# Changing a multivariate Gaussian along a dimension

I have a multivariate Gaussian parameterised by a mean vector $\mu$ and a precision matrix $\Sigma$. Now, I want to set the Gaussian along a given dimension $i$ to a point mass i.e. I set the corresponding diagonal entry i.e. $\Sigma_{ii}$ to infinity (zero variance) and also set all the entries in the ith row and column to 0.

Now, what I would like to know is how should the mean vector be adjusted to account for this? Also, if someone would be kind enough to describe an intuitive reason of how changing the covariance parameters affects the mean vector, that would be great.

• Setting a diagonal element to infinity makes no sense: perhaps you meant zero? Regardless, there is no basis for adjusting the mean: the mean, as a parameter separate from the variance matrix, remains whatever it was regardless of what changes you make to the matrix.
– whuber
Commented Feb 25, 2014 at 19:50
• The reason I said infinity is because I am parameterising it with the precision matrix, so $\Sigma$ is the precision matrix and infinity is sort of the placeholder for undefined precision. I saw some software from Microsoft that is doing some update on mean, which seemed strange to me as well as the mean and covariance are independent parameters of the distribution. I will look more into this and update this thread tomorrow.
– Luca
Commented Feb 25, 2014 at 22:22
• Thank you: I even thought you might be using a precision matrix but still overlooked your reference to it! I suspect your MS source may be updating a conditional mean.
– whuber
Commented Feb 25, 2014 at 22:25
• I am pretty sure I might have overlooked something. The code is quite complex, so it is probable that I have overlooked something. I will have access to it tomorrow and will make sure that I update the thread.
– Luca
Commented Feb 25, 2014 at 23:05

Sorry to have taken some time to come back. The answer is, as I suspected, was my own failure in understanding the code. The distribution was parameterised using precision and the $\textit{product}$ of mean and precision. I think this is so that certain calculations were easy. So, when the variance was being set to $0$ along a dimension, it was this product of mean and precision that was getting adjusted.