A Novel Kernel Correlation Coefficient with Robustness Against Nonlinear Attenuation and Impulsive Noise

Abstract

In this paper, we propose a new kernel correlation coefficient (KECC), with an emphasis on its robustness against impulsive noise and/or monotonic nonlinear transformations. To gain further insight, we compared KECC with other four correlation coefficients, namely, Pearson’s product moment correlation coefficient (PPMCC), Kendall’s tau (KT), Spearman’s rho (SR) and order statistics correlation coefficient (OSCC). Extensive simulation experiments were conducted under linear, nonlinear, normal and contaminated Gaussian models (CGM) based on seven means of performance evaluation. Theoretical analysis showed that KECC satisfies various desired properties. Numerical results suggest that KECC performs equally well with the optimal PPMCC under the bivariate normal model, and outperforms the others when impulsive noise and/or nonlinearity exist in the data. Moreover, KECC can detect accurately the time delay of signals corrupted by impulsive noise. Last but not least, KECC runs in linearithmic time, only slightly slower than the fastest PPMCC. The advantages of KECC revealed in this work might shed new light on the topic of correlation analysis, which is important in many areas including signal processing.

Appendix

Proof Proof of Theorem 2

Let $\nu _{\mathcal {X}}$ and $\nu _{\mathcal {Y}}$ be the population medians of $\mathcal {X}$ and $\mathcal {Y}$, respectively. Then, from [31], we have

$$\begin{array}{@{}rcl@{}} \nu_{\mathcal{X}}&=&\mu_{\mathcal{X}}\\ \nu_{\mathcal{Y}}&=&\mu_{\mathcal{Y}}. \end{array} $$

(27)

Define

$$\begin{array}{@{}rcl@{}} \eta_{\mathcal{X}}&\triangleq& 3.53\textit{IR}_{\mathcal{X}} \end{array} $$

(28)

$$\begin{array}{@{}rcl@{}} \eta_{\mathcal{Y}}&\triangleq& 3.53\textit{IR}_{\mathcal{Y}} \end{array} $$

(29)

$$\begin{array}{@{}rcl@{}} \mathcal{X}^{\prime}&\triangleq& \mathcal{X}e^{-\frac{\mathcal{X}^{2}}{2\eta_{\mathcal{X}}^{2}}} \end{array} $$

(30)

$$\begin{array}{@{}rcl@{}} \mathcal{Y}^{\prime}&\triangleq& \mathcal{Y}e^{-\frac{\mathcal{Y}^{2}}{2\eta_{\mathcal{Y}}^{2}}} \end{array} $$

(31)

$$\begin{array}{@{}rcl@{}} \phi_{1}(\mathcal{X},\mathcal{Y})&\triangleq& \mathcal{N}(\mu_{\mathcal{X}},\mu_{\mathcal{Y}},\sigma^{2}_{\mathcal{X}},\sigma^{2}_{\mathcal{Y}},\rho) \end{array} $$

(32)

$$\begin{array}{@{}rcl@{}} \phi_{2}(\mathcal{X},\mathcal{Y})&\triangleq& \mathcal{N}(\mu_{\mathcal{X}},\mu_{\mathcal{Y}},\sigma^{\prime 2}_{\mathcal{X}},\sigma^{\prime 2}_{\mathcal{Y}},\rho^{\prime}) \end{array} $$

(33)

$$\begin{array}{@{}rcl@{}} \phi(\mathcal{X},\mathcal{Y})&\triangleq& (1-\varepsilon)\phi_{1}(\mathcal{X},\mathcal{Y})+\varepsilon \phi_{2}(\mathcal{X},\mathcal{Y}). \end{array} $$

(34)

When the sample size $N\to \infty$, it follows that

$$ \lim_{N\to\infty} \mathbb{E}(r_{KE})=\frac{\mathbb{E}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})}{\sqrt{\mathbb{E}(\mathcal{X}^{\prime 2})E(\mathcal{Y}^{\prime 2})}}. $$

(35)

□

Therefore, we need to work out all the three expectation terms in Eq. 35.

