Timeline for Bottom to top explanation of the Mahalanobis distance?
Current License: CC BY-SA 4.0
5 events
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| Aug 10, 2022 at 5:45 | history | edited | Sycorax♦ | CC BY-SA 4.0 |
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| Aug 10, 2022 at 5:39 | history | edited | Sycorax♦ | CC BY-SA 4.0 |
added 272 characters in body
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| Aug 10, 2022 at 5:33 | comment | added | Sycorax♦ | It's just algebra. We have data $X$ with observations stored in $n$ rows and features in $p$ columns and all column means are 0. Rotating it into an orthogonal basis is done via $\tilde X = XQ$. The estimator of the covariance matrix for the orthogonal data $\tilde X$ is given by $$ \begin{align} \tilde \Sigma &= \frac{1}{n-1} \tilde X^T \tilde X \\ &= \frac{1}{n-1} Q^TX^TXQ \\ &= \frac{1}{n-1} Q^T (n-1)\Sigma Q \\ &= Q^T Q D Q^T Q \\ &= D \end{align}$$ | |
| Aug 10, 2022 at 5:08 | comment | added | IntegrateThis | Why are the entires of $D^{-\frac{1}{2}}$ equal to the inverse standard deviation of each feature in the orthogonal space. | |
| Feb 11, 2017 at 3:56 | history | answered | Sycorax♦ | CC BY-SA 3.0 |