Below is a simply mathematical proof to show that in a multivariate gaussian distribution if features are independent then probability density of a point can be computed as the product of probability density of individual features modeled as univariate gaussian distribution.

where

- K = Number of Features
- = Mean vector of size K
- = Covariance Matrix of size K X K
- = Determinant of covariance matrix
- = standard deviation of feature i
- = mean value for feature i

Covariance matrix for a dataset with independent feature is a diagonal matrix. For a diagonal matrix we can easily show that

- Inverse of a diagonal matrix is equal to the reciprocal of diagonal elements i.e

- Determinant of a diagonal matrix is equal to the product of diagonal elements i.e

Using the above two properties of the diagonal matrix we can show that equation 1 essentially same as equation 2 when features are independent. Let’s first tackle in equation 1. Since determinant of a diagonal matrix is equal to the product of diagonal elements we can rewrite

Now let’s focus on the exponential part in equation 1. Using 3, we can show that

Now can be written as . Thus

Replacing 1 with 5 and 7 we get

Hence proved.

## Published by Ritesh Agrawal

I am a data scientist @ Uber. I work on making our data infrastructure (such as Hive, Presto, Vertica, etc) intelligent and efficient using data mining and machine learning techniques.
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