Can anyone show me how this is true?
[math]\sum_{i=1}^n (x_i-\mu)^T \Sigma^{-1}(x_i-\mu) = (\bar{x}-\mu)^T(n\Sigma )^{-1} (\bar{x}-\mu)[/math]
You can use the fact that [math]\sum_{i=1}^n x_i = n \bar{x}[/math]
>>8596014
what is sigma^{-1} and (nsigma)^{-1}?
>>8596014
you got some serious notational issues to address first OP, before we can help out.
That "you can use the fact" makes no sense
>>8596018
im not op but x bar is likely the average of all x_i, so it does make sense
>>8596016
mu and x_i vectors of length d. Sigma is a matrix of dxd.
Maybe it makes more sense to write [math](n\Sigma )^{-1}[/math] as [math]\frac{Sigma^{-1}}{n}[/math]
>>8596021
using a bar for averaging when talking about vectors? nonsense! Use angle brackets or somethin
>>8596023
I suggest writing out the problem using full blown summation notation. If you are hot shit you can use Einstein notation and drop the summation sigma signs. Give me a few moments.
So far, he's the first equation written in summation notation. I took the liberty of replacing the sigma^-1 with a generic A, since it really doesn't matter to this problem. Still more to come.
[eqn]\sum_i (\vec{x}_i - \vec{\mu}_i)^T A (\vec{x}_i - \vec{\mu}_i) [/eqn]
[eqn]\sum_i \sum_j (x_{ij} - \mu_{ij}) \sum_k A_{jk} (x_{ik} - \mu_{ik}) [/eqn]
[eqn]\sum_i \sum_j \sum_k (x_{ij} - \mu_{ij}) A_{jk} (x_{ik} - \mu_{ik}) [/eqn]
ah, fuck it, I'm too tired to do your homework