For the bivariate normal distribution $\phi _{1}(\mathcal {X},\mathcal {Y})$ in Eq. 32, the pdf

$$\begin{array}{@{}rcl@{}} \phi_{1}(\mathcal{X},\mathcal{Y})=\frac{1}{2\pi\sigma_{\mathcal{X}}\sigma_{\mathcal{Y}}\sqrt{1-\rho^{2}}}e^{-\frac{1}{2}Q({\mathcal{X},\mathcal{Y}})} \end{array} $$

(36)

where

$$\begin{array}{@{}rcl@{}} Q({\mathcal{X},\mathcal{Y}})&=&\frac{1}{1-\rho^{2}} \left[ \left( \frac{\mathcal{X}-\mu_{\mathcal{X}}}{\sigma_{\mathcal{X}}} -\rho\frac{\mathcal{Y}-\mu_{\mathcal{Y}}}{\sigma_{\mathcal{Y}}} \right)^{2}\right. \\ &+&\left.(1-\rho^{2}) \left( \frac{\mathcal{Y}-\mu_{\mathcal{Y}}}{\sigma_{\mathcal{Y}}} \right)^{2} \right] \\ &=&\frac{(\mathcal{X}-a)^{2}}{(1-\rho^{2})\sigma_{\mathcal{X}}^{2}}+\frac{(\mathcal{Y}-\mu_{\mathcal{Y}})^{2}}{\sigma_{\mathcal{Y}}^{2}} \end{array} $$

(37)

with

$$\begin{array}{@{}rcl@{}} a=\mu_{\mathcal{X}}+\rho\frac{\sigma_{\mathcal{X}}}{\sigma_{\mathcal{Y}}}(\mathcal{Y}-\mu_{\mathcal{Y}})=\frac{\sigma_{\mathcal{X}}}{\sigma_{\mathcal{Y}}}\rho\mathcal{Y}. \end{array} $$

(38)

For the bivariate normal distribution $\phi _{2}(\mathcal {X},\mathcal {Y})$ in Eq. 33, the pdf is

$$\begin{array}{@{}rcl@{}} \phi_{2}(\mathcal{X},\mathcal{Y})=\frac{1}{2\pi\sigma^{\prime}_{\mathcal{X}}\sigma^{\prime}_{\mathcal{Y}}\sqrt{1-\rho^{\prime 2}}}e^{-\frac{1}{2}Q^{\prime}({\mathcal{X},\mathcal{Y}})} \end{array} $$

(39)

where

$$\begin{array}{@{}rcl@{}} &&Q^{\prime}({\mathcal{X},\mathcal{Y}})=\frac{1}{1-{\rho^{\prime 2}}}\left[ \left( \frac{\mathcal{X}-\mu_{\mathcal{X}}}{\sigma^{\prime}_{\mathcal{X}}} -\rho^{\prime}\frac{\mathcal{Y}-\mu_{\mathcal{Y}}}{\sigma{\prime}_{\mathcal{Y}}}\right)^{2}\right. \\ &&+\left.(1-\rho^{\prime 2})\left( \frac{\mathcal{Y}-\mu_{\mathcal{Y}}}{\sigma^{\prime}_{\mathcal{Y}}}\right)^{2}\right] \\ &&=\frac{(\mathcal{X}-a^{\prime 2})}{(1-{\rho^{\prime 2}}){\sigma^{\prime}_{\mathcal{X}}}^{2}}+\frac{(\mathcal{Y}-\mu_{\mathcal{Y}})^{2}}{{\sigma^{\prime}_{\mathcal{Y}}}^{2}} \end{array} $$

