What is the sum-capacity for a non-symmetric interference channel for information theorists?

Question

This question is dedicated for people who are experts in information theory.

An interesting result for a two user interference channel in information theory, is the sum-capacity to within one bit. It however was derived for the symmteric case.

I was wondering if anyone knows what the sum-capacity bound for the two user channel is, say for example user #1 wants to communicate with its own receiver at rate $R_1$ and user two communicates with receiver 2 at a rates $R_2$.

And let us say I have the following channel

$$Y_1= h_{11}X_1 + h_{12}X_2+Z_1$$ $$Y_2= h_{22}X_1 + h_{21}X_2+Z_2$$ where $h_{ij}$ for $i,j\in[1,2]^2$ are channel gains from receiver $j$ to transmitter $i$ and $Z_{i}$ for $i\in[1,2]$ are additive white gaussian noise mean zero variance one.

Can anyone help me write down what the sum-capacity for this general 'non' symmetric gaussian interference channel is?

Answer 1

For, two user non-symmetric Gaussian interference channel capacity (not just sum-capacity) is know to withing 1 bit. This result was show by Etkin, Tse, and Wang.

If you have Network Information Theory book by El Gamal look at chapter 6 they have outer bound that is tight within in 1 bit.

Answer 2

See ''Sum Capacity of the Vector Gaussian Broadcast Channel and Uplink–Downlink Duality'' by Viswanath and David, in IEEE Transactions on Information Theory v49, No. 8, Aug 2003.

These people found that for a MIMO broadcast with $N$ antennas and $K$ users, with additive iid CCS Gaussian noise $\eta$, with a flat-fading $N\times K$ channel (gain) matrix $H$,

(All this makes the reception $\vec{y}=H^{\dagger}\vec{x}+\vec{\eta}$)

the sum capacity is:

$$\underset{D\in \mathcal{A_1}}{\sup} \log \det ( I + HDH^{\dagger})$$

with $\mathcal{A_1}$ being the set of nonnegative $N \times N$ diagonal matrices, whose trace is less than or equal to your power constraint at the broadcaster.

The 'interference' in your case is cooked into $H$.