SB-robust estimation of mean direction for some new circular distributions

Abstract

The most often used distribution for modelling directional data has been the circular normal (CN) (a.k.a. von-Mises) distribution. Recently Kato and Jones (K–J) introduced a family of distribution which includes the CN distribution as a special case. We study the SB-robustness of the circular mean functional (CMF) and show that the CMF is not SB-robust at the family of all symmetric Kato–Jones distributions but is SB-robust at sub-families with bounded parameters. It is also found to be SB-robust for certain sub-families of wrapped-t (WT) distributions, mixtures of K–J distributions and mixtures of K–J and WT distributions. The SB-robustness of the circular trimmed mean functional (CTMF) is also studied and it is found that the CTMF is SB-robust for larger sub-families of symmetric Kato–Jones distributions compared to that of CMF. The SB-robustness of the CMF for asymmetric families of distributions is studied and it is shown that CMF is SB-robust at a sub-family of asymmetric Kato–Jones distributions. The performance of CTM is compared with that of circular mean (CM) through extensive simulation. It is seen that CTM has better robustness properties than the CM both theoretically and practically. Some guidelines for choice of trimming proportion for CTM is given.

Appendix

Lemma 1

Consider the family of distributions$\mathfrak {I}=\left\{ KJ\left( {0,\kappa ,r} \right) :\kappa >0,0\le r\le r_0 <1 \right\}$. Let $S(F)=\sqrt{1-\rho _F }$and $S^{*}(F)=E_F (d(\Theta ,0))$for $\mathrm{F}\in \mathfrak {I}$. Then $S\mathop \sim \limits ^\mathfrak {I}S^{*}$.

Proof

We note that

$$\begin{aligned} S(F)=\sqrt{1-\frac{1}{2\pi \mathrm{I}_0 \left( \kappa \right) }\int \limits _0^{2\pi } {\cos \theta \left( {\frac{\left( {1-\mathrm{r}^{2}} \right) }{1+\mathrm{r}^{2}-2\mathrm{r}\cos \theta }\exp \left( {\frac{\kappa \left\{ {\left( {1+\mathrm{r}^{2}} \right) \cos \theta -2\mathrm{r}} \right\} }{1+\mathrm{r}^{2}-2\mathrm{r}\cos \theta }} \right) } \right) d\theta } } \end{aligned}$$

is a continuous function of $\upkappa$ and r. Also,

$$\begin{aligned} S^{*}(F)=\frac{1}{2\pi \mathrm{I}_0 \left( \kappa \right) }\int \limits _0^{2\pi } {\left| {\pi -\left| {\pi -\theta } \right| } \right| \left( {\frac{\left( {1-\mathrm{r}^{2}} \right) }{1+\mathrm{r}^{2}-\mathrm{{2r\, cos}}\theta }\mathrm{exp}\left( {\frac{\kappa \left\{ {\left( {1+\mathrm{r}^{2}} \right) \mathrm{cos}\theta -\mathrm{2r}} \right\} }{1+\mathrm{r}^{2}-\mathrm{{2r\, cos}}\theta }} \right) } \right) d\theta } \end{aligned}$$

is a continuous function of $\kappa$ and r which is positive on the domain $\kappa >0$ and $0\le r\le r_0 <1$.

Hence, $R(F)=\frac{S(F)}{S^{*}(F)}$ is a continuous function of $\kappa$ and r on the domain $\kappa >0$ and $0\le r\le r_0 <1$. When $r=0$ we have $\mathop {\lim }\limits _{\kappa \rightarrow \infty } R(F)=\sqrt{\pi }$ and $\mathop {\lim }\limits _{\kappa \rightarrow 0} R(F)=\frac{2}{\pi }$ (see proof of Theorem 5.2 of Laha and Mahesh 2011). When $r\ne 0$ and $\upkappa \rightarrow \infty$ using the fact that $KJ\left( {0,\kappa ,r} \right)$ tends to the $N\left( {0,\kappa ^{-1}\omega _r^2 } \right)$ where $\omega _\mathrm{r} =\frac{1-r}{1+r}$, we have

$$\begin{aligned} S\left( F \right) =\sqrt{1-\exp \left( {-\frac{\omega _r^2 }{2\kappa }} \right) }. \end{aligned}$$

