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by rundigen12
2799 days ago
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"By defining \mathbb{E}\left[\mathbf{x}\right]=\muE[x]=μ, ...and using the linearity of the expectation operator \mathbb{E}E, we easily arive [sic] to the following conclusion..." Yikes. You don't define that \mathbb{E} was an 'expectation operator', or what an expectation operator even does, or the fact that it's linear. The v's disappeared somehow from inside the square brackets -- maybe you meant \muE[v]=μ? So far this "tutorial" isn't defining its terms very well. I'm lost and it's only the very beginning. |
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Hence the "E" itself is called an "operator". It can be applied to a random value in order to yield its expected value. You can read up on it here: https://en.wikipedia.org/wiki/Expected_value
The definition "E[x] = mu" is correct, though I would write it the other way, as "mu = E[x]", as it's the variable mu which is being defined.
The v's disappear because of a suppresed calculation:
sigma^2 = E[ (v^T x - E[v^T x])^2 ] = E[ (v^T x - E[v^T x]) (v^T x - E[v^T x]) ] = E[ v^T x v^T x - 2 v^T x E[v^T x] + E[v^T x] E[v^T x] ] = E[ v^T x x^T v - 2 v^T x v^T E[x] + v^T E[x] v^T E[x] ] = E[ v^T x x^T v - 2 v^T x E[x]^T v + v^T E[x] E[x]^T v ] = E[ v^T (x x^T - 2 x E[x]^T + E[x] E[x]^T) v ] = v^T E[ x x^T - 2 x E[x]^T + E[x] E[x]^T ] v = v^T E[ (x - E[x]) (x - E[x])^T ] v = v^T E[ (x - mu) (x - mu)^T ] v = v^T Sigma v.