# On Parameter Estimation for Bandlimited Optical Intensity Channels
## 3 Modified Cramer-Rao Lower Bound
### Computation of the Modified CRLB  
Using the LLF in (12), the FIM elements can be computed according to (8). Since we need not consider any nuisance parameters, the expectation is for a DA solution only with regard to the AWGN process. As a consequence, the entries would be a function of the selected synchronization sequence \(a_{k}\) with \(k\in\mathbb{K}\). In the sequel, we want to circumvent this sort of dependency by averaging the second-order derivative in (8) not only with respect to \(w(t)\), but also with respect to the i.i.d. symbols \(a_{k}\in\mathbb{A}\). This provides us with the so-called modified Cramer-Rao lower bound (MCRLB), which is in general less tight than the true limit [15; 16; 17; 18], i.e., \(\text{MCRLB}(u_{m})\leq\text{CRLB}(u_{m})\). In the next step, the FIM elements \(J_{m,n}\) will be derived for all \(m\), \(n\in\{1,2\}\). Starting with \(u_{1}=u_{2}=A\), we obtain  
\[ \begin{array}{ll}J_{1,1}&=-\text{E}\Big{[}\frac{\partial^{2}\Lambda( \mathbf{r};\mathbf{u})}{\partial A^{2}}\Big{]}\\ &=\frac{1}{\sigma_{w}^{2}}\sum\limits_{k}\sum\limits_{n}\text{E}[a_{k}a_{n}] \,g[(k-n)T]\\ &=\frac{1}{\sigma_{w}^{2}}\Big{\{}\sum\limits_{k}\text{E}[a_{k}^{2}]\,g(0)+2 \underset{k}{\sum}\sum\limits_{n>k}\text{E}[a_{k}]\,\text{E}[a_{n}]\,g[(k-n)T] \Bigg{\}}\\ &=\frac{1}{\sigma_{w}^{2}}\left\{L\,g(0)+2\,\mu_{a}^{2}\sum\limits_{i=1}^{-1} \left(L-i\right)g(iT)\right\},\end{array}  \tag{13}\]  
where, for notational brevity, it has been tacitly assumed that the summation indexes \(k\) and \(n\) are elements of K. Then, with \(u_{1}=A\), \(u_{2}=\tau\), and \(\dot{h}(t)=\partial h(t)/\partial t\), the related FIM element is given by  
\[ \begin{array}{ll}J_{1,2}&=-\text{E}\Big{[}\frac{\partial^{2}\Lambda( \mathbf{r};\mathbf{u})}{\partial A\,\partial\tau}\Big{]}\\ &=\frac{A^{2}}{\sigma_{w}^{2}}\underset{k}{\sum}\sum\limits_{n}\text{E}[a_{k}a_ {n}]\,\int\limits_{-\infty}^{\infty}\dot{h}(t-kT-\tau)h(t-nT-\tau)\,dt\\ &=\frac{A^{2}}{\sigma_{w}^{2}}\underset{k}{\sum}\sum\limits_{n}\text{E}[a_{k}a _{n}]\,\dot{g}[(k-n)T]=0\,,\end{array}  \tag{14}\]  
where we used the fact that the integral in (14) is equivalent to the first-order derivative of \(g(t)\), expressed as \(\dot{g}(t)\), which has to be evaluated at integer multiples of the symbol period within the index range determined by the observation interval. Since \(\dot{g}(t)\) exhibits an odd symmetry with respect to \(t=0\), it is not difficult to show that \(J_{1,2}=0\). Since the LLF is a steadily differentiable function with respect to \(u_{1}\) and \(u_{2}\), it is clear that \(J_{2,1}=J_{1,2}=0\). Finally, with \(u_{1}=u_{2}=\tau\) and \(\ddot{h}(t)=\partial^{2}h(t)/\partial t^{2}\), the corresponding FIM element develops as  
\[ \begin{array}{ll}J_{2,2}&=-\mathbb{E}\Big{[}\frac{\partial^{2}\Lambda(r \mathbf{u})}{\partial\tau^{2}}\Big{]}\\ &=-\frac{\Lambda^{2}}{\sigma_{m}^{2}}\sum\limits_{k}\sum\limits_{n}\mathbb{E} [a_{k}a_{n}]\int\limits_{-\infty}^{\infty}\ddot{h}(t-kT-\tau)h(t-nT-\tau)\,dt \\ &=-\frac{\Lambda^{2}}{\sigma_{m}^{2}}\sum\limits_{k}\sum\limits_{n}\mathbb{E} [a_{k}a_{n}]\,\ddot{\bar{g}}[(k-n)\,T]\\ &=-\frac{\Lambda^{2}}{\sigma_{m}^{2}}\bigg{\{}L\,\ddot{\bar{g}}(0)+2\,\mu_{a}^ {2}\sum\limits_{i=1}^{L-1}\left(L-i\right)\ddot{\bar{g}}(iT)\bigg{\}}\,,\end{array}  \tag{15}\]  
after having exploited the fact that the integral expression in (15) describes the second-order derivative of \(g(t)\), denoted as \(\ddot{\bar{g}}(t)\), which has to be evaluated at integer multiples of the symbol period for all \(k\) and \(n\in\mathrm{K}\). Taking into account that \(\ddot{\bar{g}}(t)\) is even-symmetric, the simplification in the last line of (15) is achieved straightforwardly. According to (7), it is obvious that the MCRLB of parameter \(u_{m}\) is solely determined by the reciprocal value of \(J_{m,m}\), when the off-diagonal entries are zero.