On the estimation of primary user activity statistics for long and short time scale models in cognitive radio

Abstract

Dynamic Spectrum Access (DSA)/Cognitive Radio (CR) systems access the channel in an opportunistic, non-interfering manner with the primary network. DSA/CR systems utilize spectrum sensing techniques to sense the availability of Primary user (PU). CR users can benefit from the knowledge of PU activity statistics. In this work, comprehensive analysis of estimation of distribution of PU idle and busy periods is carried out using Generalized Pareto and Pareto distributions for long and short time scale models respectively and closed form expression is derived. Moreover, the impact of sensing periods on the accuracy of estimated PU idle/busy periods is studied. Furthermore, the error in proposed estimation of distribution of PU idle and busy periods is quantified using the Kolmogorov–Smirnov test. From this study we conclude that the proposed model is better fit for the real scenarios eliminating practical limitations. Mathematical analysis is substantiated with the simulation results.

Appendices

Appendix 1: Derivation of the combined error $T_e$

$$\begin{aligned} f_{T_e}(t) = f_{T^{y}_{e}}(t) *f_{-T^{x}_{e}}(t) = \int _{-\infty }^{\infty }f_{T^{y}_{e}}(\tau ) f_{-T^{x}_{e}}(t - \tau ) \mathrm {d}\tau \end{aligned}$$

(20)

Case 1 : $t < -T_s$

No overlap in both the distributions, i.e. $f_{T^{y}_{e}}(\tau )$ and $f_{-T^{x}_{e}}(t-\tau )$.

$$\begin{aligned} \therefore \int _{-\infty }^{-T_s}f_{T^{y}_{e}}(\tau ) f_{-T^{x}_{e}}(t - \tau ) \mathrm {d}\tau = 0 \end{aligned}$$

Case 2 :$-T_s \le t < 0$

$$\begin{aligned} \int _{-T_s}^{t}f_{T^{y}_{e}}(\tau ) f_{-T^{x}_{e}}(t - \tau ) \mathrm {d}\tau&= \int _{-T_s}^{t} \Big (\frac{1}{T_s} \Big ) \Big (\frac{1}{T_s} \Big )\mathrm {d}\tau = \frac{T_s + t}{T_s^2} \end{aligned}$$

Case 3 :$0 \le t \le T_s$

$$\begin{aligned} \int _{t}^{T_s}f_{T^{y}_{e}}(\tau ) f_{-T^{x}_{e}}(t - \tau ) \mathrm {d}\tau = \int _{t}^{T_s} \Big (\frac{1}{T_s} \Big ) \Big (\frac{1}{T_s} \Big )\mathrm {d}\tau = \frac{T_s - t}{T_s^2} \end{aligned}$$

Case 4 :$t > T_s$ No overlap in both the distributions, i.e. $f_{T^{y}_{e}}(\tau )$ and $f_{-T^{x}_{e}}(t - \tau )$.

$$\begin{aligned} \therefore \int _{T_s}^{\infty }f_{T^{y}_{e}}(\tau ) f_{-T^{x}_{e}}(t - \tau ) \mathrm {d}\tau = 0 \end{aligned}$$

Appendix 2: Derivation of PDF of estimated periods for generalized Pareto Distribution

Case 1 : $t + T_s< \mu \implies t < \mu - T_s$

No overlap in both the distributions, i.e. $f_{T_i}(\tau )$ and $f_{T_e}(t-\tau )$.

$$\begin{aligned} f_{\hat{T_i}}(t) = \int _{-\infty }^{\mu - T_s} f_{T_i}(\tau ) f_{T_e}(t - \tau ) \mathrm {d}\tau = 0 \end{aligned}$$

Case 2 :$\mu - T_s \le t < \mu$

$$\begin{aligned} f_{\hat{T_i}}(t)&= \int _{\mu }^{t + T_s}\frac{(\sigma )^{\frac{1}{\xi }}}{\Big (\sigma + \xi (\tau - \mu )\Big )^{\frac{1}{\xi } + 1}} \Big (\frac{T_s + t - \tau }{T_s^2}\Big ) \mathrm {d}\tau \\&= \frac{(\sigma )^{\frac{1}{\xi }}}{T_s^2}\Big [\int _{\mu }^{t + T_s}({T_s + t}){\Big (\sigma + \xi (\tau - \mu )\Big )^{-\frac{1}{\xi } - 1}} \\&\quad - \int _{\mu }^{t + T_s}{({\tau }) \bigg (\sigma + \xi (\tau - \mu )\Big )^{-\frac{1}{\xi } - 1}} \mathrm {d}\tau \bigg ] \end{aligned}$$

