Consider a binary-input additive white Gaussian noise channel.

Let $\mathbf{x}_0 = (\sqrt{Es},\sqrt{E_s},⋯,\sqrt{E_s})$ and $\mathbf{x}_1 = (-\sqrt{E_s},-\sqrt{E_s},⋯,-\sqrt{E_s})$, be two codewords of length $d$. Suppose $\mathbf{x}_0$ is transmitted and the received vector is given by $\mathbf{y} = \mathbf{x}_0 + \mathbf{n}$, where the noise vector $\mathbf{n}$ is an i.i.d. Gaussian random vector with mean zero and variance $\frac{N_0}{2}$ . Suppose that maximum-likelihood decoding is employed. The pairwise error probability $P_d$ is the probability that the decoder chooses the incorrect codeword $\mathbf{x}_1$ as the decision instead of the originally transmitted $\mathbf{x}_0$. Show that $P_d=Q(\sqrt{\frac{2dE_s}{N_0}})$, where $Q(x) = (\frac{1}{\sqrt{2 \pi}})\int ^{\infty}_{x}e^{-\frac{t^2}{2}}dt$

I know how to calculate the error probability for bpsk in the other type,but not codewords type,does anyone know how to calculate the probability for codeword type?


The probability of error of the ML detector is equal to the probability that the received vector $\mathbf{y}$ is closer to $\mathbf{x}_1$ than to $\mathbf{x}_0$, which is equal to the probability that the noise component in the direction of $\mathbf{x}_0-\mathbf{x}_1$ is greater than half of the Euclidean distance between $\mathbf{x}_0$ and $\mathbf{x}_1$. The Euclidean distance between $\mathbf{x}_0$ and $\mathbf{x}_1$ is


The noise variance in the direction of $\mathbf{x}_0-\mathbf{x}_1$ equals $N_0/2$ (as in any other direction). The probability that a zero mean Gaussian noise variable with variance $N_0/2$ assumes a value greater than $D/2$ is given by


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  • $\begingroup$ Is it $\sqrt{\sum\limits_{i=1}^{d}(x_{0,i}-x_{1,i})^2}=\sqrt{\sum\limits_{i=1}^{d}4E_s}$? $\endgroup$ – electronic component Aug 19 '18 at 10:24
  • $\begingroup$ @electroniccomponent: Yes, that was a typo, corrected. $\endgroup$ – Matt L. Aug 19 '18 at 10:27

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