned}$$

(10.93)

Thus, we want to show that $\int _0^\infty g(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty g(\gamma ,w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w\ge 0$. Ignoring the positive factor $\varGamma (\nu /2+1)\frac{\alpha \nu /2}{(\gamma +\alpha )^{\nu /2+1}}$, we are thus lead to the function $h(\gamma ,w)=\frac{w^{\nu /2-1}\, _1F_1(1+\nu /2,2,\gamma /w)}{e^{\gamma /w}}$, and we show that $\int _0^\infty h(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty h(\gamma ,w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w\ge 0$ as follows. In Lemma 6, we show that $h(\gamma ,w)$ is decreasing in the variable $w$, if $\nu /2<1$. Then, in Lemma 7, we show that for any decreasing positive function $H(w)$, we have $\int _0^\infty H(w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty H(w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w\ge 0$.

Next, let $\nu /2>1$. Then, we obtain the strict upper bound for $\frac{\partial }{\partial \, \gamma } \, f(\gamma ,w)$

$$\begin{aligned} \frac{\alpha \nu /2}{e^{\gamma /w}(\gamma +\alpha )^{\nu /2+1}w} \Bigl \{ \, _1F_1(1+\nu /2,2,\gamma /w)- \frac{w}{\alpha }\, _1F_1(1+\nu /2,2,\gamma /w) \Bigr \}. \end{aligned}$$

(10.94)

The same argument as above (but with reversed inequalities) leads to

$$\begin{aligned} \frac{\partial }{\partial \, \gamma } M_{\text{ HK }}^{(\nu )}(\gamma ,\alpha )< \int _0^\infty g(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty g(\gamma ,w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w, \end{aligned}$$

(10.95)

where the function $g(\gamma ,w)$ is defined as above. So, in this case, we want to show that $\int _0^\infty h(\gamma ,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty h(\gamma ,w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w\le 0$, where $h(\gamma ,w)$ is defined as above. But, this is implied by Lemmas 6 and 7 (case $\nu /2>1$). This completes the proof of the theorem. $\blacksquare$

Lemma 6

a) If $\nu /2<1$, the function $h(\gamma ,w)=\frac{w^{\nu /2-1}\, _1F_1(1+\nu /2,2,\gamma /w)}{e^{\gamma /w}}$ is decreasing in the variable $w$.

b) If $\nu /2>1$, the function $h(\gamma ,w)$ is increasing in the variable $w$.

Proof

Using the change of variable $x=\gamma /w$, we consider the function $F(x)=\frac{\, _1F_1(1+\nu /2,2,x)}{e^{x}x^{\nu /2-1}}$. So, we want to show that $F(x)$ is increasing if $\nu /2<1$ and $F(x)$ is decreasing if $\nu /2>1$ (the function $x=\gamma /w$ is decreasing in the variable $w$).

We compute

$$\begin{aligned} \frac{d}{d\, x} \, F(x)= \frac{\frac{d}{d\, z}\, _1F_1(1+\nu /2,2,x)- \, _1F_1(1+\nu /2,2,x)\Bigl ( 1+(\frac{\nu }{2}-1)x^{-1}\Bigr )}{e^{x}x^{\nu /2-1}}.\end{aligned}$$

(10.96)

Using (Gradshteyn and Ryshik 1994, 9.213, p. 1086) and (Gradshteyn and Ryshik 1994, 9.212(3), p. 1086), we have $\frac{d}{d\, z}\, _1F_1(1+\nu /2,2,x)=\frac{(1+\nu /2)}{2} \, _1F_1(2+\nu /2,3,x)= \frac{(\nu /2-1)}{2}\, _1F_1(1+\nu /2,3,x)+\frac{2}{2}\, _1F_1(1+\nu /2,2,x)$.

So, we obtain after algebraic simplifications

$$\begin{aligned} \frac{(\nu /2-1)}{e^{x}x^{\nu /2}} \Bigl \{ \frac{x}{2}\, _1F_1(1+\nu /2,3,x)- \, _1F_1(1+\nu /2,2,x) \Bigr \}.\end{aligned}$$

(10.97)

Using (Gradshteyn and Ryshik 1994, 9.212(2), p. 1086), we finally obtain

$$\begin{aligned} - \frac{(\nu /2-1)}{e^{x}x^{\nu /2}} \, _1F_1(1+\nu /2,1,x).\end{aligned}$$

(10.98)

The result is now clear. $\blacksquare$

Lemma 7

a) Let $H(w)$ be a decreasing positive function. Then, one has $\int _0^\infty H(w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty H(w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w\ge 0$.

b) Let $H(w)$ be an increasing positive function. Then, one has

$\int _0^\infty H(w)\mathcal{G}(w\,|\,\alpha ,1)\, d w-\int _0^\infty H(w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w\le 0$.