(40)

with

$$\begin{array}{@{}rcl@{}} a^{\prime}=\mu_{\mathcal{X}}+\rho^{\prime}\frac{\sigma^{\prime}_{\mathcal{X}}}{\sigma^{\prime}_{\mathcal{Y}}}(\mathcal{Y}-\mu_{\mathcal{Y}})=\frac{\sigma^{\prime}_{\mathcal{X}}}{\sigma^{\prime}_{\mathcal{Y}}}\rho^{\prime}\mathcal{Y}. \end{array} $$

(41)

From Eqs. 30–41, the numerator in Eq. 35 becomes

$$\begin{array}{@{}rcl@{}} \mathbb{E}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})&=&\iint\limits_{-\infty}^{+\infty}\mathcal{X}e^{-\frac{\mathcal{X}^{2}}{2\eta_{\mathcal{X}}^{2}}} \mathcal{Y}e^{-\frac{\mathcal{Y}^{2}}{2\eta_{\mathcal{Y}}^{2}}}\phi(\mathcal{X},\mathcal{Y})d\mathcal{X}d\mathcal{Y}\\ &=&(1-\varepsilon)\mathbb{E}_{1}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})+\varepsilon \mathbb{E}_{2}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime}) \end{array} $$

(42)

where

$$\begin{array}{@{}rcl@{}} \mathbb{E}_{1}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})&=&\iint\limits_{-\infty}^{+\infty}\mathcal{X}^{\prime}\mathcal{Y}^{\prime}\phi_{1}(\mathcal{X},\mathcal{Y})d\mathcal{X}d\mathcal{Y} \end{array} $$

(43)

$$\begin{array}{@{}rcl@{}} &=&\frac{1}{2\pi\sigma_{\mathcal{X}}\sigma_{\mathcal{Y}}\sqrt{1-\rho^{2}}}\int\limits_{-\infty}^{+\infty}\mathcal{Y^{\prime}}e^{-\frac{\mathcal{Y}^{2}}{2\sigma_{\mathcal{Y}}^{2}}}H_{1}(\mathcal{Y})d\mathcal{Y} \\ \end{array} $$

(44)

with

$$ H_{1}(\mathcal{Y})=\int\limits_{-\infty}^{+\infty}\mathcal{X}e^{-\frac{\mathcal{X}^{2}}{2\eta_{\mathcal{X}}^{2}}}e^{-\frac{(\mathcal{X}-a)^{2}}{2(1-\rho^{2})\sigma_{\mathcal{X}}^{2}}}d\mathcal{X} $$

(45)

and

$$\begin{array}{@{}rcl@{}} \mathbb{E}_{2}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})&=&\iint\limits_{-\infty}^{+\infty}\mathcal{X}^{\prime}\mathcal{Y}^{\prime}\phi_{2}(\mathcal{X},\mathcal{Y})d\mathcal{X}d\mathcal{Y} \end{array} $$

(46)

$$\begin{array}{@{}rcl@{}} &=&\frac{1}{2\pi\sigma_{\mathcal{X}}^{\prime}\sigma_{\mathcal{Y}}^{\prime}\sqrt{1-\rho^{\prime 2}}}\int\limits_{-\infty}^{+\infty}\mathcal{Y^{\prime}}e^{-\frac{\mathcal{Y}^{2}}{2\sigma_{\mathcal{Y}}^{\prime 2}}}H_{2}(\mathcal{Y})d\mathcal{Y} \end{array} $$

(47)

with

$$ H_{2}(\mathcal{Y})=\int\limits_{-\infty}^{+\infty}\mathcal{X}e^{-\frac{\mathcal{X}^{2}}{2\eta_{\mathcal{X}}^{2}}}e^ {-\frac{(\mathcal{X}-a^{\prime})^{2}}{2(1-\rho^{\prime 2})\sigma_{\mathcal{X}}^{\prime 2}}}d\mathcal{X}. $$

(48)

Now we need to evaluate H ₁ and H ₂. It follows that

$$\begin{array}{@{}rcl@{}} \frac{\mathcal{X}^{2}}{2\eta_{\mathcal{X}}^{2}}+\frac{(\mathcal{X}-a)^{2}}{2(1-\rho^{2})\sigma_{\mathcal{X}}^{2}}=\frac{(\mathcal{X}-Z)^{2}+A}{2B} \end{array} $$

