# Covariance of a bivariate Gaussian given identity matrix

I have a homework problem about finding an optimal decision boundary. I know the formula (not really the process) for calculating one, so that may be another question entirely, but I do know I need the mean, covariance, and priors. The question is shown below

Suppose points in R^2 are being obtained from two classes, C1 and C2, both of which
are well described by bivariate Gaussians with means at (0,0) and (1,3) and covariances
I and 2I respectively. I is the (2x2) identity matrix. If the priors of C1 and C2 are
0.4 and 0.6 respectively, what is the ideal (i.e. Bayes Optimal) decision boundary
(derive the equation for this boundary)?


I'm puzzled when it says "covariances I and 2I... I is the (2x2) identity matrix."

Is there a way to calculate a numerical value for covariance based on this information? If so, how? I've never heard it phrased this way.

Conditioned on being from class C1, the observation $(X,Y)$ is a pair of independent $N(0,1)$ random variables, while conditioned on being from class C2, the observation $(X,Y)$ consists of independent $N(1,2)$ and $N(3,2)$ random variables. How do we know this? The random variables are given to be (conditionally) independent because their covariance matrix is $I$ or $2I$ and so we know that $\text{cov}(X,Y) = I_{1,2} = 0$ in one case, and $2I_{1,2} = 0$ in the other case. As has been discussed repeatedly on this stackexchange, uncorrelated jointly normal random variables are independent random variables.

So, you can write down the conditional joint pdfs $f_1(x,y)$ and $f_2(x,y)$ under the two hypotheses as bivariate normal densities of independent random variables. The Bayesian decision boundary is the set of all points $(x,y)$ for which $$0.4f_1(x,y) = 0.6f_2(x,y).$$

• Thank you I appreciate the help. Would you mind letting me know how I would solve this if they were not uncorrelated? Feb 22, 2013 at 3:22
• It is exactly the same: $0.4f_1(x,y) = 0.6f_2(x,y)$ but $f_1$ and $f_2$ will be different functions (bivariate correlated normal instead of bivariate independent normal). Feb 22, 2013 at 3:27
• Alright thanks. So my last step for this is to solve for those two bivariate independent normals? Feb 22, 2013 at 3:40