Expanding and simplifying we get,

$$\begin{aligned} S(F)=\frac{\omega _r }{\sqrt{2\kappa }}\sqrt{1-\frac{1}{2!}\left( {\frac{\omega _r^2 }{2\kappa }} \right) +\frac{1}{3!}\left( {\frac{\omega _r^2 }{2\kappa }} \right) ^{2}+o\left( {(2\kappa )^{-4}} \right) }. \end{aligned}$$

Straightforward calculations show that $S^{*}\left( F \right) =\omega _r \sqrt{\frac{2}{\pi \kappa }}$ . Then we have $\mathop {\lim }\limits _{\kappa \rightarrow \infty } R(F)=\frac{\sqrt{\pi }}{2}.$ Again, when $r\ne 0$ and $\kappa \rightarrow 0$, using the fact $\mathrm{KJ}\left( {\mathrm{0,}\,\kappa \mathrm{,r}} \right)$ tends to $WC\left( {0,\mathrm{r}} \right)$, we have $S(F)=\sqrt{1-r}$ and $S^{*}(F)=\frac{1-r^{2}}{2\pi }\int \limits _0^{2\pi } {\frac{\left| {\pi -|\pi -\theta |} \right| }{1+r^{2}-2r\cos \theta }d\theta }$. Using the series expansion of $\frac{1-r^{2}}{1+r^{2}-2r\cos \theta }=1+2\sum \limits _{p=1}^\infty {(r)^{p}} \cos p\theta$ (see Gradshteyn and Ryzhik 2000), we get $S^{*}(F)=\frac{\pi }{2}-\frac{4}{\pi }\sum \limits _{n=0}^\infty {\frac{r^{(2n+1)}}{\left( {2n+1} \right) ^{2}}}$.

Therefore, $\mathop {\lim }\limits _{\kappa \rightarrow 0} R(F)=\sqrt{1-r}\left( {\frac{\pi }{2}-\frac{4}{\pi }\sum _{n=0}^\infty {\frac{r^{(2n+1)}}{\left( {2n+1} \right) ^{2}}} } \right) ^{-1}$.

From the above we can conclude that $\mathop {\mathrm{sup}}\limits _\mathfrak {I}\mathrm{R}\left( \mathrm{F} \right) <\infty$. Also, it can be seen that $\mathop {\mathrm{sup}}\limits _\mathfrak {I}\mathrm{R}^{-1}\left( \mathrm{F} \right) <\infty$. Hence $S\mathop \sim \limits ^\mathfrak {I}S^{*}$.

Hence the lemma is established. $\square$

Lemma 2

Consider the family of distributions$\mathfrak {I}_1^*=\left\{ \alpha \mathrm{KJ}(0,\kappa _1 ,r_1 )+ (1-\alpha )\mathrm{{KJ}}(0,\,\kappa _2 ,r_2 )\mathrm{; 0}<\mathrm{m}\le \kappa _1 ,\kappa _2 \le M,\mathrm{0}\le \mathrm{r}_1 \mathrm{,r}_2 \le \mathrm{r}_0<1,0\le \alpha <1 \right\} .$Then $S\mathop \sim \limits ^{\mathfrak {I}_1^*} S^{*}$.

