Error probability of binary PAM

In binary (polar) PAM bit 1 is sent as a pulse of amplitude $+A$ and bit 0 as $-A$. White Gaussian noise is added; the receiver samples once per bit and decides with threshold $\lambda=0$.

The sample is $y=\pm A+n$, $n\sim\mathcal{N}(0,\sigma^2)$:

$$f(y\mid1)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-A)^2/2\sigma^2},\qquad f(y\mid0)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+A)^2/2\sigma^2}$$

    f(y|0)       f(y|1)
     .--.    |    .--.
    /    \   |   /    \
 --/------\..|../------\---> y
  -A         0        +A

Error when 0 is sent ($y>0$): let $u=\frac{y+A}{\sqrt2\sigma}$

$$\begin{aligned} P_{e0} &= \int_0^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+A)^2/2\sigma^2}dy=\frac{1}{\sqrt\pi}\int_{A/\sqrt2\sigma}^{\infty}e^{-u^2}du \\ &= \frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\,\sigma}\right) \end{aligned}$$

By symmetry $P_{e1}=P_{e0}$. With equal priors:

$$P_e=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\,\sigma}\right)=Q\left(\frac{A}{\sigma}\right)$$

With a matched filter, $\frac{A^2}{\sigma^2}=\frac{2E_b}{N_0}$, so $P_e=\frac12\operatorname{erfc}\sqrt{E_b/N_0}$. For unipolar PAM ($A$, 0, threshold $A/2$): $P_e=\frac12\operatorname{erfc}\left(\frac{A}{2\sqrt2\sigma}\right)$, which needs twice the average power for the same $P_e$.