(49)

where

$$\begin{array}{@{}rcl@{}} A&=\frac{\eta_{\mathcal{X}}^{2}\sigma_{\mathcal{X}}^{2}a^{2}(1-\rho^{2})}{{\big[(1-\rho^{2})\sigma_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{2}\big]}^{2}} \end{array} $$

(50)

$$\begin{array}{@{}rcl@{}} B&=\frac{\eta_{\mathcal{X}}^{2}\sigma_{\mathcal{X}}^{2}(1-\rho^{2})}{(1-\rho^{2})\sigma_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{2}} \end{array} $$

(51)

$$\begin{array}{@{}rcl@{}} Z&=\frac{a\eta_{\mathcal{X}}^{2}}{(1-\rho^{2})\sigma_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{2}}. \end{array} $$

(52)

Then, from Eqs. 38, 50 and 51, we have

$$\begin{array}{@{}rcl@{}} \frac{A}{B}=\frac{\sigma_{\mathcal{X}}^{2}\rho^{2}\mathcal{Y}^{2}}{\sigma_{\mathcal{Y}}^{2}\big[(1-\rho^{2})\sigma_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{2}\big]} \end{array} $$

(53)

and hence

$$\begin{array}{@{}rcl@{}} H_{1}(\mathcal{Y})&=&\int\limits_{-\infty}^{+\infty}\mathcal{X}e^{-\frac{(\mathcal{X}-Z)^{2}}{2B}}e^{-\frac{A}{2B}}d\mathcal{X} \\ &=&e^{-\frac{A}{2B}}\int\limits_{-\infty}^{+\infty}\left[(\mathcal{X}-Z)e^{-\frac{(\mathcal{X}-Z)^{2}}{2B}}+Ze^{-\frac{(\mathcal{X}-Z)^{2}}{2B}}\right]d\mathcal{X} \\ &=&e^{-\frac{A}{2B}}\int\limits_{-\infty}^{+\infty}Ze^{-\frac{(\mathcal{X}-Z)^{2}}{2B}}d\mathcal{X} \\ &=&\sqrt{2\pi B}Ze^{-\frac{A}{2B}} \\ &=&C\mathcal{Y}e^{-\frac{\mathcal{Y}^{2}}{2D^{2}}} \end{array} $$

(54)

where

$$\begin{array}{@{}rcl@{}} C&=&\frac{\sigma_{\mathcal{X}}^{2}\eta_{\mathcal{X}}^{3}\rho\sqrt{2\pi(1-\rho^{2})}}{\sigma_{\mathcal{Y}}\sqrt{[(1-\rho^{2})\sigma_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{2}]^{3}}} \end{array} $$

(55)

$$\begin{array}{@{}rcl@{}} D^{2}&=&\frac{\sigma_{\mathcal{Y}}^{2}[(1-\rho^{2})\sigma_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{2}]}{\sigma_{\mathcal{X}}^{2}\rho^{2}}. \end{array} $$

(56)

Write

$$\begin{array}{@{}rcl@{}} \frac{1}{J^{2}}&=&\frac{1}{\eta_{\mathcal{Y}}^{2}}+\frac{1}{\sigma_{\mathcal{Y}}^{2}}+\frac{1}{D^{2}}\\ &=&\frac{W}{\sigma_{\mathcal{Y}}^{2}\eta_{\mathcal{Y}}^{2}[(1-\rho^{2})\sigma_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{2}]} \end{array} $$

(57)

where

$$\begin{array}{@{}rcl@{}} W=\sigma_{\mathcal{X}}^{2}\sigma_{\mathcal{Y}}^{2}(1-\rho^{2})+\sigma_{\mathcal{Y}}^{2}\eta_{\mathcal{X}}^{2}+\sigma_{\mathcal{X}}^{2}\eta_{\mathcal{Y}}^{2}+\eta_{\mathcal{X}}^{2}\eta_{\mathcal{Y}}^{2}. \end{array} $$

