on 01-Jan-2024 (Mon)

Using the resultant \(n − k\) = 3-bit syndrome \(\mathbf{s}\), the Hamming decoder decides if it thinks there are any bit errors in \(\hat{\mathbf{y}}\):
- If the syndrome is \(\mathbf{s} = \begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then the Hamming decoder thinks there are no bit errors in \(\hat{\mathbf{y}}\) (it may be wrong though). In this case, it outputs \(\hat{\mathbf{x}} = \begin{array}{cccc}[ \hat{y}_{3} &\hat{y}_{5} &\hat{y}_{6} &\hat{y}_{7} ]\end{array}^{T}\) since \(y_{3} = x_{1}\), \(y_{5} = x_{2}\), \(y_{6} = x_{3}\) and \(y_{7} = x_{4}\) in \(\mathbf{G}\).
- If the syndrome \(\mathbf{s}\) is not equal to \(\begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then its 3-bit number is converted into a decimal number \(i ∈ [1, 7]\). In this case, the Hamming decoder thinks that the ith bit in \(\hat{\mathbf{y}}\) has been flipped by a bit error (it may be wrong though). The Hamming decoder flips the \(i\)th bit in \(\hat{\mathbf{y}}\) before outputting \(\hat{\mathbf{x}}=\left[\begin{array}{llll}\hat{y}_3 & \hat{y}_5 & \hat{y}_6 & \hat{y}_7\end{array}\right]^T\). If there are multiple bit errors in the received codeword \(\hat{\mathbf{y}}\), the syndrome \(\mathbf{s}\) identifies which bit of \(\hat{\mathbf{y}}\) can be toggled to give the legitimate permutation of \(\mathbf{y}\) that is most similar.

Law of Total Probability: If \(M_{i}\), \(i\) = 1, 2, 3, . . . , is a partition of \(Ω\) then \(P(A) = \sum\limits_{i} P(A ∩ M_{i}) = \sum\limits_{i} P(A|M_{i})P(M_{i})\)

Bayes’ Theorem: For any two events \(A\) and \(B\) with \(P(B) > 0\), \(P(A|B) = P(B|A) \frac{P(A)}{P(B)}\)

The variance of \(X\) is:

\(\begin{aligned}\operatorname{Var}[X] & =\mathbf{E}\left[(X-\mathbf{E}[X])^2\right] \\& =\int_{\Omega}(X(\omega)-\mathbf{E}[X])^2 P(d \omega)=\int_{\mathbb{R}}(x-\mathbf{E}[X])^2 \mu(d x)\end{aligned}\)