Impact of Uncertainty About a User to be Active on OFDM Transmission Strategies

Abstract

In this paper we investigate the impact that incomplete knowledge regarding user activity can have on the equilibrium transmission strategy for an OFDM-based communication system. The problem is formulated as a two user non-zero sum game for independent fading channel gains, where the equilibrium strategies are derived in closed form. This allows one to show that a decrease in uncertainty about the user activity could reduce the number of subcarriers jointly used by the users. For the boundary case (with complete information, which reflects a classical water-filling game) the equilibrium strategies are given explicitly. The necessary and sufficient conditions, when channels sharing strategies are optimal, is established as well as the set of shared subcarriers is identified. The stability of the upper bound of the size of this set with respect to power budgets is derived.

A Appendix

Proof of Theorem 1:

Since $v^j({\varvec{P}}^1,{\varvec{P}}^2)$ is concave on ${\varvec{P}}^j$, (a) follows [2]. The KKT Theorem straightforward implies (b).

(c) It is clear that the sets of equilibrium coincide for the games with payoffs scaled by positive multiplies. That is why, to find equilibrium instead of the original game with payoffs $(v^1,v^2)$ we can consider equivalent game with payoffs $(V^1,V^2)=(v^1,q^1v^2)$. The last game is an exact potential game [14], and so, the best response algorithm converges. Recall that the game with payoffs $(V^1,V^2)$ is an exact potential game if and only if there is a function $V({\varvec{P}}^1,{\varvec{P}}^2)$ such that for any strategies $({\varvec{P}}^1,{\varvec{P}}^2)$ and $({\varvec{P}}^1_*,{\varvec{P}}^2_*)$ the following conditions hold:

$$\begin{aligned} \begin{aligned} V({\varvec{P}}^1_*,{\varvec{P}}^2)-V({\varvec{P}}^1,{\varvec{P}}^2)&=V^1({\varvec{P}}^1_*,{\varvec{P}}^2)-V^1({\varvec{P}}^1,{\varvec{P}}^2), \\ V({\varvec{P}}^1,{\varvec{P}}^2_*)-V({\varvec{P}}^1,{\varvec{P}}^2)&=V^2({\varvec{P}}^1,{\varvec{P}}^2_*)-V^2({\varvec{P}}^1,{\varvec{P}}^2). \end{aligned} \end{aligned}$$

(17)

It is clear for the function

$$ V({\varvec{P}}^1,{\varvec{P}}^2)=q^1\sum \limits _{i=1}^n\ln (\sigma ^2+ h^1_iP^1_i+ h^2_iP^2_i)+q^0\sum \limits _{i=1}^n\ln (\sigma ^2+ h^1_iP^1_i) $$

the condition (17) holds, and the result follows. $\quad \blacksquare$

Proof of Theorem 2:

Since $v^j({\varvec{P}}^1,{\varvec{P}}^2)$ is concave on ${\varvec{P}}^j$, by KKT Theorem, $({\varvec{P}}^1,{\varvec{P}}^2)$ is an equilibrium if and only if there are $\omega ^1$ and $\omega ^2$ (Lagrangian multipliers) such that the following conditions hold:

$$\begin{aligned} \frac{q^1 h^1_i}{\sigma ^2+ h^1_iP^1_i + h^2_iP^2_i}+\frac{q^0 h^1_i}{\sigma ^2+ h^1_iP^1_i} {\left\{ \begin{array}{ll} =\omega ^1,&{}P^1_i>0,\\ \le \omega ^1,&{}P^1_i=0, \end{array}\right. } \end{aligned}$$

(18)

$$\begin{aligned} \frac{h^2_i}{\sigma ^2+ h^1_iP^1_i + h^2_iP^2_i} {\left\{ \begin{array}{ll} =\omega ^2,&{}P^2_i>0,\\ \le \omega ^2,&{}P^2_i>0. \end{array}\right. } \end{aligned}$$

(19)

Thus, by (18) and (19), we have that

(a) if $P^1=0$ and $P^2=0$ then $h^1_i/\sigma ^2\le \omega ^1$ and $h^2_i/\sigma ^2\le \omega ^2$,

(b) if $P^1>0$ and $P^2=0$ then $P^1_i=1/\omega ^1-\sigma ^2/h^1_i$ with $h^1_i/\sigma ^2> \omega ^1$ and $h^2_i/h^1_i\le \omega ^2/\omega ^1$.

(c) if $P^1=0$ and $P^2>0$ then $P^2_i=1/\omega ^2-\sigma ^2/h^2_i$ with $h^2_i/\sigma ^2> \omega ^2$ and $q^1h^1_i/h^2_i\omega ^2 + q^0h^1_i/\sigma ^2 \le \omega ^1$.