(58)

Then

$$\begin{array}{@{}rcl@{}} \mathbb{E}_{1}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})&=&\frac{C}{2\pi\sigma_{\mathcal{X}}\sigma_{\mathcal{Y}}\sqrt{1-\rho^{2}}}\int\limits_{-\infty}^{+\infty}\mathcal{Y}^{2}e^{-\frac{\mathcal{Y}^{2}}{2J^{2}}}d\mathcal{Y}\\ &=&\frac{\sqrt{2\pi}CJ^{3}}{2\pi\sigma_{\mathcal{X}}\sigma_{\mathcal{Y}}\sqrt{1-\rho^{2}}}\\ &=&\frac{\eta_{\mathcal{X}}^{3}\eta_{\mathcal{Y}}^{3}\sigma_{\mathcal{X}}\sigma_{\mathcal{Y}}\rho}{\sqrt{W^{3}}} . \end{array} $$

(59)

In a parallel manner, we have

$$\begin{array}{@{}rcl@{}} H_{2}(\mathcal{Y})=C^{\prime}\mathcal{Y}e^{-\frac{\mathcal{Y}^{2}}{2D^{\prime 2}}} \end{array} $$

(60)

where

$$\begin{array}{@{}rcl@{}} C^{\prime}&=&\frac{\sigma_{\mathcal{X}}^{\prime 2}\eta_{\mathcal{X}}^{3}\rho^{\prime}\sqrt{2\pi(1-\rho^{\prime 2})}}{\sigma_{\mathcal{Y}}^{\prime}\sqrt{[(1-\rho^{\prime 2})\sigma_{\mathcal{X}}^{\prime 2}+\eta_{\mathcal{X}}^{2}]^{3}}} \end{array} $$

(61)

$$\begin{array}{@{}rcl@{}} D^{\prime 2}&=&\frac{\sigma_{\mathcal{Y}}^{\prime 2}[(1-\rho^{\prime 2})\sigma_{\mathcal{X}}^{\prime 2}+\eta_{\mathcal{X}}^{2}]}{\sigma_{\mathcal{X}}^{\prime 2}\rho^{\prime 2}}. \end{array} $$

(62)

Write

$$\begin{array}{@{}rcl@{}} \frac{1}{J^{\prime 2}}&=&\frac{1}{\eta_{\mathcal{Y}}^{2}}+\frac{1}{\sigma_{\mathcal{Y}}^{\prime 2}}+\frac{1}{D^{\prime 2}}\\ &=&\frac{W^{\prime}}{\sigma_{\mathcal{Y}}^{\prime 2}\eta_{\mathcal{Y}}^{2}[(1-\rho^{\prime 2})\sigma_{\mathcal{X}}^{\prime 2}+\eta_{\mathcal{X}}^{2}]} \end{array} $$

(63)

where

$$\begin{array}{@{}rcl@{}} W^{\prime}=\sigma_{\mathcal{X}}^{\prime 2}\sigma_{\mathcal{Y}}^{\prime 2}(1-\rho^{\prime 2})+\sigma_{\mathcal{Y}}^{\prime 2}\eta_{\mathcal{X}}^{2}+\sigma_{\mathcal{X}}^{\prime 2}\eta_{\mathcal{Y}}^{2}+\eta_{\mathcal{X}}^{2}\eta_{\mathcal{Y}}^{2}. \end{array} $$

(64)