Proof

a) Since $H(w)$ is a positive decreasing function, we can approximate it by functions of the form $\sum \nolimits_{n=1}^N a_n \, B(b_n,w)$, where $a_n\ge 0$, $b_n>0$, and $B(b,w)$ is equal to $1$, if $w\in [0,b]$, and $B(b,w)=0$, if $w>b$. Now, $\int _0^\infty B(b,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w=\int _0^b \mathcal{G}(w\,|\,\alpha ,1)\, d w=1-\frac{\varGamma (\alpha ,b)}{\varGamma (\alpha )}$, where $\varGamma (\alpha ,b)$ is the incomplete Euler gamma function. But the function $1-\frac{\varGamma (\alpha ,b)}{\varGamma (\alpha )}$ is decreasing. Therefore, $\int _0^\infty B(b,w)\mathcal{G}(w\,|\,\alpha ,1)\, d w>\int _0^\infty B(b,w)\mathcal{G}(w\,|\,\alpha +1,1)\, d w$, and we are done.

b) Since $H(w)$ is a positive increasing function,, we can approximate $H(w)$ by functions of the form $\sum\nolimits _{n=1}^N a_n \, (1-B(b_n,w))$, where $a_n\ge 0$, $b_n>0$. Now, $\int _0^\infty (1-B(b,w))\mathcal{G}(w\,|\,\alpha ,1)\, d w=\frac{\varGamma (\alpha ,b)}{\varGamma (\alpha )}$, and we are done. $\blacksquare$

Proof of Corollary 2

Proof of Corollary 2. Let $\nu /2<1$. Since $0<M<1$, we conclude from Theorem 26, using the Intermediate Value Theorem, that for any $\alpha >0$ such that $M_\mathrm{K }(\alpha ) \le M$, there is a unique value of $\gamma \ge 0$ for which $M_\mathrm{HK }^{(\nu )}(\gamma ,\alpha )=M$. Thus, if $M \ge \Gamma (\nu /2+1)$, $\alpha$ has no restrictions, because $M_\mathrm{K }^{(\nu )}(\alpha )<\Gamma (\nu /2+1)$ for any $\alpha >0$ (Theorem 21). On the other hand, if $M_\mathrm{K }(\alpha ) < M$, let $\alpha _0$ be the unique solution to the equation $M_\mathrm{K }^{(\nu )}(\alpha _0)=M$ (Theorem 21). Then, using once more Theorem 21, we obtain that $M_\mathrm{K }^{(\nu )}(\alpha )< M$ if and only if $\alpha \le \alpha _0$. Henceforth, if $M_\mathrm{K }^{(\nu )}(\alpha ) < M$, the domain of the function $\gamma _M^{(\nu )}(\alpha )$ is the interval $(0,\alpha _0]$

The case $\nu /2>1$ is handled similarly, but with reversed inequalities. $\blacksquare$

Proof of Lemma 4

Proof of Lemma 4. Part a). From the definition in Eq. (10.6), the distribution $P_{\text{ HK }}(A\,|\,\varepsilon ,\sigma ^2,\alpha )$ is equal to $\int _0^\infty P_{\text{ Ri }}(A\,|\,\varepsilon ,\sigma ^2 w) \mathcal{G}(w\,|\,\alpha ,1) d\, w$. Using the identity $I_0(z)=\frac{1}{\pi }\int _{0}^{\pi }e^{z \cos \theta }d\, \theta$ from Abramowitz and Stegun (1972, Eq. (9.6.16), p. 376) and the definition of the Rice distribution (10.2) , we can express $P_{\text{ Ri }}(A\,|\,\varepsilon ,\sigma ^2 w)$ in the form $\frac{1}{\pi }\int _{0}^{\pi } \frac{A}{\sigma ^2w} \exp \Bigl (\frac{\varepsilon }{\sigma ^2w} A \cos \theta \Bigr ) \exp \Bigl ( -\frac{(\varepsilon ^2+A^2)}{2 \sigma ^2w} \Bigr )d\, \theta$. It follows that $P_{\text{ HK }}(A\,|\,\varepsilon ,\sigma ^2,\alpha )$ can be written as

