In the paper "Directed Information, Causal Estimation, and Communication in Continuous Time" the author show an example of continuous Gaussian Channel:

Let $\{B_t\}$ be a standard Brownian motion and $A ∼ N(0,1)$ be independent of $\{B_t\}$. Let $X_t ≡ A$ for all t and $dY_t = X_tdt + dB_t$. Letting $J(P, N) = (1/2) ln((P + N)/N)$ denote the mutual information between a Gaussian random variable of variance $P$ and its corrupted version by an independent Gaussian noise of variance $N$, we have for every $t \in [0, T)$:

$I(Y_t^{t+δ};X_0^{t+δ}|Y_0^t) = J(\frac{1/t}{1+1/t},1/δ)$


It looks that the author have $P(X_t^{t+\delta})=\frac{1/t}{1+1/t}$. How does this come into being?

Some useful backgrounds: in Information Theory,under the case of discrete Gaussian channel $Y = X + n$ where $X ∼ N(0,\sigma_x^2)$ and $n ∼ N(0,\sigma_0^2)$, the mutual information $I(X;Y) = \frac{1}{2} ln(\frac{\sigma_x^2 + \sigma_0^2}{\sigma_0^2})$. The paper mentioned above could be accessible at :http://arxiv.org/abs/1109.0351 (Page 7,example 1). A simplified one could be found at http://circuit.ucsd.edu/~yhk/pdfs/isit2009c.pdf


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.