-2
$\begingroup$

How do you construct a Bayes classifier for a binary target where it is assumed: $p(y=1)=\alpha$ and $p(x|y)$ both multivariate gaussian?

$\endgroup$
  • 2
    $\begingroup$ This does not make sense. $\endgroup$ – Michael Chernick Apr 17 '17 at 3:51
  • $\begingroup$ What do you mean that "... a and p(x|y) both multivariate gaussian"? a seems like a scalar. Do you mean that you want a normal prior for a? What do you mean by "multivariate"? You seem to have only 2 variables. Etc. $\endgroup$ – gung Apr 17 '17 at 12:32
  • 2
    $\begingroup$ I vote to re-open this question. It is not well-phrased, but I interpret it as asking for a decision rule to decide between two hypotheses: $Y=1$ and $Y=0$ with prior probabilities $\alpha$ and $1-\alpha$ respectively, based on the observation $X$ where the conditional distribution of $X$ given $Y$ is a multivariate Gaussian distribution, both for $Y=1$ and $Y=0$. The decision rule is, of course, easy to state, but implementation is messy. The case when $X$ is bivariate Gaussian under both hypotheses is discussed in my answer to this question $\endgroup$ – Dilip Sarwate Apr 17 '17 at 14:35
1
$\begingroup$

Assumption: I assume that this question is asking for the Bayes classifier given that $Y$ is a Bernoulli random variable with parameter $\alpha$ and the conditional distributions of the observation $X$ are multivariate Gaussian distributions.

If both conditional distributions of $X$ given $Y$ are non-degenerate multivariate Gaussian distributions, that is, they possess densities $f_i(x) = f_{X\mid Y=i}(x\mid Y=i), i = 0, 1$, then the Bayes classifier can be expressed as $$\frac{\alpha}{1-\alpha}\cdot \frac{f_1(x)}{f_0(x)}~~\begin{array}{c}\hat Y=1\\\gtrless\\\hat Y=0\end{array}~~ 1.$$ The decision boundary can also be stated as the hypersurface defined by the equation $${\alpha}\cdot{f_1(x)} - {1-\alpha}\cdot{f_0(x)} = 0$$ which does not have a simple solution in general. The two-dimensional case is treated in this question where the hypersurface is found to be a conic section.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.