Proof

Let $D^{*}=\left\{ (\kappa _1 ,\kappa _2 \mathrm{,r}_1 \mathrm{,r}_2 \mathrm{,}\alpha \mathrm{):0}<\mathrm{m}\le \kappa _1 ,\kappa _2 \le M,\mathrm{0}\le \mathrm{r}_1 \mathrm{,r}_2 \le \mathrm{r}_0<1,0\le \alpha <1 \right\}$. Since the p.d.f of $\alpha ~\mathrm{KJ}\,(0,\kappa _1 ,r_1 )+ (1-\alpha )~\mathrm{KJ}\,(0,\kappa _2 ,r_2 )$ is continuous in $\kappa _1 ,\kappa _2 ,\mathrm{r}_1 \mathrm{,r}_2$ and $\alpha$ we can conclude that both $S(F)=\sqrt{1-\left| {E_F \left( {\cos \Theta } \right) } \right| }$ and $S^{*}(F)=E_F (d(\Theta ,0))$ are continuous functions of $\kappa _1 ,\kappa _2 ,\mathrm{r}_1 \mathrm{,r}_2$ and $\alpha$ on the set $D^{*}$ where $\mathrm{F}=\alpha \mathrm{F}_1 +\left( {1-\alpha } \right) \mathrm{F}_2$, $\mathrm{F}_1 =\mathrm{KJ}\left( {0,\kappa _1 \mathrm{,r}_1 } \right)$ and $\mathrm{F}_2 =\mathrm{KJ}\left( {0,\kappa _2 \mathrm{,r}_2 } \right)$. Since for all $F\in \mathfrak {I}_1^*$ we have $R(F)=S(F)S^{*-1}(F)$ is a continuous function of $\kappa _1 ,\kappa _2 ,\mathrm{r}_1 \mathrm{,r}_2$ and $\alpha$. Therefore, $\mathop {\mathrm{sup}}\limits _{\mathfrak {I}_1^*}\, \mathrm{R}\left( \mathrm{F} \right) <\infty$. Similarly we can prove $\mathop {\mathrm {sup}}\limits _{\mathfrak {I}_1^{*} }\, {\mathrm R}^{-1}\left( {\mathrm F} \right) <\infty$. Hence $S\mathop \sim \limits ^{\mathfrak {I}_1^*} S^{*}$. Hence the lemma is established. $\square$

Lemma 3

Consider the family of distributions

Then$S\mathop \sim \limits ^{\mathfrak {I}_1 } S^{*}$.

Proof

We have $S\left( F \right) =\sqrt{1+C\left( \nu \right) -\alpha \left( {C\left( \nu \right) +\int \limits _0^{2\pi } {\cos \theta \, \mathrm{f}_1 \left( \theta \right) \mathrm{d}\theta } } \right) }$ and

$$\begin{aligned} S^{*}(F)= & {} \int \limits _0^{2\pi } {\left| {\pi -\left| {\pi -\theta } \right| } \right| \left( {\alpha f_1 \left( \theta \right) +\left( {1-\alpha } \right) f_2 \left( \theta \right) } \right) } d\theta \\= & {} \alpha \int \limits _0^{2\pi } {\left| {\pi -\left| {\pi -\theta } \right| } \right| f_1 \left( \theta \right) } d\theta +\left( {1-\alpha } \right) \int \limits _0^{2\pi } {\left| {\pi -\left| {\pi -\theta } \right| } \right| f_2 \left( \theta \right) } d\theta \\= & {} I_1 +I_2 \end{aligned}$$

Note that $I_1$ is a continuous function of $\kappa$ and r which is positive on the domain $\left\{ {\kappa >0} \right\} \cap \{0\le r\le r_0 <1\}$. Also, following arguments given in the proof of theorem 3 it can be seen that, as $\nu \rightarrow \infty$, $I_2$ is finite. Hence we can conclude that $\mathop {\mathrm{sup}}\limits _{\mathfrak {I}_1 }\, \mathrm{R}\left( \mathrm{F} \right) <\infty$. Following arguments similar to above, it can be seen that $\mathop {\mathrm{sup}}\limits _{\mathfrak {I}_1 }\, \mathrm{R}^{-1}\left( \mathrm{F} \right) <\infty$. Hence $S\mathop \sim \limits ^{\mathfrak {I}_1 } S^{*}$.

Hence the lemma is established. $\square$

Lemma 4

Consider the family of distributions

$\mathfrak {I}_2 =\left\{ {\mathrm{{WT}}\left( {0,\uplambda ,\upnu } \right) :0<\uplambda _0 \le \uplambda \le \uplambda _1 <\infty ,\upnu \ge \upnu _0 } \right\} .$Then $S\mathop \sim \limits ^{\mathfrak {I}_2 } S^{*}$.

