channel $P_{Y|X,S}$ with state $S$. The decoder observes $Y$ and recovers $\hat{M}$ with error probability $P_{e}=\mathbf{P}\{M\neq\hat{M}\}$.

We show a one-shot version of the Gelfand-Pinsker theorem [@gelfand1980coding]. This is the first one-shot bound attaining the best known second order result in [@scarlett2015dispersions] (which considers a finite-blocklength, not one-shot scenario). Our bound is stronger than the one-shot bounds in [@verdu2012nonasymp; @yassaee2013oneshot; @watanabe2015nonasymp] (in the second order), and significantly simpler to state and prove than all the aforementioned results. Unlike previous approaches, our proof does not require sub-codebooks.

\[thm:channel\_state\]Fix any $P_{U|S}$ and function $x:\mathcal{U}\times\mathcal{S}\to\mathcal{X}$. There exists a code for the channel $P_{Y|X,S}$ with state distribution $P_{S}$ with message $M\sim\mathrm{Unif}[1:\mathsf{L}]$, with error probability $$P_{e}\le\mathbf{E}\left[1-(1+\mathsf{L}2^{\iota_{U;S}(U;S)-\iota_{U;Y}(U;Y)})^{-1}\right]$$ if $P_{US}\ll P_{U}\times P_{S}$ and $P_{UY}\ll P_{U}\times P_{Y}$, where $(S,U,X,Y)\sim P_{S}P_{U|S}\delta_{x(U,S)}P_{Y|X,S}$.

Let $\{(\bar{U}_{i},\bar{M}_{i}),T_{i}\}_{i\in\mathbb{N}}$ be the points of a Poisson process with intensity measure $P_{U}\times P_{M}\times\lambda_{\mathbb{R}_{\ge0}}$ independent of $M,S$. The encoding function is $(m,s)\mapsto x(\tilde{U}_{P_{U|S}(\cdot|s)\times\delta_{m}},s)$ (let $U=\tilde{U}_{P_{U|S}(\cdot|S)\times\delta_{M}}$, $X=x(U,S)$), and the decoding function is $y\mapsto\tilde{M}_{P_{U|Y}(\cdot|y)\times P_{M}}$ (i.e., $\hat{M}=\tilde{M}_{P_{U|Y}(\cdot|Y)\times P_{M}}$). Note that $(M,S,U,X,Y)\sim P_{M}\times P_{S}P_{U|S}\delta_{x(U,S)}P_{Y|X,S}$. We have $$\begin{aligned}
 & \mathbf{P}\{M\neq\tilde{M}_{P_{U|Y}(\cdot|Y)\times P_{M}}\}\\
 & \le\mathbf{P}\{(U,M)\neq(\tilde{U},\tilde{M})_{P_{U|Y}(\cdot|Y)\times P_{M}}\}\\
 & =\mathbf{E}\left[\mathbf{P}\left\{ \left.(U,M)\neq(\tilde{U},\tilde{M})_{P_{U|Y}(\cdot|Y)\times P_{M}}\,\right|\,M,S,U,Y\right\} \right]\\
 & \stackrel{(a)}{\le}\mathbf{E}\left[1-\left(1+\frac{dP_{U|S}(\cdot|S)\times\delta_{M}}{dP_{U|Y}(\cdot|Y)\times P_{M}}(U,M)\right)^{-1}\right]\\
 & =\mathbf{E}\left[1-(1+\mathsf{L}2^{\iota_{U;S}(U;S)-\iota_{U;Y}(U;Y)})^{-1}\right].\end{aligned}$$ where (a) is by the conditional Poisson matching lemma on $((M,S),\,(U,M),\,Y,\,P_{U|Y}\times P_{M})$ (note that $P_{U,M|M,S}=P_{U|S}\times\delta_{M}$). Therefore there exists a 