Solving both the above integrals separately and combining them, we get,

$$\begin{aligned} f_{\hat{T_i}}(t)&= \frac{(T_s + t - \mu )}{T_s^2} \\&\quad - \frac{1}{(1 - \xi )(T_s^2)}\\&\quad \bigg [\sigma - {(\sigma + \xi (t + T_s - \mu ))}^2 f_{T_i}(t + T_s)\bigg ] \end{aligned}$$

Case 3 :$\mu \le t \le \mu + T_s$

$$\begin{aligned} f_{\hat{T_i}}(t)&= \int _{\mu }^{t + T_s} f_{T_i}(\tau ) f_{T_e}(t - \tau ) \mathrm {d}\tau \\&= \int _{\mu }^{t} \frac{(\sigma )^{\frac{1}{\xi }}}{\Big (\sigma + \xi (\tau - \mu )\Big )^{\frac{1}{\xi } + 1}} \Big (\frac{T_s - t + \tau }{T_s^2}\Big ) \mathrm {d}\tau \\&\quad + \int _{t}^{t + T_s} \frac{(\sigma )^{\frac{1}{\xi }}}{\Big (\sigma + \xi (\tau - \mu )\Big )^{\frac{1}{\xi } + 1}} \Big (\frac{T_s + t - \tau }{T_s^2}\Big ) \mathrm {d}\tau \end{aligned}$$

Solving the above integrals separately and combining them, we get,

$$\begin{aligned} \begin{aligned} f_{\hat{T_i}}(t)&= \frac{T_s - t + \mu }{T_s^2} + \frac{1}{(1 - \xi )(T_s^2)}\bigg [\sigma - 2(\sigma + \xi (t - \mu ))^2 f_{T_i}(t) \\&\quad + (\sigma + \xi (t + T_s - \mu ))^2 f_{T_i}(t + T_s) \bigg ] \end{aligned} \end{aligned}$$

Case 4 :$t > \mu + T_s$

$$\begin{aligned} f_{\hat{T_i}}(t)&= \int _{t - T_s}^{t + T_s} f_{T_i}(\tau ) f_{T_e}(t - \tau ) \mathrm {d}\tau \\&= \int _{t - T_s}^{t} \frac{(\sigma )^{\frac{1}{\xi }}}{\Big (\sigma + \xi (\tau - \mu )\Big )^{\frac{1}{\xi } + 1}} \Big (\frac{T_s - t + \tau }{T_s^2}\Big ) \mathrm {d}\tau \\&\quad + \int _{t}^{t + T_s} \frac{(\sigma )^{\frac{1}{\xi }}}{\Big (\sigma + \xi (\tau - \mu )\Big )^{\frac{1}{\xi } + 1}} \Big (\frac{T_s + t - \tau }{T_s^2}\Big ) \mathrm {d}\tau \end{aligned}$$

Solving the above two integrals and combining them, we get,

$$\begin{aligned} \begin{aligned} f_{\hat{T_i}}(t)&= \frac{1}{(1 - \xi )T_s^2} \bigg [(\sigma + \xi (t + T_s - \mu ))^2 f_{T_i}(t + T_s) \\&\quad + (\sigma + \xi (t - T_s - \mu ))^2 f_{T_i}(t - T_s) \\&\quad - 2(\sigma + \xi (t - \mu ))^2 f_{T_i}(t) \bigg ] \end{aligned} \end{aligned}$$

Appendix 3 : Derivation of CDF of estimated periods for generalized Pareto Distribution

$$\begin{aligned} F_{\hat{T_i}}(t) = \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau \end{aligned}$$

(21)

Case 1 : $t < \mu - T_s$

Substituting the PDF derived in Case 1 of “Appendix 8”, we get,

$$\begin{aligned} F_{\hat{T_i}}(t) = \int _{-\infty }^{t} 0 \mathrm {d}\tau = 0 \end{aligned}$$

Case 2 :$\mu - T_s \le t < \mu$

$$\begin{aligned} F_{\hat{T_i}}(t)&= \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau = \int _{-\infty }^{\mu - T_s} 0 \mathrm {d}\tau + \int _{\mu - T_s}^{t} f_{T_i}(\tau ) \mathrm {d}\tau \end{aligned}$$

