Is it possible to give an expression for the conditional differential entropy $h(A+B\mid C+D),$ where $A,B,C,D$ are normally distributed with known standard deviations $σ_A,\ldots,σ_D$ and where all but $A$ and $C$ are conditinally independent? For $A$ and $C,$ the Pearson’s correlation value $C$ is known. The resulting expression should depend on $σ_A,\ldots,σ_D$ and $C.$ Thank you very much for your help!
1 Answer
$\newcommand\si\sigma$The question is stated very poorly. In particular, it is unclear
- how you define the conditional differential entropy
- if $A, B, C, D$ are jointly normal
- if they are zero-mean
- what you mean by "all but A and C are conditinally independent"
Assuming that $A, B, C, D$ are jointly normal and zero-mean, $A+B$ and $C+D$ will be jointly normal and zero-mean, and then simple calculations yield
$$h(A+B|C+D)=\ln\si_{A+B}+\frac12(1+\ln(2\pi(1-\rho^2))),$$
where $\si_{A+B}$ is the standard deviation of $A+B$, $\rho$ is the correlation coefficient of $A+B$ and $C+D$,
$$h(X|Y):=-\int_{\mathbb R^2} f_{X,Y}(x,y)\ln f_{X|Y}(x|y)\,dx\,dy,$$
$f_{X,Y}$ is the joint pdf of random variables $X$ and $Y$, and $f_{X|Y}$ is the conditional pdf of $X$ given $Y$.
In turn, it is straightforward to express $\si_{A+B}$ and $\rho$ in terms of the variances and covariances of $A, B, C, D$.
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$\begingroup$ Thank you very much! Apologies for being very imprecise with my question. Indeed, your assumptions of zero mean and jointly normal distributions are correct. With conditionally independent, I meant that $h(A|B) = h(A)$, $h(B|A) = h(B)$, etc... And finally, I guess you have a different definition of conditional differential entropy, since I would assume $h(X|Y)$ to be $h(X|Y)=ln(\sigma_X \sqrt{2 \pi e}) + \frac{1}{2} ln(1-\rho^2)$ for normal distributions. $\endgroup$ Commented Dec 20, 2022 at 12:24
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$\begingroup$ @ChristianWagner : Your expression for the conditional differential entropy in the jointly normal case is the same as mine, with an easy rearrangement. $\endgroup$ Commented Dec 20, 2022 at 19:26
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$\begingroup$ @ChristianWagner : Do you have a further response to the answer and my comment? $\endgroup$ Commented Dec 21, 2022 at 21:00