Proof

Let $S(F)=Q_1 \left( {\lambda ,\nu } \right) =\sqrt{1-C\left( {\lambda ,\nu } \right) }$ and $S^{*}(F)=Q_2 (\lambda ,\nu )=\int \limits _0^{2\pi } {\left| {\pi -\left| {\pi -\theta } \right| } \right| } f\left( \theta \right) d\theta$ where $f(\theta )$ is density of $WT\left( {0,\lambda ,\nu } \right)$ distribution. Note that both are continuous functions of $\lambda$ and $\nu$, $\mathop {\lim }\limits _{\nu \rightarrow \infty } Q_1 (\lambda ,\nu )<\infty$ and $\mathop {\lim }\limits _{\nu \rightarrow \infty } Q_2 (\lambda ,\nu )<\infty$. Hence $\mathop {\mathrm{sup}}\limits _{\mathfrak {I}_2 } \mathrm{R}\left( \mathrm{F} \right) <\infty$ and $\mathop {\mathrm{sup}}\limits _{\mathfrak {I}_2 } \mathrm{R}^{\mathrm{-1}}\left( \mathrm{F} \right) <\infty .$ Thus $S\mathop \sim \limits ^{\mathfrak {I}_2 } S^{*}$.

Hence the lemma is established. $\square$

Lemma 5

Let $F\in \mathfrak {I}_1^{**}$. In the expansion $\rho _{\gamma ,\mathrm{F}} =C_1 +\frac{C_2 }{\kappa }+\frac{C_3 }{\kappa ^{2}}+\cdots$, we have $C_1 >0$.

Proof

Consider $\rho _{\gamma ,F} =\frac{1}{1-2\gamma }\int \limits _{B_\gamma }^{A_\gamma } {\cos \theta ~~\mathrm{f}\left( \theta \right) } d\theta =\frac{2}{1-2\gamma }\int \limits _0^{{\omega _r }/{\sqrt{\kappa }}} {\cos \theta ~\mathrm{f}~\left( \theta \right) } d\theta$. Now, since $\cos \theta ~~\mathrm{f}\left( \theta \right)$ is a monotonically decreasing function in the interval $\left( {0,A_\gamma } \right)$ we have

$$\begin{aligned} \rho _{\gamma ,F} \ge \frac{2}{1-2\gamma }\frac{\omega _r }{\sqrt{\kappa }}\left[ {\cos \left( {\frac{\omega _r }{\sqrt{\kappa }}} \right) \mathrm{f}\left( {\frac{\omega _r }{\sqrt{\kappa }}} \right) } \right] . \end{aligned}$$

Using the series expansion of $\mathrm{cos }\left( {\frac{\omega _r }{\sqrt{\kappa }}} \right)$ and replacing $f\left( \theta \right)$ by $\mathrm{f}\left( {\frac{\omega _r }{\sqrt{\kappa }}} \right)$ we get, $\rho _{\gamma ,F} \ge \frac{1}{\pi \left( {1-2\gamma } \right) }\mathop {\lim }\limits _{\kappa \rightarrow \infty } \frac{\exp \left( {\tau \left( {\kappa ,\mathrm{r}} \right) } \right) }{\sqrt{\kappa }I_0 \left( \kappa \right) }$ where $\tau \left( {\kappa ,\mathrm{r}} \right) =\left\{ {\frac{\kappa \left\{ {\left( {1+r^{2}} \right) \left( {1-\frac{\omega _r^2 }{\kappa }} \right) -2r} \right\} }{1+r^{2}-2r\left( {1-\frac{\omega _r^2 }{\kappa }} \right) }-\kappa } \right\}$.

Using $I_0 \left( \kappa \right) =\frac{\exp \left( \kappa \right) }{\sqrt{2\pi \kappa }}$ for large $\kappa$(see Oldham et al. 2009) and simplifying $\exp \left( {\tau \left( {\kappa ,\mathrm{r}} \right) } \right)$ as $\kappa \rightarrow \infty$ we get $\rho _{\gamma ,F} \ge \frac{\sqrt{2}}{\sqrt{\pi }\left( {1-2\gamma } \right) }\frac{1}{\omega _r }\exp \left( {-\frac{1}{2\omega _r^2 }} \right) =\tau _1 \left( r \right)$. Since $0\le \mathrm{r}_0 \le r\le r_1 <1$ we get $\tau _1 \left( {r_1 } \right) <1$. Also we know that $\tau _1 \left( r \right) >0$ and hence $0<\tau _1 \left( {r_1 } \right) <1$. Thus, $C_1 >0$.

