Mutual Information, Relative Entropy, and Estimation in the Poisson Channel

Abstract

Let $X$ be a non-negative random variable and let the conditional distribution of a random variable $Y$, given $X$, be ${Poisson}(\gamma \cdot X)$, for a parameter $\gamma \geq 0$. We identify a natural loss function such that: 1) The derivative of the mutual information between $X$ and $Y$ with respect to $\gamma$ is equal to the \emph{minimum} mean loss in estimating $X$ based on $Y$, regardless of the distribution of $X$. 2) When $X \sim P$ is estimated based on $Y$ by a mismatched estimator that would have minimized the expected loss had $X \sim Q$, the integral over all values of $\gamma$ of the excess mean loss is equal to the relative entropy between $P$ and $Q$. For a continuous time setting where $X^T = \{X_t, 0 \leq t \leq T \}$ is a non-negative stochastic process and the conditional law of $Y^T=\{Y_t, 0\le t\le T\}$, given $X^T$, is that of a non-homogeneous Poisson process with intensity function $\gamma \cdot X^T$, under the same loss function: 1) The minimum mean loss in \emph{causal} filtering when $\gamma = \gamma_0$ is equal to the expected value of the minimum mean loss in \emph{non-causal} filtering (smoothing) achieved with a channel whose parameter $\gamma$ is uniformly distributed between 0 and $\gamma_0$. Bridging the two quantities is the mutual information between $X^T$ and $Y^T$. 2) This relationship between the mean losses in causal and non-causal filtering holds also in the case where the filters employed are mismatched, i.e., optimized assuming a law on $X^T$ which is not the true one. Bridging the two quantities in this case is the sum of the mutual information and the relative entropy between the true and the mismatched distribution of $Y^T$. Thus, relative entropy quantifies the excess estimation loss due to mismatch in this setting. These results parallel those recently found for the Gaussian channel.