## 3 Derivation of the Lower Capacity Bounds for UWOC with APD  
In order to simplify the math description, we set a scalar symbol \(H\) as \(H=GL\tau\). Thus, (1) is reduced to  
\[Y=Hx+\sqrt{Hx}n^{\prime}+n. \tag{14}\]  
Then, the probability density function (PDF) of the received current signal \(Y\), conditioned on the transmitted OOK symbol \(x\), which is recovered by the APD receiver, can be given by  
\[W(Y|x)=\frac{1}{\sqrt{2\pi\sigma^{2}(1+Hx\varsigma^{2})}}e^{-\frac{(y-Hx)^{2 }}{2\sigma^{2}(1+Hx\varsigma^{2})}}. \tag{15}\]  
According to the complicated constraints given in (10) to (12), and the biggest differential entropy conditions shown in [19], the classic Shannon channel capacity formula cannot be utilized in (14) anymore. That is, we need to re-derive the channel capacity of the UWOC system described by (14) under the constraints of (10) to (12). Generally speaking, it is a very hard issue to give an exact capacity formula to the channel mentioned above [15], so we circumambulate the issue and try to give an alternative solution, i.e., a tight lower  
Figure 2: Illustration of projection of \(\overset{\rightarrow}{V}_{n}\) on the first quadrant of X-Y plane. bound on the channel capacity, by using the knowledge of Shannon information theory. From the definition of channel capacity, we have the following inequality, i.e.,  
\[C\geq I(Q,W)\big{|}_{any\ specifid\ Q(\cdot)}\]  
\[=h(Y)-h(Y|x), \tag{16}\]  
where \(C\) is the target channel capacity; \(I(Q,W)\) stands for the mutual information between the input \(x\) and the output \(Y\) of the channel with PDF \(W(\cdot|\cdot)\) shown in (15) when the input \(x\) has distribution \(Q(\cdot);h(Y)\) is the output entropy and \(h(Y|x)\) is the conditional entropy. By the definition of differential entropy [17] and (15), \(h(Y|x)\) can be obtained as  
\[h(Y|x)=\frac{1}{2}E_{Q}\big{[}log2\pi e\sigma^{2}\Big{(}1+Hx\varsigma^{2} \Big{)}\big{]}\]  
\[=\frac{1}{2}log(2\pi e\sigma^{2}\varsigma^{2}H)+\frac{1}{2}E_{Q}[logx]+\frac{1 }{2}E_{Q}[log(1+\frac{1}{H\varsigma^{2}\chi})], \tag{17}\]  
where, \(E_{Q}[\cdot]\) means calculating the mathematical expectation of the expression consisting of variable \(x\) in brackets through the variable's probability distribution \(Q(\cdot)\). (17) shows us that the entropy \(h(Y|x)\) is a function of distribution \(Q(\cdot)\). In order to solve the lower bound on capacity in (16), we need to further manipulate \(h(Y)\) into the function of \(Q(\cdot)\). According to the work in [15], we can similarly get the lower bound of \(h(Y)\) in terms of \(Q(\cdot)\). To this goal, we first rewrite (14) as  
\[Y=\underbrace{Hx+\sqrt{Hx}n^{\prime}}_{\dot{Y}_{1}}+n=Y_{1}+n, \tag{18}\]  
where \(Y_{1}\) and \(n\) are independent variables, satisfying \(Y_{1}\sim N\big{(}Hx,Hx\sigma^{2}\varsigma^{2}\big{)}\) and \(n\sim N_{R}\big{(}0,\sigma^{2}\big{)}\). By the fact that conditioning will reduce entropy, we have  
\[h(Y)=h(Y_{1}+n)\geq h(Y_{1}+n|n)=h(Y_{1}|n)=h(Y_{1}). \tag{19}\]  
Hence, the problem of (16) is reduced to finding a lower bound to \(h(Y_{1})\). According to the average light power constraint given in (11), we have an exponential distribution \(Q_{Exp}(\cdot)\) on input \(x\) to maximize the entropy \(h(Y_{1})\)[23], i.e.,  
\[Q_{Exp}(x)\triangleq\frac{1}{\varepsilon}e^{-\frac{x}{\varepsilon}},\ \ \ \ \ \ \ \ \ x\geq 0, \tag{20}\]  
where \(\varepsilon\) is the input power constraint value.