Solving the above integrals by substituting the appropriate PDFs derived for Case 1 and Case 2 in “Appendix 8” and combining them, we get,

$$\begin{aligned} \begin{aligned} F_{\hat{T_i}}(t)&= \frac{1}{2} - \frac{(\mu - t)}{T_s} + \frac{(\mu - t)^2}{2T_s^2} - \frac{\sigma (t - \mu + T_s)}{(1-\xi )T_s^2} \\&\quad + \frac{(\sigma )^{\frac{1}{\xi }}}{(2\xi - 1)(1 - \xi )T_s^2}\\&\quad \Big [(\sigma + \xi (t + T_s- \mu ))^{2 - \frac{1}{\xi }} - (\sigma )^{2 - \frac{1}{\xi }} \Big ] \end{aligned} \end{aligned}$$

Case 3 :$\mu \le t \le \mu + T_s$

$$\begin{aligned} F_{\hat{T_i}}(t)&= \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau \\&= \int _{-\infty }^{\mu - T_s} 0 \mathrm {d}\tau + \int _{\mu - T_s}^{\mu } f_{T_i}(\tau ) \mathrm {d}\tau + \int _{\mu }^{t}f_{T_i}(\tau ) \mathrm {d}\tau \end{aligned}$$

Solving the above integrals by substituting the appropriate PDFs derived for Case 1, Case 2 and Case 3 in “Appendix 8” and combining them, we get,

$$\begin{aligned} \begin{aligned} F_{\hat{T_i}}(t)&= \frac{1}{2} - \frac{\mu - t}{T_s} - \frac{{(\mu - t)}^2}{2T_s^2} + \frac{(\sigma )(t - \mu - T_s)}{(1 - \xi )T_s^2} \\&\quad + \frac{(\sigma )^{\frac{1}{\xi }}}{(2\xi - 1)(1 - \xi )(T_s^2)} \Big [(\sigma + \xi (t + T_s - \mu ))^{2 - \frac{1}{\xi }} \\&\quad + (\sigma )^{2-\frac{1}{\xi }} - 2(\sigma + \xi (t - \mu ))^{2 - \frac{1}{\xi }}\Big ] \end{aligned} \end{aligned}$$

Case 4 :$t > \mu + T_s$

$$\begin{aligned} F_{\hat{T_i}}(t)&= \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau \\&= \int _{-\infty }^{\mu - T_s} 0 \mathrm {d}\tau + \int _{\mu - T_s}^{\mu } f_{T_i}(\tau ) \mathrm {d}\tau \\&\quad + \int _{\mu }^{\mu +T_s}f_{T_i}(\tau ) \mathrm {d}\tau + \int _{\mu +T_s}^{t}f_{T_i}(\tau ) \mathrm {d}\tau \end{aligned}$$

Solving the above integrals by substituting the appropriate PDFs derived for Case 1, Case 2, Case 3 and Case 4 in “Appendix 8” and combining them, we get,

$$\begin{aligned} \begin{aligned} F_{\hat{T_i}}(t)&= 1 + \frac{(\sigma )^{\frac{1}{\xi }}}{(2\xi - 1)(1 - \xi )(T_s^2)} \Big [(\sigma + \xi (t + T_s - \mu ))^{2 - \frac{1}{\xi }} \\&\quad - 2(\sigma + \xi (t - \mu ))^{2 - \frac{1}{\xi }} + (\sigma + \xi (t - T_s - \mu ))^{2 - \frac{1}{\xi }} \Big ] \end{aligned} \end{aligned}$$

Appendix 4: Derivation of PDF of estimated periods for Pareto distribution

Case 1 : $t + T_s< \mu \implies t < \mu - T_s$

No overlap in both the distributions, i.e. $f_{T_i}(\tau )$ and $f_{T_e}(t-\tau )$.

$$\begin{aligned} f_{\hat{T_i}}(t) = \int _{-\infty }^{\mu - T_s} f_{T_i}(\tau ) f_{T_e}(t - \tau ) \mathrm {d}\tau = 0 \end{aligned}$$

Case 2 :$\mu - T_s \le t < \mu$

$$\begin{aligned} f_{\hat{T_i}}(t)&= \int _{\mu }^{t + T_s}\frac{\alpha }{\lambda } \left( 1+\frac{t-\mu }{\lambda }\right) ^{-(\alpha +1)} \left( \frac{T_s+t-\tau }{T_s^2} \right) d\tau \end{aligned}$$

On solving the above integral, we get

$$\begin{aligned} f_{\hat{T_i}}(t)&= \frac{\lambda \left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{1-\alpha }+(\alpha -1) (t+T_s-\mu )-\lambda }{(\alpha -1) T_s^2} \end{aligned}$$

