We model White Gaussian noise (having power $\sigma^2$) as a Gaussian process $n$, whereby for every interval $(t_0,~ t_0 + L)$ the integral $\int_{t_0}^{t_0 + L}n~dt$ is a Gaussian random variable with distribution $\mathcal N(0, L \sigma^2)$ representing the energy in the signal over the interval. For white noise, we may also assume that the resulting new variables for non-overlapping intervals are independent.
Letting $n$ be such a signal and $s_0 = - \sqrt{\frac{2~E_b}{T_b}}\cos(\omega t) $ be our BPSK signal over one bit, we take the sum through a matched filter to find
\begin{align*} q &= \int_0^{T_b} (s_0(t) + n(t)) \cos(\omega t)~dt\\ &= -\sqrt{\frac{2~ E_b}{T_b}} \int_0^{T_b} \cos(\omega t)^2~dt + \int_0^{T_b} n(t) \cos(\omega t)~dt\\ &= -\sqrt{\frac{E_b~T_b}{2}} + X \end{align*}
where we define a new random variable
$$X = \int_0^{T_b} n(t) \cos(\omega t)~dt$$
and our decision criterion is $q < 0$.
Assuming white noise, I can show that $X$ is Gaussian with distribution $\mathcal N \left (0 , \frac{1}{2} T_b \sigma^2 \right )$
It follows that
\begin{align*} \mathbb{P}~ \{ \text{Error}\} &= \mathbb{P}~ \{ q > 0 \}\\ &= \mathbb{P} \left \{ X > \sqrt{\frac{E_b~T_b}{2}} \right \}\\ &= Q \left ( \frac{\sqrt{\frac{E_b~T_b}{2}}}{\sqrt{\frac{1}{2} T_b \sigma^2}} \right )\\ &= Q \left ( \sqrt{\frac{E_b}{\sigma^2}}\right ) \end{align*}
From here I want to get to the expected form
$$\mathbb{P}~ \{ \text{Error}\} = Q \left ( \sqrt{\frac{2~E_b}{N_0}}\right )$$
which suggests that
$$\sigma^2 = \frac{N_0}{2}$$
This is indeed the assumption made here, here, here, here... all without justification for why this is taken to be the signal "variance".
But this $\frac{N_0}{2}$ is really only a tiny fraction of my signal power $\sigma^2$ (as defined in the beginning). Where does this assertion that $\sigma^2 = \frac{N_0}{2}$ come from? I would have thought that if we are to have a (single-sided) bandwidth of at least $\frac{1}{2~T_b}$, then we should lower-bound the resulting power (variance) by at least $\frac{N_0}{2~T_b} \leq \sigma^2$ (which is agreed with here).
In this case I would put
$$\mathbb{P}~ \{ \text{Error}\} = Q \left ( \sqrt{\frac{2~E_b~T_b}{N_0}}\right )$$
So why do we take the variance to be $\frac{N_0}{2}$?
Follow-up 1: Calculation of $\int_0^T n(t) \cos(\omega t)$
I will show here that this is a Gaussian RV with variance $\frac{1}{2} T \sigma^2$ when $\sigma^2$ is the variance of $n$. I use the two defining properties I am assuming of noise from the outset;
- that $\int_0^T n~ dt$ is a Gaussian RV with distribution $\mathcal N(0, T \sigma^2)$
- If intervals $I_1$ and $I_2$ are such that $I_1 \cap I_2 = \emptyset$ then $\text{Cov} \left ( \int_{I_1} n~dt, \int_{I_2} n~dt\right ) = 0$
First, take $X = \lim_{K \rightarrow \infty} X_K$ where
\begin{align*} X_K &= \sum_{k=1}^K \int_{\frac{(k-1)T}{K}}^{\frac{kT}{K}} n(t) \cos \left (\omega~ T~ \frac{k}{K} \right ) dt\\ &= \sum_{k=1}^K \cos \left (\omega~ T ~\frac{k}{K} \right ) \int_{\frac{(k-1)T}{K}}^{\frac{kT}{K}} n(t) dt\\ &=: \sum_{k=1}^K Y_k \end{align*}
whereby
$$Y_k = \cos \left (\omega~ T ~\frac{k}{K} \right ) \int_{\frac{(k-1)T}{K}}^{\frac{kT}{K}} n(t) dt\\\\$$