Hence the lemma is established. $\square$

Lemma 6

Let $\mathfrak {I}_1^{**} =\left\{ {KJ\left( {0,\kappa ,r} \right) \mathrm{; }\kappa >M,\mathrm{0}\le \mathrm{r}_0 \le r\le r_1 <1} \right\}$and $\mathfrak {I}_2^{**} =\left\{ {KJ\left( {0,\kappa ,r} \right) \mathrm{; }\kappa<m,\mathrm{0}\le \mathrm{r}_0 \le r\le r_1 <1} \right\}$. Then (a) $S_1 \mathop \sim \limits ^{\mathfrak {I}_1^{**} } S_2$and (b) $S_1 \mathop \sim \limits ^{\mathfrak {I}_2^{**} } S_2$.

Proof

(a) Note that as $\kappa \rightarrow \infty$, the $\mathrm{{KJ}}\left( {0,\upkappa , \mathrm{r}} \right)$ tends to $\mathrm{N}\left( {0, \upomega _\mathrm{r}^2 } \right)$ distribution where $\omega _\mathrm{r} =\frac{1-\mathrm{r}}{1+\mathrm{r}}$. Using this we get $\mathrm{A}_\upgamma =\frac{\upomega _r \Phi ^{\mathrm{-1}}\left( {1-\upgamma } \right) }{\sqrt{\upkappa }}$. Since $\mathrm{{S_2}} \left( \mathrm{F} \right) =\sqrt{\mathrm{E}_{\upgamma ,\mathrm{F}} \left( {1-\cos \Theta } \right) }$, using Jensen’s inequality we get

$$\begin{aligned} \sqrt{\mathrm{E}_{\upgamma ,\mathrm{F}} \Big ( {1-\cos \Theta } \Big )}\ge & {} \mathrm{E}_{\upgamma ,\mathrm{F}} \left( {\sqrt{1-\cos \Theta }} \right) \cong \mathrm{E}_{\upgamma ,\mathrm{F}} \left( {\sqrt{\frac{\Theta ^{2}}{2}}} \right) \\= & {} \displaystyle \frac{1}{\sqrt{2}}\mathrm{E}_{\upgamma ,\mathrm{F}} \left( {\left| \Theta \right| } \right) =\frac{\sqrt{2}}{\left( {1-2\upgamma } \right) }\int \limits _0^{\mathrm{A}_\upgamma } {\uptheta \mathrm{f}\left( \uptheta \right) } \mathrm{d}\uptheta . \end{aligned}$$

Note that ${S}_1 ({F})=\frac{2}{\left( {1-2\gamma } \right) }\int \limits _0^{{ A}_\gamma } {\theta {f}\left( \theta \right) } {d}\theta$ and hence $\mathop {\sup }\limits _{\mathfrak {I}_1^{**} } \frac{\mathrm{S}_1 (F)}{\mathrm{S}_2 (F)}\le \sqrt{2}<\infty$. Further,

$$\begin{aligned} \mathrm{S}_1 (\mathrm{F})\cong & {} \sqrt{\frac{2}{\uppi }}\frac{\sqrt{\upkappa }}{\left( {1-2\upgamma } \right) \upomega _r }\int \limits _0^{\mathrm{A}_\upgamma } \uptheta \mathrm{exp} \left( {-\frac{\uptheta ^{2}\upkappa }{2\upomega _\mathrm{r} }} \right) \mathrm{d}\uptheta \cong \sqrt{\frac{2}{\uppi }}\frac{\sqrt{\upkappa }}{\left( {1-2\upgamma } \right) \upomega _\mathrm{r} }\int \limits _0^{\mathrm{A}_\upgamma } \uptheta \left( {1-\frac{\uptheta ^{2}\upkappa }{2\upomega _\mathrm{r} }} \right) \mathrm{d}\uptheta \\\cong & {} \frac{\left( {\Phi ^{-1}\left( {1-\upgamma } \right) } \right) ^{2}\upomega _\mathrm{r} }{\sqrt{2\uppi \upkappa }\left( {1-2\upgamma } \right) }\left[ {1-\frac{\left( {\Phi ^{-1}\left( {1-\upgamma } \right) } \right) ^{2}}{4}} \right] . \end{aligned}$$

