Show that if $\Sigma_0 = \Sigma_1$ this classifier is linear?

Question

I have $Y \in \{ 0;1 \} $ $$ P(Y=1 \mid X = x)= \frac{1}{ \left[ 1 + \frac{\det(\Sigma_0)}{\det(\Sigma_1)} \cdot e^{-\frac{1}{2} \left[ (x-\mu_1)^T\Sigma^{-1}_1(x-\mu_1) - (x-\mu_0)^T \Sigma^{-1}_0(x-\mu_0)\right]} \right] } $$ $P(Y=0 \mid X = x)$ is the same as $P(Y=1 \mid X = x)$ but where you switch the indices $0$ and $1$

If $ \Sigma_0 =\Sigma_1$ how can I show that the classifier defined by $\operatorname{argmax} \{P(Y=y\mid X=x)\}$ is linear in $x$?

Answer 1

\begin{align} & (x-\mu_1)^T \Sigma^{-1} (x-\mu_1) - (x-\mu_0)^T \Sigma^{-1}(x-\mu_0) \\[8pt] = {} & (x-\mu_1)^T \Sigma^{-1} (\mu_0-\mu_1) + (x-\mu_1)^T\Sigma^{-1} (x-\mu_0) - (x-\mu_0)^T \Sigma^{-1} (x-\mu_0) \\[8pt] = {} & (x-\mu_1)^T \Sigma^{-1} (\mu_0-\mu_1) + \Big((x-\mu_1) - (x-\mu_0) \Big) \Sigma^{-1} (x-\mu_0) \\[8pt] = {} & (x-\mu_1)^T \Sigma^{-1} (\mu_0-\mu_1) + (\mu_0 - \mu_1)\Sigma^{-1}(x-\mu_0) \end{align}

Each term above is a scalar and thus equal to its transpose, and also $\Sigma^{-1}$ is its own transpose; thus this is equal to $$ (\mu_0-\mu_1)^T \Sigma^{-1} (x-\mu_1) + (\mu_0-\mu_1)\Sigma^{-1} (x-\mu_0). $$ This is a scalar-valued function that is linear affine in $x$ (i.e. linear plus constant).