0$ if $T < 2^{nI(\rvu;\rvyr)}$.

### Secrecy Analysis

We need to show that for the proposed encoder and decoder, the equivocation at the eavesdropper satisfies $$\label{eq:equivCondn}\frac{1}{n}H({\mathsf{\kappa}}|\rvye^n)= I(\rvu;\rvyr)-
I(\rvu;\rvye) + o_n(1),$$where $o_n(1)$ is a term that goes to zero as $n\rightarrow
\infty$.

Note that while the key ${\mathsf{\kappa}}$ in general can be a function of $(\rvs^n,\rvm_\rvx)$ as indicated in , in our coding scheme the secret key is a deterministic functino of $\rvu^n$ and hence we have

$$\begin{aligned}
 \frac{1}{n}H({\mathsf{\kappa}}|\rvye^n)&=
\frac{1}{n}H({\mathsf{\kappa}},\rvu^n|\rvye^n) - \frac{1}{n}H(\rvu^n|\rvye^n,
{\mathsf{\kappa}})\\
&= \frac{1}{n}H(\rvu^n|\rvye^n) - \frac{1}{n}H(\rvu^n|\rvye^n,
{\mathsf{\kappa}})\\
&= \frac{1}{n}H(\rvu^n|\rvye^n) - \eps_n\\\end{aligned}$$

where the last step follows from the fact that there are $T_0 = 2^{n(I(\rvu;\rvye)-\eps_n)}$ sequences in each bin. Again applying the packing lemma we can show that with high probability the eavesdropper uniquely finds the codeword $\rvu^n(L)$ jointly typical with $\rvye^n$ in this set and hence Fano’s Inequality implies that $$\frac{1}{n}H(\rvu^n|\rvye^n,
{\mathsf{\kappa}}) \le \eps_n.$$ It remains to show that $$\frac{1}{n}H(\rvu^n|\rvye^n) \ge I(\rvu;\rvyr)-
I(\rvu;\rvye) - o_n(1).$$ Using the chain rule of the joint entropy we have $$\begin{aligned}
&\frac{1}{n}H(\rvu^n|\rvye^n) = \frac{1}{n}H(\rvu^n) +
\frac{1}{n}H(\rvye^n|\rvu^n)- \frac{1}{n}H(\rvye^n)\\
&= \frac{1}{n}H(\rvu^n) + \frac{1}{n}H(\rvye^n|\rvu^n,\rvs^n)-
\frac{1}{n}H(\rvye^n) + \frac{1}{n}I(\rvs^n;\rvye^n|\rvu^n).
\label{eq:condEntLB}\end{aligned}$$ We now appropriately bound each term in . First note that since the sequence $\rvu^n$ is uniformly distributed among 