$$\begin{aligned} \begin{array}{ll} \mathrm{S}_2 \left( \mathrm{F} \right) &{}=\sqrt{\frac{\sqrt{\mathrm{k}}}{\left( {1-2\upgamma } \right) \upomega _\mathrm{r} \uppi }\int \limits _0^{\mathrm{A}_\upgamma } {\left( {1-\cos \theta } \right) \exp \left( {-\frac{\uptheta ^{2}\upkappa }{2\upomega _\mathrm{r}^2 }} \right) \mathrm{d}\uptheta } }\cong \sqrt{\frac{\sqrt{\mathrm{k}}}{\left( {1-2\upgamma } \right) \upomega _\mathrm{r} \uppi }\int \limits _0^{\mathrm{A}_\upgamma } {\frac{\uptheta ^{2}}{2}\left( {1-\frac{\uptheta ^{2}\upkappa }{2\upomega _\mathrm{r}^2 }} \right) \mathrm{d}\uptheta } } \\ \quad &{}\cong \frac{\left( {\Phi ^{-1}\left( {1-\upgamma } \right) } \right) ^{3/2}\upomega _\mathrm{r} }{\sqrt{2\uppi \upkappa }\sqrt{\left( {1-2\upgamma } \right) }}\left[ {\frac{1}{3}-\frac{\left( {\Phi ^{-1}\left( {1-\upgamma } \right) } \right) ^{2}}{10}} \right] ^{1/2}. \\ \end{array} \end{aligned}$$

Hence $\frac{\mathrm{S}_2 \left( \mathrm{F} \right) }{\mathrm{S}_1 \left( \mathrm{F} \right) }\cong \frac{\left( {\Phi ^{-1}\left( {1-\upgamma } \right) } \right) ^{{-1}/2}\sqrt{\left( {1-2\upgamma } \right) }\left[ {\frac{1}{3}-\frac{\left( {\Phi ^{-1}\left( {1-\upgamma } \right) } \right) ^{2}}{10}} \right] ^{1/2}}{\left[ {1-\frac{\left( {\Phi ^{-1}\left( {1-\upgamma } \right) } \right) ^{2}}{4}} \right] }$. Therefore, $\mathop {\sup }\limits _{\mathfrak {I}_1^{**} } \frac{\mathrm{S}_2 (\mathrm{F})}{\mathrm{S}_1 (\mathrm{F})}<\infty$.

Thus,$S_1 \mathop \sim \limits ^{\mathfrak {I}_1^{**} } S_2$.

(b) We have $S_1 \left( F \right) =\frac{1}{2\pi \left( {1-2\gamma } \right) }\left\{ {\alpha ^{2}\left( r \right) +4\sum \limits _{p=1}^\infty {r^{p}} \left( {\frac{\alpha \left( r \right) S_{p,\mathrm{0}} }{\mathrm{p}}+\frac{C_{p,\mathrm{0}} -1}{\mathrm{p}^{2}}} \right) } \right\}$ where $S_{p,\nu } =\sin \left[ {p\alpha \left( r \right) -\nu } \right]$ and $C_{p,\nu } =\cos \left[ {p\alpha \left( r \right) -\nu } \right]$ and $S_2 \left( F \right) =\sqrt{1-\rho _{\gamma ,\mathrm{F}} }$. Now a straight forward calculation yields $\mathop {\sup }\limits _{\mathfrak {I}_2^{**} } R(F)<\infty$ and $\mathop {\sup }\limits _{\mathfrak {I}_2^{**} } R^{-1}(F)<\infty$. Thus, $S_1 \mathop \sim \limits ^{\mathfrak {I}_2^{**} } S_2$.

Hence the lemma is established. $\square$

Lemma 7

Let $H\left( r \right) =\int \limits _0^\pi {\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }} d\theta$, then $\mathop {\lim }\limits _{r\rightarrow 1^{-}} H\left( r \right) =0$.

Proof

Let $g\left( \theta \right) =\frac{\theta }{1+r^{2}-2r\cos \theta }$, $g^{{\prime }}\left( \theta \right) =\frac{\left( {1+r^{2}-2r\cos \theta } \right) -\theta \left( {2r\sin \theta } \right) }{\left( {1+r^{2}-2r\cos \theta } \right) ^{2}}$

$$\begin{aligned} g^{{\prime }}\left( \theta \right) =0\Rightarrow \frac{1+\mathrm{r}^{2}}{2r}=\cos \theta +\theta \sin \theta . \end{aligned}$$

Now let $k\left( \theta \right) =\cos \theta +\theta \sin \theta$. Then $k^{{\prime }}\left( \theta \right) =\theta \cos \theta$. Note that $k^{{\prime }}\left( \theta \right) \ge 0$ for all $\theta \in \left[ {0,\frac{\pi }{2}} \right]$ and $k^{{\prime }}\left( \theta \right) <0$ for all $\theta \in \left[ {\frac{\pi }{2},\pi } \right]$. Therefore $k\left( \theta \right)$ is increasing in $\left[ {0,\frac{\pi }{2}} \right]$ and decreasing in $\left( {\frac{\pi }{2},\pi } \right]$. Hence $k\left( \theta \right)$ has a maximum value at $\theta =\frac{\pi }{2}$.

