that the eavesdropper’s channel is less noisy compared to the legitimate receiver’s channel. Then the secret-key capacity with $\rvs^n$ revealed to both the legitimate terminals is $$C = \max_{p_{\rvx|\rvs}} H(\rvs|\rvye).\label{eq:LessNoisySymCap}$$

Intuitively, when the wiretap channel cannot contribute to the secrecy,  states that transmitter should select an input that masks the state from the output as much as possible.

Symmetric CSI: Numerical Example {#subsec:NumericalExample}
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![image](dataRates4)

It can be easily seen that for the dirty paper coding example in section \[subsec:Gaussian\], the secret-key capacity when $\rvs$ is also revealed to the legitimate receiver is infinity. More generally higher the entropy of $\rvs$, higher will be the gains in the secret-key capacity with symmetric CSI. In this section illustrate the secret-key rate for an on-off channel for the receivers: $$\begin{aligned}
\rvyr&= \rvs_{r} \rvx + \rvzr\\
\rvye&= \rvs_{e} \rvx + \rvze,
\end{aligned}\label{eq:fading}$$ where both $\rvs_{r}, \rvs_{e} \in \{0,1\}$, the random variables are mutually independent and $\Pr(\rvs_{r}=0)= \Pr({\mathsf{s}_{e}}= 0) =
0.5$. Furthermore we assume that ${\mathsf{s}_{r}}$ is revealed to the legitimate terminals, whereas the eavesdropper is revealed $\tilde{\rvye}=({\mathsf{s}_{e}},\rvye)$. The noise random variables are mutually independent, zero mean and unit variance Gaussian random variables and the power constraint is that $E[\rvx^2] \le P$.

We evaluate 