MMSE estimation and lattice encoding/decoding

C = 1 2 log2 (1 + SNR) where SNR = PX σ N . Shannon showed that in the limit of using long block length n, generating 2 i.i.d. N (0, PX) codewords and averaging across all codebooks is a capacity-achieving strategy: with high likelihood the power constraint will be satisfied and the average probability of error under ML decoding tends to 0 as n → ∞. This result can be derived geometrically as well. With very high likelihood, from the law of large numbers, a length-n i.i.d. N (0, σ) vector lies on the boundary shell of an n-sphere of radius √ nσ2. Since we’ve generated X ∼ N (0, PX) and N ∼ N (0, σ N ), the received signal is N (0, PX + σ N) and thus with very high likelihood lies on the boundary shell of an n-sphere of radius √