A substitution of Eq. 60 into Eq. 47 gives

$$\begin{array}{@{}rcl@{}} \mathbb{E}_{2}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})&=&\frac{C^{\prime}}{2\pi\sigma_{\mathcal{X}}^{\prime}\sigma_{\mathcal{Y}}^{\prime}\sqrt{1-\rho^{\prime 2}}}{\int}_{-\infty}^{+\infty}\mathcal{Y}e^{-\frac{\mathcal{Y}^{2}}{2J^{\prime 2}}}d\mathcal{Y}\\ &=&\frac{\sqrt{2\pi}C^{\prime}J^{\prime 3}}{2\pi\sigma_{\mathcal{X}}^{\prime}\sigma_{\mathcal{Y}}^{\prime}\sqrt{1-\rho^{\prime 2}}}\\ &=&\frac{\rho^{\prime}\sigma_{\mathcal{X}}^{\prime}\sigma_{\mathcal{Y}}^{\prime}\eta_{\mathcal{X}}^{3}\eta_{\mathcal{Y}}^{3}} {\sqrt{W^{\prime 3}}}. \end{array} $$

(65)

Similarly we can also obtain

$$\begin{array}{@{}rcl@{}} \mathbb{E}(\mathcal{X}^{\prime 2})&=&\frac{(1-\varepsilon)\eta_{\mathcal{X}}^{6}\sigma^{2}_{\mathcal{X}}}{\sqrt{(2\sigma_{\mathcal{X}}^{2}\eta_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{4})^{3}}}+ \frac{\varepsilon \eta_{\mathcal{X}}^{6}\sigma_{\mathcal{X}}^{\prime 2}}{\sqrt{(2\sigma_{\mathcal{X}}^{\prime 2}\eta_{\mathcal{X}}^{2}+\eta_{\mathcal{X}}^{4})^{3}}} \end{array} $$

(66)

$$\begin{array}{@{}rcl@{}} \mathbb{E}(\mathcal{Y}^{\prime 2})&=&\frac{(1-\varepsilon)\eta_{\mathcal{Y}}^{6}\sigma^{2}_{\mathcal{Y}}}{\sqrt{(2\sigma_{\mathcal{Y}}^{2}\eta_{\mathcal{Y}}^{2}+\eta_{\mathcal{Y}}^{4})^{3}}}+ \frac{\varepsilon \eta_{\mathcal{Y}}^{6}\sigma_{\mathcal{Y}}^{\prime 2}}{\sqrt{(2\sigma_{\mathcal{Y}}^{\prime 2}\eta_{\mathcal{Y}}^{2}+\eta_{\mathcal{Y}}^{4})^{3}}} . \end{array} $$

(67)

Given Eqs. 42, 59, 65–67 along with the fact that ρ∈[−1,+1], we arrive at

$$\begin{array}{@{}rcl@{}} \lim_{\begin{array}{c} N\to\infty\\ \sigma^{\prime}_{\mathcal{X}} \rightarrow \infty\\ \sigma^{\prime}_{\mathcal{Y}} \rightarrow \infty \end{array}} \mathbb{E}(r_{KE})&=&\frac{\mathbb{E}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})}{\sqrt{\mathbb{E}(\mathcal{X}^{\prime 2})E(\mathcal{Y}^{\prime 2})}}\\ &=&\frac{(1-\varepsilon)\mathbb{E}_{1}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})+\varepsilon \mathbb{E}_{2}(\mathcal{X}^{\prime}\mathcal{Y}^{\prime})}{\sqrt{\mathbb{E}(\mathcal{X}^{\prime 2})\mathbb{E}(\mathcal{Y}^{\prime 2})}}\\ &=&\frac{\rho}{\sqrt{(\frac{1-\rho^{2}}{S}+\frac{1}{2}T)^{3}}} \end{array} $$

(68)

where S and T are defined in Eqs. 9 and 10, respectively.

