# On Optimal Cooperative Sensing with Energy Detection in Cognitive Radio
## 3 Cooperative Sensing with Optimal Thresholds
### Convex Optimization
Since (8) is not convex optimization, we again define \(v_{i}\triangleq\log\mathcal{Q}\left(x_{i}\right)\) to make the inequality constraint of \(v_{i}\) form a convex set. Using \(x_{i}\ =\ (\log\mathcal{Q})^{-1}(v_{i})\), the optimization in (8) becomes

\[\max_{v_{i},i\in\{1,\cdots,N\}} \sum_{i=1}^{N}\log\mathcal{Q}\left(a_{i}\left((\log\mathcal{Q})^{- 1}(v_{i})-\tilde{\gamma}_{i}\right)\right),\] \[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]0

Here, we define a function \(f(t)\) as

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]1

Then, for the objective function of (A), the pointwise sum \(\sum_{i=1}^{N}\log f(v_{i})\) is concave if \(f(v_{i})\) is a log-concave function of \(v_{i}\) for all \(i\)[29]. The log-concavity property of \(f(v_{i})\) needs to hold for all \(i\), so we now omit the subscript \(i\) for simplicity. To check whether \(f(v)\) is log-concave or not, we use the log-concave condition, e.g., the inequality is \(f(v)f^{\prime\prime}(v)\leq{f^{\prime}}^{2}(v)\)[29]. Although one may directly obtain the first and second derivatives of \(f(v)\) with respect to \(v\), it is quite difficult to deal with \((\log\mathcal{Q})^{-1}(v)\) in \(f(v)\). Interestingly, we notice that it is possible to express both \(f^{\prime}(v)\) and \(f^{\prime\prime}(v)\) as _a function of \(x\)_. Once we use a function defined as \(z(t)\triangleq\mathcal{Q}(a(t-\tilde{\gamma}))\), then \(f(v)=z(x(v))=\mathcal{Q}(a(x(v)-\tilde{\gamma}))\), where \(x(v)=(\log\mathcal{Q})^{-1}(v)\). To obtain \(f^{\prime}(v)\) and \(f^{\prime\prime}(v)\), based on the chain rule \(\frac{df(v)}{d\mathcal{Q}}=\frac{dz(x(v))}{d\mathcal{Q}}=\frac{dz(x)}{d \mathcal{Q}}\cdot\frac{dx}{d\mathcal{Q}}\) as well as the derivative formula \((u^{-1})^{\prime}(v)=\frac{1}{u^{\prime}(u^{-1}(v))}\) for any analytic function \(u:\mathbf{R}\rightarrow\mathbf{R}\)[30], we have

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]2

which is _only a function of \(x\)_. Thus, we have \(f^{\prime}(v)\) as follows:

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]3

In a similar way, we also have \(f^{\prime\prime}(v)\) as follows:

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]4Applying the log-concave condition \(f^{\prime\prime}(v)f(v)\leq{f^{\prime}}^{2}(v)\) with (A2) and (A3), the following long inequality condition is given by

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]5 \[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]6 (A4)

After several steps of manipulations with \(\mathcal{Q}^{\prime\prime}(t)=\frac{t}{\sqrt{2\pi}}\exp\left(\frac{-t^{2}}{2 }\right)=-t\mathcal{Q}^{\prime}(t)\) by (3), the inequality condition in (A4) boils down to:

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]7

Finally, using a function \(R(t)=\frac{\mathcal{Q}^{\prime}(t)}{\mathcal{Q}(t)}\), the inequality condition is given by

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]8

With notations of \(x_{i}\) and \(y_{i}\), we obtain

\[\sum_{i=1}^{N}v_{i}^{*}-\log\overline{P^{f}} \leq 0,\]9

which completes the proof. ## References

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