Case 3 :$\mu \le t \le \mu + T_s$

$$\begin{aligned} f_{\hat{T_i}}(t)&= \int _{\mu }^{t + T_s} f_{T_i}(\tau ) f_{T_e}(t - \tau ) \mathrm {d}\tau \\&= \int _{\mu }^{t} \frac{\alpha }{\lambda } \left( 1+\frac{t-\mu }{\lambda }\right) ^{-(\alpha +1)} \left( \frac{T_s-t+\tau }{T_s^2} \right) d\tau \\&\quad + \int _{t}^{t + T_s} \frac{\alpha }{\lambda } \left( 1+\frac{t-\mu }{\lambda }\right) ^{-(\alpha +1)} \left( \frac{T_s+t-\tau }{T_s^2} \right) d\tau \end{aligned}$$

Solving the above integrals separately and combining them, we get,

$$\begin{aligned} f_{\hat{T_i}}(t)&= {} \frac{\lambda }{T_s^2 (\lambda -\alpha \lambda )} \bigg [\left( \frac{\lambda +t-\mu }{\lambda }\right) ^{-\alpha } ((\alpha -1) T_s \\&+\lambda +t-\mu )+\left( \frac{\lambda +t-\mu }{\lambda }\right) ^{-\alpha }\\&(-\alpha T_s+\lambda +t+T_s-\mu ) \\&- (\lambda +t+T_s-\mu ) \left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{-\alpha }\\&+(\alpha -1) (t-T_s-\mu )-\lambda \bigg ] \end{aligned}$$

Case 4 :$t > \mu + T_s$

$$\begin{aligned} f_{\hat{T_i}}(t)&= \int _{t - T_s}^{t + T_s} \frac{\alpha }{\lambda } \left( 1+\frac{t-\mu }{\lambda }\right) ^{-(\alpha +1)} \left( \frac{T_s-t+\tau }{T_s^2} \right) d\tau \\&\quad + \int _{t}^{t + T_s} \frac{\alpha }{\lambda } \left( 1+\frac{t-\mu }{\lambda }\right) ^{-(\alpha +1)} \left( \frac{T_s+t-\tau }{T_s^2} \right) d\tau \end{aligned}$$

Solving the above two integrals and combining them, we get,

$$\begin{aligned} f_{\hat{T_i}}(t)&= {} \frac{1}{T_s^2 (\lambda -\alpha \lambda )} \bigg [-\lambda ^2 \left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{1-\alpha } \\&-\lambda \bigg (\lambda \left( \frac{\lambda +t-T_s-\mu }{\lambda }\right) ^{1-\alpha }-\left( \frac{\lambda +t-\mu }{\lambda }\right) ^{-\alpha }\\&((\alpha -1) T_s+\lambda +t-\mu )\bigg ) + \lambda \left( \frac{\lambda +t-\mu }{\lambda }\right) ^{-\alpha }\\&\times (-\alpha T_s+\lambda +t+T_s-\mu )\bigg ]. \end{aligned}$$

Appendix 5: Derivation of CDF of estimated periods for Pareto distribution

$$\begin{aligned} F_{\hat{T_i}}(t) = \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau \end{aligned}$$

(22)

Case 1 :$t < \mu - T_s$ Substituting the PDF derived in Case 1 of “Appendix 10”, we get,

$$\begin{aligned} F_{\hat{T_i}}(t) = \int _{-\infty }^{t} 0 \mathrm {d}\tau = 0 \end{aligned}$$

Case 2 :$\mu - T_s \le t < \mu$

$$\begin{aligned} F_{\hat{T_i}}(t)&= \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau = \int _{-\infty }^{\mu - T_s} 0 \mathrm {d}\tau + \int _{\mu - T_s}^{t} f_{\hat{T_i}}(\tau ) \mathrm {d}\tau \end{aligned}$$

Solving the above integrals by substituting the appropriate PDFs derived for Case 1 and Case 2 in “Appendix 10” and combining them, we get,

$$\begin{aligned} \begin{aligned} F_{\hat{T_i}}(t)&= \frac{1}{2 (\alpha -1) T_s^2}\bigg [\frac{2 \lambda \left( \lambda \left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{2-\alpha }-\lambda \right) }{2-\alpha }\\&\quad +(t+T_s-\mu ) ((\alpha -1) (t+T_s-\mu )-2 \lambda )\bigg ] \end{aligned} \end{aligned}$$

