Threshold for Fisher linear classifier The fisher linear classifier for two classes is a classifier with this discriminant function:
$h(x) = V^{T}X + v_0$
where 
$V = \left[ \frac{1}{2}\Sigma_1 + \frac{1}{2}\Sigma_2\right]^{-1}(M_2-M_1)$
and $M_1$, $M_2$ are means and $\Sigma_1$,$\Sigma_2$ are covariances of the classes.
$V$ can be calculated easily but the fisher criterion cannot give us the optimum $v_0$. 
Then, how we can find the optimum $v_0$ analytically if we are using fisher classifier?
 A: When $X$ is normally distributed with known mean $M_1$ and covariance $\Sigma_1$ or with mean $M_2$ and covariance $\Sigma_2$, as indicated in comments to the question, then $V^{\ '}X$ is normally distributed either with mean $\mu_1 = V^{\ '} M_1$ and covariance $\sigma_1^2 = V^{\ '} \Sigma_1 V$ or with mean $\mu_2 = V^{\ '} M_2$ and covariance $\sigma_2^2 = V^{\ '} \Sigma_2 V$; $\mu_2 \gt \mu_1$.  We might then care to optimize the chance of correct classification.  This can be done provided we stipulate a prior distribution for the two classes.  Letting $\pi_1$ be the chance of class 1 and $\pi_2$ the chance of class 2 and $\phi$ the standard normal pdf, then the posterior probabilities of the classes are equal (and therefore $x$ is at the threshold) when
$$f(x) = \pi_1 \phi(\frac{x - \mu_1}{\sigma_1}) - \pi_2 \phi(\frac{x - \mu_2}{\sigma_2}) = 0.$$
There will be at most one zero of $f$ between $x = \mu_1$ and $x = \mu_2$.  (When the zeros lie outside this interval we might question the utility of this classifier.)  Assuming one exists and choosing $v_0$ to be the negative of this zero gives a linear classifier $X \to V^{\ '}X + v_0$ that, when negative, indicates class 1 is more likely than class 2 and, when positive, indicates class 2 is more likely than class 1.
A simple case arises when the two classes are taken to be equally likely, $\pi_1 = \pi_2 = 1/2,$ for then it is clear from the symmetry and unimodality of $\phi$ that $v_0 = -(\mu_1 + \mu_2)/2$.  Note, though, that in general it is not the case that the zero equals $\pi_1 \mu_1 + \pi_2 \mu_2$ (although that might be a good starting guess in a systematic search for the zero).