$$\begin{aligned} \frac{1}{\pi } \int _{0}^{\pi } \Bigl \{ \int _0^\infty \frac{A}{\sigma ^2w} \exp \Bigl (\frac{\varepsilon }{\sigma ^2w} A \cos \theta \Bigr ) \exp \Bigl ( -\frac{(\varepsilon ^2+A^2)}{2 \sigma ^2w}\Bigr ) \mathcal{G}(w\,|\,\alpha ,1) d\, w \Bigr \} d\ \theta , \end{aligned}$$

(10.99)

which yields Eq. (10.56) after evaluation of the inner integral.

Part b). Using Eq. (10.56), the partial derivative of the homodyned K-distribution with respect to $\varepsilon$ is equal to

$$\begin{aligned}&\frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \frac{\partial }{\partial \varepsilon } \Bigl \{ \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) \Bigr \} d\ \theta \\&= \frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl \{ \frac{(\alpha -1)}{2}\Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -2} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) \nonumber \\&\qquad + \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} \frac{d}{d\, z}K_{\alpha -1}\bigl ( X(\theta ) \bigr ) \Bigr \} \frac{\partial }{\partial \varepsilon } X(\theta ) d\ \theta \nonumber \\&= -\frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} K_{\alpha -2}\bigl ( X(\theta ) \bigr ) 2\frac{(\varepsilon - A \cos \theta )}{\sigma ^2 X(\theta )} d\ \theta .\nonumber \end{aligned}$$

(10.100)

Here, we have used the identity $z\frac{d}{d\, z} K_{\alpha -1}(z)+(\alpha -1)K_{\alpha -1}(z)=-z\, K_{\alpha -2}(z)$ (Abramowitz and Stegun 1972, Eq. (9.6.26), 2nd identity, p. 376) and algebraic simplifications.

Part c). Using Eq. (10.56), the partial derivative of the homodyned K-distribution with respect to $\sigma ^2$ is equal to

$$\begin{aligned}&\frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\varGamma (\alpha )} \frac{\partial }{\partial \sigma ^2} \Bigl \{\frac{1}{\sigma ^2}\Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) \Bigr \} d\ \theta \\&\quad =-\frac{1}{\sigma ^2}\frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) d\ \theta \nonumber \\&\qquad+ \frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl \{ \frac{(\alpha -1)}{2}\Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -2} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) \nonumber \\&\qquad+ \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} \frac{d}{d\, z}K_{\alpha -1}\bigl ( X(\theta ) \bigr ) \Bigr \} \frac{\partial }{\partial \sigma ^2} X(\theta ) d\ \theta \nonumber \\&\quad= -\frac{1}{\sigma ^2}\frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) d\ \theta \nonumber \\&\qquad-\frac{1}{\sigma ^2} \frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl \{ (\alpha -1)\Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -2} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) \nonumber \\&\qquad - \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1}K_{\alpha }\bigl ( X(\theta ) \bigr ) \Bigr \} \Bigl ( \frac{X(\theta )}{2} \Bigr ) d\ \theta \nonumber \\&\quad= -\frac{\alpha }{\sigma ^2}\frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha -1} K_{\alpha -1}\bigl ( X(\theta ) \bigr ) d\ \theta \nonumber \\&\qquad+ \frac{1}{\sigma ^2}\frac{1}{\pi } \int _{0}^{\pi } \frac{2A}{\sigma ^2\varGamma (\alpha )} \Bigl ( \frac{X(\theta )}{2} \Bigr )^{\alpha } K_{\alpha }\bigl ( X(\theta ) \bigr ) d\ \theta \nonumber .\end{aligned}$$

(10.101)

Here, we have used the identity $\frac{z}{2}\frac{d}{d\, z} K_{\alpha -1}(z)=-\frac{z}{2} K_{\alpha }(z)+\frac{(\alpha -1)}{2}K_{\alpha -1}(z)$ (Abramowitz and Stegun 1972, Eq. (9.6.26), 4th identity, p. 376) and algebraic simplifications.

Part d). Eq. (10.59) follows from part a) upon taking the logarithmic derivative of the integrand in Eq. (10.56). $\blacksquare$