Case 3 :$\mu \le t \le \mu + T_s$

$$\begin{aligned} F_{\hat{T_i}}(t)&= \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau \\&= \int _{-\infty }^{\mu - T_s} 0 \mathrm {d}\tau + \int _{\mu - T_s}^{\mu } f_{\hat{T_i}}(\tau ) \mathrm {d}\tau + \int _{\mu }^{t}f_{\hat{T_i}}(\tau ) \mathrm {d}\tau \end{aligned}$$

Solving the above integrals by substituting the appropriate PDFs derived for Case 1, Case 2 and Case 3 in “Appendix 10” and combining them, we get,

$$\begin{aligned} F_{\hat{T_i}}(t)&= {} \frac{1}{2 (\alpha -1) T_s^2}\bigg [\frac{2 \lambda \left( \lambda \left( \frac{\lambda +T_s}{\lambda }\right) ^{2-\alpha }-\lambda \right) }{2-\alpha }+T_s ((\alpha -1) T_s\\&-2 \lambda )\bigg ] -\frac{1}{(\alpha -1) T_s^2}\bigg [-\frac{\lambda ^2 \left( \left( \frac{\lambda +T_s}{\lambda }\right) ^{-\alpha }-2\right) }{\alpha -2}\\&+\frac{\lambda \left( \frac{\lambda +t-\mu }{\lambda }\right) ^{1-\alpha } (\alpha T_s-\lambda -t-2 T_s+\mu )}{\alpha -2} \\&- \frac{\lambda \left( \frac{\lambda +t-\mu }{\lambda }\right) ^{1-\alpha } (\alpha T_s+\lambda +t-2 T_s-\mu )}{\alpha -2} \\&+\frac{(\lambda +t+T_s-\mu )^2 \left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{-\alpha }}{\alpha -2}+\frac{T_s^2 \left( \frac{\lambda +T_s}{\lambda }\right) ^{-\alpha }}{2-\alpha } \\&+\lambda \left( \mu -\frac{2 T_s \left( \frac{\lambda +T_s}{\lambda }\right) ^{-\alpha }}{\alpha -2}\right) +\frac{\alpha t^2}{2}-\alpha t T_s-\alpha t \mu \\&+(\alpha -1) T_s \mu +\frac{1}{2} (\alpha -1) \mu ^2-\lambda t-\frac{t^2}{2}+t T_s+t \mu \bigg ] \end{aligned}$$

Case 4 :$t > \mu + T_s$

$$\begin{aligned} F_{\hat{T_i}}(t)&= \int _{-\infty }^{t} f_{\hat{T_i}} (\tau ) \mathrm {d}\tau \\&= \int _{-\infty }^{\mu - T_s} 0 \mathrm {d}\tau + \int _{\mu - T_s}^{\mu } f_{\hat{T_i}}(\tau ) \mathrm {d}\tau \\&\quad + \int _{\mu }^{\mu +T_s}f_{\hat{T_i}}(\tau ) \mathrm {d}\tau + \int _{\mu +T_s}^{t}f_{\hat{T_i}}(\tau ) \mathrm {d}\tau \end{aligned}$$

Solving the above integrals by substituting the appropriate PDFs derived for Case 1, Case 2, Case 3 and Case 4 in “Appendix 10” and combining them, we get,

$$\begin{aligned} F_{\hat{T_i}}(t)&= {} \frac{1}{2 (2-\alpha ) (\alpha -1)^2 T_s^2}\bigg [2 T_s^2 \bigg (-\left( \frac{\lambda +t-T_s-\mu }{\lambda }\right) ^{-\alpha }\\&-\left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{-\alpha } +\alpha \bigg (\left( \frac{\lambda +t-T_s-\mu }{\lambda }\right) ^{-\alpha }\\&+\left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{-\alpha }- (\alpha -4) \alpha -5\bigg )+2\bigg ) \\&+ 4 (\alpha -1) T_s (\lambda +t-\mu ) \bigg (\left( \frac{\lambda +t+T_s-\mu }{\lambda }\right) ^{-\alpha }\\&-\left( \frac{\lambda +t-T_s-\mu }{\lambda }\right) ^{-\alpha }\bigg ) +2 (\alpha -1) (\lambda +t-\mu )^2 \\&\times \bigg (\left( \frac{\lambda +t-T_s-\mu }{\lambda }\right) ^{-\alpha } +\left( \frac{\lambda +t+T_s-\mu }{l}\right) ^{-\alpha } \\&-2 \left( \frac{\lambda +t-\mu }{\lambda }\right) ^{-\alpha }\bigg )\bigg ] \end{aligned}$$