[proofplan] We derive the Bayes rule by comparing the two equal-prior Gaussian class-conditional densities. Because the covariance matrices are equal, the quadratic terms cancel and the decision rule reduces to thresholding the one-dimensional Fisher score $(\mu_1-\mu_2)^\top\Sigma^{-1}X$ at the midpoint. Under either class this centered score is a univariate normal random variable with mean $\Delta^2/2$ or $-\Delta^2/2$ and variance $\Delta^2$, so each conditional misclassification probability equals $\Phi(-\Delta/2)$. The equal prior weights then leave the same value as the total Bayes error. [/proofplan] [step:Handle the case where the two Gaussian laws coincide] Assume first that $\Delta=0$. Since $\Sigma$ is positive definite, the quadratic form $v \mapsto v^\top \Sigma^{-1}v$ is positive definite, so $\Delta=0$ implies $\mu_1=\mu_2$. Hence the two conditional laws of $X$ are identical. Let $\delta:\mathbb{R}^p\to\{1,2\}$ be any deterministic classifier. Denote the common conditional law of $X$ by $\nu$. Its error probability is \begin{align*} \mathbb{P}(\delta(X)\neq Y) &= \frac{1}{2}\mathbb{P}(\delta(X)=2\mid Y=1) + \frac{1}{2}\mathbb{P}(\delta(X)=1\mid Y=2)\\ &= \frac{1}{2}\nu(\{x\in\mathbb{R}^p:\delta(x)=2\}) + \frac{1}{2}\nu(\{x\in\mathbb{R}^p:\delta(x)=1\})\\ &= \frac{1}{2}. \end{align*} Thus the Bayes error is $1/2$. Since $\Phi(0)=1/2$, this equals $\Phi(-\Delta/2)$ when $\Delta=0$. [/step] [step:Compute the equal-prior Bayes decision rule when the means differ] Assume now that $\Delta>0$. Define the mean difference vector $d\in\mathbb{R}^p$, the Fisher direction $a\in\mathbb{R}^p$, and the midpoint $m\in\mathbb{R}^p$ by \begin{align*} d := \mu_1-\mu_2, \qquad a := \Sigma^{-1}d, \qquad m := \frac{\mu_1+\mu_2}{2}. \end{align*} For $k\in\{1,2\}$, let $f_k:\mathbb{R}^p\to(0,\infty)$ be the density of $\mathcal{N}_p(\mu_k,\Sigma)$ with respect to $\mathcal{L}^p$: \begin{align*} f_k(x) := \frac{1}{(2\pi)^{p/2}(\det\Sigma)^{1/2}} \exp\left( -\frac{1}{2}(x-\mu_k)^\top\Sigma^{-1}(x-\mu_k) \right). \end{align*} With equal priors, the [Bayes classifier](/theorems/1941) chooses class $1$ exactly where $f_1(x)\ge f_2(x)$ and class $2$ otherwise. Taking logarithms, this condition is equivalent to \begin{align*} (x-\mu_1)^\top\Sigma^{-1}(x-\mu_1) \le (x-\mu_2)^\top\Sigma^{-1}(x-\mu_2). \end{align*} Expanding both quadratic forms and cancelling the common term $x^\top\Sigma^{-1}x$ gives \begin{align*} -2\mu_1^\top\Sigma^{-1}x+\mu_1^\top\Sigma^{-1}\mu_1 \le -2\mu_2^\top\Sigma^{-1}x+\mu_2^\top\Sigma^{-1}\mu_2. \end{align*} Rearranging yields \begin{align*} d^\top\Sigma^{-1}x \ge \frac{1}{2}\left(\mu_1^\top\Sigma^{-1}\mu_1-\mu_2^\top\Sigma^{-1}\mu_2\right). \end{align*} Since $\Sigma^{-1}$ is symmetric, the right-hand side is $d^\top\Sigma^{-1}m$. Therefore the Bayes classifier $\delta_*:\mathbb{R}^p\to\{1,2\}$ is \begin{align*} \delta_*(x) = \begin{cases} 1, & a^\top(x-m)\ge 0,\\ 2, & a^\top(x-m)<0. \end{cases} \end{align*} [guided] The equal-prior Bayes rule compares posterior probabilities. Because the priors are equal, comparing posterior probabilities is the same as comparing the class-conditional densities $f_1$ and $f_2$. For each $k\in\{1,2\}$, the conditional density of $X$ given $Y=k$ is the map $f_k:\mathbb{R}^p\to(0,\infty)$ defined by \begin{align*} f_k(x) := \frac{1}{(2\pi)^{p/2}(\det\Sigma)^{1/2}} \exp\left( -\frac{1}{2}(x-\mu_k)^\top\Sigma^{-1}(x-\mu_k) \right). \end{align*} The normalizing constants are identical because the covariance matrix is the same in both classes. Therefore $f_1(x)\ge f_2(x)$ holds exactly when \begin{align*} (x-\mu_1)^\top\Sigma^{-1}(x-\mu_1) \le (x-\mu_2)^\top\Sigma^{-1}(x-\mu_2). \end{align*} The key cancellation is the common quadratic term in $x$. Expanding gives \begin{align*} x^\top\Sigma^{-1}x -2\mu_1^\top\Sigma^{-1}x +\mu_1^\top\Sigma^{-1}\mu_1 \le x^\top\Sigma^{-1}x -2\mu_2^\top\Sigma^{-1}x +\mu_2^\top\Sigma^{-1}\mu_2. \end{align*} After cancelling $x^\top\Sigma^{-1}x$ and rearranging, we obtain \begin{align*} (\mu_1-\mu_2)^\top\Sigma^{-1}x \ge \frac{1}{2}\left(\mu_1^\top\Sigma^{-1}\mu_1-\mu_2^\top\Sigma^{-1}\mu_2\right). \end{align*} Now define \begin{align*} d := \mu_1-\mu_2, \qquad a := \Sigma^{-1}d, \qquad m := \frac{\mu_1+\mu_2}{2}. \end{align*} Since $\Sigma^{-1}$ is symmetric, \begin{align*} d^\top\Sigma^{-1}m &= \frac{1}{2}(\mu_1-\mu_2)^\top\Sigma^{-1}(\mu_1+\mu_2)\\ &= \frac{1}{2}\left(\mu_1^\top\Sigma^{-1}\mu_1-\mu_2^\top\Sigma^{-1}\mu_2\right). \end{align*} Thus the decision rule is the hyperplane rule \begin{align*} \delta_*(x) = \begin{cases} 1, & a^\top(x-m)\ge 0,\\ 2, & a^\top(x-m)<0. \end{cases} \end{align*} This is the Fisher linear discriminant rule: project $x$ onto the direction $a=\Sigma^{-1}(\mu_1-\mu_2)$ and compare with the projected midpoint. [/guided] [/step] [step:Compute the class one misclassification probability from the projected score] Define the score map $S:\mathbb{R}^p\to\mathbb{R}$ by \begin{align*} S(x) := a^\top(x-m). \end{align*} Conditionally on $Y=1$, the random variable $S(X):\Omega\to\mathbb{R}$ is normally distributed with mean \begin{align*} a^\top(\mu_1-m) &= a^\top\left(\frac{\mu_1-\mu_2}{2}\right) = \frac{1}{2}d^\top\Sigma^{-1}d = \frac{\Delta^2}{2} \end{align*} and variance \begin{align*} a^\top\Sigma a &= d^\top\Sigma^{-1}\Sigma\Sigma^{-1}d = d^\top\Sigma^{-1}d = \Delta^2. \end{align*} Therefore, under $Y=1$, \begin{align*} \frac{S(X)-\Delta^2/2}{\Delta}\sim \mathcal{N}(0,1). \end{align*} The classifier assigns class $2$ exactly when $S(X)<0$, so \begin{align*} \mathbb{P}(\delta_*(X)=2\mid Y=1) &= \mathbb{P}(S(X)<0\mid Y=1)\\ &= \mathbb{P}\left(\frac{S(X)-\Delta^2/2}{\Delta}<-\frac{\Delta}{2}\,\middle|\,Y=1\right)\\ &= \Phi\left(-\frac{\Delta}{2}\right). \end{align*} [/step] [step:Compute the class two misclassification probability by the same score] Conditionally on $Y=2$, the same score $S(X)$ is normally distributed with mean \begin{align*} a^\top(\mu_2-m) &= a^\top\left(\frac{\mu_2-\mu_1}{2}\right) = -\frac{1}{2}d^\top\Sigma^{-1}d = -\frac{\Delta^2}{2} \end{align*} and variance \begin{align*} a^\top\Sigma a=\Delta^2. \end{align*} The classifier assigns class $1$ exactly when $S(X)\ge 0$. Hence \begin{align*} \mathbb{P}(\delta_*(X)=1\mid Y=2) &= \mathbb{P}(S(X)\ge 0\mid Y=2)\\ &= \mathbb{P}\left(\frac{S(X)+\Delta^2/2}{\Delta}\ge \frac{\Delta}{2}\,\middle|\,Y=2\right)\\ &= 1-\Phi\left(\frac{\Delta}{2}\right)\\ &= \Phi\left(-\frac{\Delta}{2}\right), \end{align*} where the final equality uses the symmetry identity $\Phi(-t)=1-\Phi(t)$ for the standard normal distribution. [/step] [step:Average the two conditional errors using the equal priors] The Bayes error rate is the error probability of $\delta_*$. Since $\mathbb{P}(Y=1)=\mathbb{P}(Y=2)=1/2$, the [law of total probability](/theorems/1113) gives \begin{align*} \mathbb{P}(\delta_*(X)\neq Y) &= \frac{1}{2}\mathbb{P}(\delta_*(X)=2\mid Y=1) + \frac{1}{2}\mathbb{P}(\delta_*(X)=1\mid Y=2)\\ &= \frac{1}{2}\Phi\left(-\frac{\Delta}{2}\right) + \frac{1}{2}\Phi\left(-\frac{\Delta}{2}\right)\\ &= \Phi\left(-\frac{\Delta}{2}\right). \end{align*} Together with the already handled case $\Delta=0$, this proves that the Bayes error rate is $\Phi(-\Delta/2)$ for all $\mu_1,\mu_2\in\mathbb{R}^p$. [/step]