Now I will use Lindeberg Central Limit Theorem via Feller's theorem to show convergence of the sequence of partial sums, to a Gaussian RV with distribution $\mathcal N(0, \frac{1}{2} T \sigma^2)$. I will adopt the following abuse of notation: for some $x$ being a variable with distribution $\mathcal N(\mu, \sigma^2)$, I will write $x \in \mathcal N(\mu, \sigma^2)$
So take
\begin{align*} s_K^2 &= \sum_{k=1}^K \text{Var}(Y_k)\\ &= \sum_{k=1}^K \text{Var} \left (\cos \left (\omega~ T ~\frac{k}{K} \right ) \int_{\frac{(k-1)T}{K}}^{\frac{kT}{K}} n(t) dt \right )\\ &= \sum_{k=1}^K \cos \left (\omega~ T ~\frac{k}{K} \right )^2 \text{Var} \left ( \int_{\frac{(k-1)T}{K}}^{\frac{kT}{K}} n(t) dt \right )\\ &= \sum_{k=1}^K \cos \left (\omega~ T ~\frac{k}{K} \right )^2 \frac{T}{K} \sigma^2\\ &= \sigma^2 \sum_{k=1}^K \cos \left (\omega~ T ~\frac{k}{K} \right )^2 \frac{T}{K} \\ &\xrightarrow{K \rightarrow \infty} \sigma^2 \int_0^T \cos \left (\omega~ t \right )^2 dt\\ &= \sigma^2 \int_0^T \left (\frac{1}{2} + \frac{1}{2} \cos \left (2 \omega~ t \right ) \right )dt\\ &\approx \sigma^2 \int_0^T \frac{1}{2} dt\\ &= \frac{1}{2} T \sigma^2 \end{align*}
The plan is to use Feller's conditions to show that Lindeberg's condition is satisfied, hence we shall have proven
\begin{align*} \frac{1}{\sqrt{\frac{1}{2} T \sigma^2}} X &= \lim_{K \rightarrow \infty} \frac{1}{s_K} \sum_{k=1}^K Y_k\\ & \in \mathcal N(0, 1) \end{align*}
Feller's Condition 1
We require $\forall \epsilon > 0$, in the limit as $K \rightarrow \infty$, $\forall 1 \leq k \leq K$ that $\mathbb P \{|Y_k| > \epsilon~s_K\} = 0$.
Now (pretty much) by assumption, we have
$$Y_k \in \mathcal N\left (0, \frac{T}{K} \cos \left ( \omega~T~\frac{k}{K}\right )^2 \sigma^2 \right )$$
Let $\epsilon > 0$ and let $M \in \mathbb N$ be arbitrary. Then let $K > \max \{M, \left ( \frac{\sqrt{T}~M~\sigma}{\epsilon~s_M}\right ) ^2 \}$. Then $s_K \geq s_M$ and
\begin{align*} \frac{\epsilon~ s_K}{\sqrt{\frac{T}{K}} \left \lvert \cos \left ( \omega T \frac{k}{K} \right ) \right \rvert \sigma} &\geq \frac{\sqrt{K}~\epsilon~s_M}{\sqrt{T}~\sigma}\\ &= M \end{align*}
Interpret the above as saying, given $K$ large enough, we can make the fraction $\frac{\epsilon s_K}{\sqrt{\frac{T}{K}} \left \lvert \cos \left ( \omega~t~\frac{k}{K} \right ) \right \rvert \sigma}$ arbitrarily large.
Hence, taking $K$ as arbitrary and letting $1 \leq k \leq K$
\begin{align*} \mathbb P \{ \lvert Y_k \rvert > \epsilon~ s_K\} &= Q \left ( \frac{\epsilon s_K}{\sqrt{\frac{T}{K}} \left \lvert \cos \left ( \omega~t~\frac{k}{K} \right ) \right \rvert \sigma} \right )\\ &\xrightarrow{K \rightarrow \infty} 0 \end{align*}
$\square$
Feller's Condition 2
We require that the sequence $K \mapsto \frac{X_K}{s_K}$ converges weakly to a standard normal distribution as $K \rightarrow \infty$. This will follow from the assumption of whiteness. Since each $\frac{X_K}{s_K}$ is a sum of zero mean Gaussians, it is a zero-mean Gaussian with variance
\begin{align*} \text{Var} \left (\frac{X_K}{s_K} \right ) &= \frac{1}{\sum_{k=1}^K \text{Var}(Y_k)} \text{Var} \left ( \sum_{k=1}^K Y_k\right )\\ &=\frac{1}{\sum_{k=1}^K \text{Var}(Y_k)} \sum_{i, j=1}^K \mathbb E \left ( Y_i Y_j\right )\\ &=\frac{1}{\sum_{k=1}^K \text{Var}(Y_k)} \sum_{i, j=1}^K \text{Var}(Y_i) \delta_{i, j}\\ &=\frac{1}{\sum_{k=1}^K \text{Var}(Y_k)} \sum_{k=1}^K \text{Var}(Y_k)\\ &= 1 \end{align*}
Thus the sequence $K \mapsto \frac{S_K}{s_K}$ has a constant distribution, so the distribution does converge weakly to $\mathcal N(0, 1)$.
$\square$