(d) if $P^1>0$ and $P^2>0$ then

$$ P^1_i=\frac{q^0}{\omega ^1 -q^1 h^1_i \omega ^2/h^2_i}-\frac{\sigma ^2}{h^1_i} \text { and } P^2_i=\frac{1}{\omega ^2} - \frac{h^1_i}{h^2_i}\frac{q^0}{\omega ^1 -q^1 h^1_i \omega ^2/h^2_i}, $$

and the result follows. $\quad \blacksquare$

Proof of Theorem 3:

The set of the channels jointly used by both users is $I_{11}(\omega ^1,\omega ^2)=\left\{ i: \omega ^2/h^2_i<\omega ^1/h^1_i< q^1\omega ^2/h^2_i+q^0/\sigma ^2\right\}$. First, note that due to $q^1\omega ^2/h^2_i+q^0/\sigma ^2>\omega ^2/h^2_i$ and $q^0+q^1=1$ yield that $\sigma ^2>\omega ^2/h^2_i$. Then, $q^1\omega ^2/h^2_i+q^0/\sigma ^2= q^0(1/\sigma ^2-\omega ^2/h^2_i)+\omega ^2/h^2_i$ is increasing on $q^0$, and the result follows. $\quad \blacksquare$

Proof of Theorem 5:

Since $v^j({\varvec{P}}^1,{\varvec{P}}^2)$ is concave on ${\varvec{P}}^j$, by KKT Theorem, $({\varvec{P}}^1,{\varvec{P}}^2)$ is an equilibrium if and only if there are $\omega ^1$ and $\omega ^2$ (Lagrangian multipliers) such that the following conditions hold for $m=1,2:$

$$\begin{aligned} \frac{h^m_i}{\sigma ^2+ h^1_iP^1_i + h^2_iP^2_i} {\left\{ \begin{array}{ll} =\omega ^m,&{}P^m_i>0,\\ \le \omega ^m,&{}P^m_i=0. \end{array}\right. } \end{aligned}$$

(20)

Then, by (20),

$$\begin{aligned} (P^1_i,P^2_i)\!=\! {\left\{ \begin{array}{ll} (0,0),&{}h^1_i/\omega ^1\le \sigma ^2 \text { and } h^2_i/\omega ^2\le \sigma ^2,\\ \left( \frac{\displaystyle 1}{\displaystyle \omega ^1}-\frac{\displaystyle \sigma ^2}{\displaystyle h^1_i},0\right) ,&{}h^1_i/\omega ^1> \sigma ^2 \text { and } h^2_i/\omega ^2 \le h^1_i /\omega ^1,\\ \left( 0,\frac{\displaystyle 1}{\displaystyle \omega ^2}-\frac{\displaystyle \sigma ^2}{\displaystyle h^2_i}\right) ,&{}h^2_i/\omega ^2> \sigma ^2 \text { and } h^1_i/\omega ^1 \le h^2_i /\omega ^2,\\ \sigma ^2+ h^1_iP^1_i + h^2_iP^2_i=\frac{\displaystyle h^1_i}{\displaystyle \omega ^1}=\frac{\displaystyle h^2_i}{\displaystyle \omega ^2},&{} h^1_i/\omega ^1=h^2_i/\omega ^2> \sigma ^2. \end{array}\right. } \end{aligned}$$

(21)

By (21), if $P^1_i>0$ and $P^2_i>0$ then $h^2_i/h^1_i=\omega ^2/\omega ^1$. Thus, by (8), both strategies can employ only at most one channel for joint use. Moreover, there is a k such that

$$\begin{aligned} P^1_i{\left\{ \begin{array}{ll}>0,&{}i<k-1,\\ \ge 0,&{}i=k,\\ =0,&{}i>k, \end{array}\right. } \text { and } P^2_i{\left\{ \begin{array}{ll} =0,&{}i<k-1,\\ \ge 0,&{}i=k,\\>0,&{}i>k, \end{array}\right. } \end{aligned}$$

(22)

where (a) $P^1_k>0$ and $P^2_k>0$ if $h^2_k/h^1_k=\omega ^2/\omega ^1$, (b) $P^1_k>0$ and $P^2_k=0$ if $h^2_k/h^1_k< \omega ^2/\omega ^1$, and (c) $P^1_k=0$ and $P^2_k>0$ if $h^2_k/h^1_k>\omega ^2/\omega ^1$.

Thus, by assumption (8), we have to consider separately two cases: (A) there is a k such that $h_k<\omega ^2/\omega ^1<h_{k+1}$, (B) there is a k such that $h_k=\omega ^2/\omega ^1$.

