Help with interpretation of entropy and conditional entropy with multiple random variables.

Question

I am working in a problem where I need to calculate the Mutual Information between two random variables $m$ and $z$ given a linear transformation $\mathbf{G}$. In this problem, $m$ is turned into $x$ through a non-linear transformation and then $z$ is given by $$ z = \Gamma\left[ \mathbf{G} x + w \right], $$ where w is additive white Gaussian noise and $\Gamma \left[ \cdot \right] $ is a floor function.

The Mutual Information is then, $$ I(m; z | \mathbf{G}) = h(m| \mathbf{G}) - h(m | z, \mathbf{G}). $$ As $m$ and $\mathbf{G}$ are independent, $$ I(m; z | \mathbf{G}) = h(m) - h(m | z, \mathbf{G}). $$

Now I need to find a way to compute $ h(m | z, \mathbf{G}) $, which I tried by using the following property: $$ h(m | z, \mathbf{G}) = h(z, m | \mathbf{G}) - h(z, m| \mathbf{G}). $$

My questions is: how can interpret the entropies $ h(m| z,\mathbf{G}) $ and $ h(z, m| \mathbf{G}) $?

To be more specific:

1. Is $ h(m | z, \mathbf{G}) $ the entropy of $m$ given $ z $ and $ \mathbf{G} $?

2. In this case could I make $ h(m | z, \mathbf{G}) = h(m|z) $ even if $z$ depends on $ \mathbf{G} $?

3. Is $ h(z, m | \mathbf{G}) $ the entropy of a random variable originated from $z$ and $ m | \mathbf{G} $, or it is the conditional entropy of a random variable originated from $m$ and $z$, given $\mathbf{G}$? I do believe it is the later case, but I am not sure. This is the most confusing one for me.

Thanks for any help and for understanding this confusing question (at least for me). The main goal is to achieve an expression without any conditional if it is possible, even if it is an upper/lower bound, so feel free to provide any further directions if you wish. Also, any material that could be of help in solving this is also very much appreciated. I am available to provide any further information if needed.

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Updating the question with additional information:

The noise is $ w \sim \mathcal{CN}(0, \sigma^2)$. The random variable $m$ is discrete uniformly distributed, for instance, drawn from a QPSK modulation set $ \left\{ \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \right\} $.

The variable $x$ can be obtained as $$ x(n)=\frac{1}{h_0}\Gamma\left[m(n)-\sum_l^{ }h_lx(n-l)\right] $$, where $n$ is a time index, and $h_l \sim \mathcal{NC}(0, \sigma_l^{2})$ are channel taps, and $$\Gamma \left[ \alpha \right ] = \alpha - 2A \left\lfloor\frac{\alpha + A +jA}{2A} \right\rfloor$$ and $A$ is a constant.

The transformation $\mathbf{G}$ is actually a convolution matrix containing the taps of a channel $g_l$. Thus, $ \mathbf{G} x $ is the output of filtering $x$ by a filter with inpulse responde $g$. Both $g$ and $h$ are finite impulse responses.

My goal is to compute the mutual information for when $g \neq h $ and independent from each other, while the solution for the case $ h = g $ is well known and is given by $$ I(m; z|H) = I(m; z) = h\left(\Gamma[m+w]\right)-h\left(\Gamma[w]\right) $$

Answer 1

1. Yes, but be careful that here it is differential entropy.
2. No, you can't. For example suppose $X$ and $Y$ are independent and $Z=X \oplus Y$. Then, $$H(X|Z,Y)=0 \neq H(X|Z). $$ Note that for simplicity I gave an example on continuous random variables. 3.To interpret $h(z,m|G)$, let $Y=(z,m)$, and then you have $h(Y|G)$. Hence, as you mentioned, it's the entropy of random variables $(z,m)$ given random variable $G$.

The question is not well defined at all. Particularly, the non-linear transformation from $m$ to $x$ is not defined, which is the key step. Also, the distribution of input $m$ and also the variance and mean of noise is not mentioned. Moreover, further information about $G$ is needed, like if it's invertible or not.