Thus $k\left( 0 \right) \le \theta \sin \theta +\cos \theta \le k\left( {\frac{\pi }{2}} \right) \Rightarrow$

$$\begin{aligned} 1\le \theta \sin \theta +\cos \theta \le \frac{\pi }{2}\,\,\mathrm{for~ all}\,\, \theta \in \left[ {0,\frac{\pi }{2}} \right] \end{aligned}$$

(9)

Now since $\frac{1+r^{2}}{2r}>1$ for $0<r<1$ and $\frac{1+r^{2}}{2r}\rightarrow 1$as $r\rightarrow 1^{-}$. Using (9) we get $\frac{1+\mathrm{r}^{2}}{2r}=\cos \theta +\theta \sin \theta$ has a solution $\theta ^{*}$for r close to 1. Observe that $g^{{\prime }}\left( \theta \right) >0$ when $\theta <\theta ^{*}$and $g^{{\prime }}\left( \theta \right) <0$ for $\theta >\theta ^{*}$. Therefore $g\left( \theta \right)$ has a maximum value at $\theta ^{*}$.

Further note that $\theta ^*=\theta _{r} \rightarrow 0$ as $r\rightarrow 1^{-}$. Using the approximation $\sin \theta \approx \theta$ for small $\theta$ and $\cos \theta \approx 1-\frac{\theta ^{2}}{2}$ for small $\theta$, we have

$$\begin{aligned} \theta ^{*}\sin \theta ^{*}+\cos \theta ^{*}=\theta ^{{*}{2}}+\left( {1-\frac{\theta ^{{*}{2}}}{2}} \right) =1+\frac{\theta ^{{*}{2}}}{2}=\frac{1+r^{2}}{2r}\; \mathrm{for}\,\, r\,\, \mathrm{close\,\, to\,\, 1.} \end{aligned}$$

Solving we get $\theta ^{*}=\frac{\left( {1-r} \right) }{\sqrt{r}}$ for r close to 1.

Thus $g\left( {\theta ^{*}} \right) =\frac{\theta ^{*}}{1+r^{2}-2r\cos \theta ^{*}}\approx \frac{1}{\sqrt{r}\left( {1-r} \right) }$ for r close to 1.

Now $\int \limits _0^\pi {\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }d\theta } =\int \limits _0^\delta {\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }d\theta +} \int \limits _\delta ^\pi {\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }d\theta }$ for any small $\delta >0$.

Now $\mathop {\lim }\limits _{r\rightarrow 1^{-}} \int \limits _\delta ^\pi {\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }d\theta =\int \limits _\delta ^\pi {\mathop {\lim }\limits _{r\rightarrow 1^{-}} } } \frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }d\theta =0$ as $\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }$ is a continuous function of r. Again, $\int \limits _0^\delta {\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }d\theta } =\left( {1-r^{2}} \right) \int \limits _0^\delta {g\left( \theta \right) } d\theta \le \left( {1-r^{2}} \right) \delta g\left( {\theta ^{*}} \right) \approx \frac{1-r^{2}}{\sqrt{r}\left( {1-r} \right) }\delta =\frac{\left( {1+r} \right) \delta }{\sqrt{r}}$ for r close to 1. Therefore $\int \limits _0^\delta {\frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }} d\theta \le \frac{\left( {1+r} \right) }{\sqrt{r}}\delta \le 3\delta$ for r close to 1. Since $\delta$ is arbitrary, we conclude$\mathop {\lim }\limits _{r\rightarrow 1^{-}} \int \limits _0^\pi \frac{\left( {1-r^{2}} \right) \theta }{1+r^{2}-2r\cos \theta }d\theta =0$. $\square$