(A) Let there exist a k such that $h_k<\omega ^2/\omega ^1<h_{k+1}$. Then, by (21), (22) and the fact that ${\varvec{P}}^1\in \varPi ^1$ and ${\varvec{P}}^2\in \varPi ^2$, we have that ${\varvec{P}}^1$ and ${\varvec{P}}^2$ have to be given by (12), and $\omega ^1=\omega ^1_k$ and $\omega ^2=\omega ^2_k$. Thus, (11) also has to hold.

(B) Let there exist a k such that $h_k=\omega ^2/\omega ^1$. Thus, (15) holds. Also, by (21), (22) and the fact that ${\varvec{P}}^1\in \varPi ^1$ and ${\varvec{P}}^2\in \varPi ^2$, we have that ${\varvec{P}}^1$ and ${\varvec{P}}^2$ have to be given by (14) and also the following condition has to hold:

$$\begin{aligned} \sigma ^2+ h^1_kP^1_k + h^2_kP^2_k=\frac{\displaystyle h^1_k}{\displaystyle \omega ^1}=\frac{\displaystyle h^2_k}{\displaystyle \omega ^2} \end{aligned}$$

(23)

By (23) with right side $h^1_k/\omega ^1$, $P^1_k=1/\omega ^1- (\sigma ^2+P^2_k)/h^1_k<1/\omega ^1- \sigma ^2/h^1_k$. Substituting this $P^1_k$ into (14) and taking into account that ${\varvec{P}}^1\in \varPi ^1$ yield that

$$\begin{aligned} \omega ^1\le \omega ^1_k. \end{aligned}$$

(24)

Similarly, dealing with strategy ${\varvec{P}}^2$ in condition (23) with right side $h^2_k/\omega ^2$ implies that

$$\begin{aligned} \omega ^2\le \omega ^2_{k-1}. \end{aligned}$$

(25)

By (14), the condition (23) with right side $h^1_k/\omega ^1$ is equivalent to

$$\begin{aligned} \sigma ^2+ h^1_k\left( \overline{P}^1-\sum \limits _{j=1}^{k-1}\left\lfloor \frac{\displaystyle 1}{\displaystyle \omega ^1}-\frac{\displaystyle \sigma ^2}{\displaystyle h^1_i}\right\rfloor _+\right) + h^2_k\left( \overline{P}^2-\sum \limits _{j=k+1}^{n}\left\lfloor \frac{\displaystyle 1}{\displaystyle \omega ^2}-\frac{\displaystyle \sigma ^2}{\displaystyle h^2_i}\right\rfloor _+ \right) =\frac{\displaystyle h^1_k}{\displaystyle \omega ^1}. \end{aligned}$$

(26)

Substituting (15) into (26) implies (16).

Since the left side of Eq. (16) is decreasing on $\omega ^1$, by (24), it has a root if and only of $F(\omega ^1_k)<\overline{P}^1+h_k \overline{P}^2$. This condition is equivalent to

$$ \overline{P}^2>\sum \limits _{j=k+1}^{n}\left\lfloor \frac{\displaystyle 1}{\displaystyle \omega ^1 h_k}-\frac{\displaystyle \sigma ^2}{\displaystyle h^2_i}\right\rfloor _+\,=\,(\text {by (15)})\,=\, \sum \limits _{j=k+1}^{n}\left\lfloor \frac{\displaystyle 1}{\displaystyle \omega ^2}-\frac{\displaystyle \sigma ^2}{\displaystyle h^2_i}\right\rfloor _+. $$

Thus, $\omega ^2>\omega ^2_k$. Substituting (15) in the last inequality and taking into account (24) implies that

$$\begin{aligned} \xi _k<h_k. \end{aligned}$$

(27)

By (14) and (15), the condition (23) with right side $h^2_k/\omega ^2$ is equivalent to

$$\begin{aligned} G(\omega ^2):=\sum \limits _{j=1}^{k-1}\left\lfloor \frac{\displaystyle 1}{\displaystyle \omega ^2}-\frac{\displaystyle \sigma ^2}{\displaystyle h^1_i h_k}\right\rfloor _+ + \sum \limits _{j=k}^{n}\left\lfloor \frac{\displaystyle 1}{\displaystyle \omega ^2}-\frac{\displaystyle \sigma ^2 }{\displaystyle h^2_i}\right\rfloor _+= \overline{P}^1/h_k+\overline{P}^2. \end{aligned}$$

(28)

Thus, by (25), this equation has a positive root if and only if $G(\omega ^2_{k-1})<\overline{P}^1/h_k+\overline{P}^2$. By (15) and (28), this is equivalent to $\omega ^1_{k-1}<\omega ^1$. This, jointly with (15) and (25), implies that $\xi _{k-1}>h_k.$ Then, taking into account (27) yields (13), and the result follows. $\quad \blacksquare$