Timeline for What is the distribution of the Euclidean distance between two normally distributed random variables?
Current License: CC BY-SA 3.0
6 events
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| Mar 3, 2014 at 23:57 | comment | added | kyle | Actually, based on their definition of $b_j$, the numerator in the exponential does reduce to $\mu_j^2$ in the symmetric (independent dimensions with common variance) case. | |
| Feb 28, 2014 at 20:48 | comment | added | kyle | @NRH I've worked through the MGF myself in the symmetric case ($\lambda_j = \sigma^2$) where $p=2$ and instead of $b_j^2 \lambda_j$ in the summation, I have $\mu_j^2$. Simulation verifies the first moment. It's possible that this is the "linear function" you mention and that this is peculiar to the symmetric case, but I thought I'd point it out in case there's an error. | |
| Jun 21, 2011 at 23:58 | comment | added | Nick | Thanks for the reference, I found the book and am slowly trying to make my way through it | |
| Jun 21, 2011 at 23:56 | vote | accept | Nick | ||
| May 12, 2011 at 23:43 | history | edited | NRH | CC BY-SA 3.0 |
deleted 7 characters in body
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| May 12, 2011 at 18:48 | history | answered | NRH | CC BY-SA 3.0 |