To show that the last term in Eq. 68 is approximately equal to ρ, we need to evaluate $\eta _{\mathcal {X}}$ and $\eta _{\mathcal {Y}}$ contained in S and T. Let Φ₁(⋅) and Φ₂(⋅) be the cdf of $\mathcal {N}(\mu _{\mathcal {X}},\sigma ^{2}_{\mathcal {X}})$ and $\mathcal {N}(\mu _{\mathcal {X}},\sigma ^{\prime 2}_{\mathcal {X}})$, respectively. Let q _U and q _L be the upper and lower quartiles of $\mathcal {X}$, respectively. Since the marginal distribution of $\mathcal {X}$ is

$$(1-\varepsilon)\mathcal{N}(\mu_{\mathcal{X}},\sigma^{2}_{\mathcal{X}})+\varepsilon\mathcal{N}(\mu_{\mathcal{X}},\sigma^{\prime 2}_{\mathcal{X}}), $$

we have

$$ (1-\varepsilon){\Phi}_{1}(q_{U})+\varepsilon{\Phi}_{2}(q_{U})=\frac{3}{4}. $$

(69)

By the assumption of $\sigma ^{\prime }_{\mathcal {X}}\gg \sigma _{\mathcal {X}}$, it follows that 0<Φ₂(q _U )<Φ₁(q _U ), and hence, from Eq. 69,

$$ (1-\varepsilon){\Phi}_{1}(q_{U}) < \frac{3}{4} <{\Phi}_{1}(q_{U}). $$

(70)

Let ${\Phi }_{1}^{-1}(\cdot )$ be the inversion function of Φ₁(⋅). Due to the monotonic property of ${\Phi }_{1}^{-1}$, we have, after imposing the transform ${\Phi }_{1}^{-1}$ to Eq. 70,

$$ {\Phi}_{1}^{-1}[(1-\varepsilon){\Phi}_{1}(q_{U})]<{\Phi}_{1}^{-1}\left( \frac{3}{4}\right)=0.6745\sigma_{\mathcal{X}} < q_{U}. $$

(71)

An application of Taylor’s theorem to the leftmost term in Eq. 71 gives

$${\Phi}_{1}^{-1}\left[(1-\varepsilon){\Phi}_{1}(q_{U})\right]=q_{U}-\frac{d{\Phi}_{1}^{-1}(t)}{dt}\bigg|_{t=\beta}\varepsilon {\Phi}_{1}(q_{U}) $$

where β lies in between (1−ε)Φ₁(q _U ) and Φ₁(q _U ). This means that

$${\Phi}_{1}^{-1}[(1-\varepsilon){\Phi}_{1}(q_{U})]=q_{U}-\mathcal{O}(\varepsilon) $$

which along with Eq. 71 leads to

$$q_{U}=0.6745\sigma_{\mathcal{X}}+\mathcal{O}(\varepsilon). $$

Similarly, we also have

$$q_{L}=-0.6745\sigma_{\mathcal{X}}-\mathcal{O}(\varepsilon). $$

Then

$$\text{IR}_{\mathcal{X}}\triangleq q_{U}-q_{L}=1.3490\sigma_{\mathcal{X}}+\mathcal{O}(\varepsilon) $$

and

$$ \eta_{\mathcal{X}}=3.53\text{IR}_{\mathcal{X}}= 4.7620\sigma_{\mathcal{X}}+\mathcal{O}(\varepsilon). $$

(72)

In a similar way, we can also obtain

$$ \eta_{\mathcal{Y}}=3.53\text{IR}_{\mathcal{Y}}= 4.7620\sigma_{\mathcal{Y}}+\mathcal{O}(\varepsilon). $$

(73)

Substituting Eqs. 72 and 73 into Eqs. 10 and 9, respectively, we have

$$T=2+\mathcal{O}(\varepsilon) $$

and

$$S=559.57+\mathcal{O}(\varepsilon). $$

Applying these results to the last term in (68) with the assumption of ε being small, we finally have

$$\begin{array}{@{}rcl@{}} \lim_{\begin{array}{c} N\to\infty\\ \sigma^{\prime}_{\mathcal{X}} \rightarrow \infty\\ \sigma^{\prime}_{\mathcal{Y}} \rightarrow \infty \end{array}} \mathbb{E}(r_{KE})\approx \rho \end{array} $$

thus completing the proof.