3 ms·

"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 follo

"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.

doublerebel 8y ago
That's code for KaTeX/MathJax. It should be rendered, check your script blocker.
IngoBlechschmid 8y ago
"E[foo]" is syntax to mean the expected value of the random variable foo, roughly speaking the mean value. (For instance the expected value of a dice roll is 3.5. The terminology is slightly suboptimal, since we will never expect a dice to come up 3.5.) 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 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.
- bunderbunder 8y ago
  Also, the "E[foo]" notation is something you'd pick up in an introductory statistics course. Which, IMO, means it's perfectly appropriate to use it without further explanation in this sort of context. It's not really reasonable to expect technical subjects like this to always be presented in a way that's easily digestible to people who lack any background in the subject area. This article is clearly aimed at people who are studying machine learning, and anyone who is studying machine learning should already have a good command of basic statistics in linear algebra.
- johndough 8y ago
  Slightly more formated: http://mathb.in/28658 http://mathb.in/28658
em500 8y ago
Rewind to the start of the paragraph: "Let \mathbf{x} be a random vector with N dimensions." You're assumed to already know what a "random vector with N dimensions" is. It's very resonable then to also assume you know what expectations and covariances of random vectors are, and some of their basic properties, such as linearity of expectations and quadratic forms of covariance matrices, since all of these are typically taught in the same course.

doublerebel 8y ago

That's code for KaTeX/MathJax. It should be rendered, check your script blocker.

IngoBlechschmid 8y ago

"E[foo]" is syntax to mean the expected value of the random variable foo, roughly speaking the mean value. (For instance the expected value of a dice roll is 3.5. The terminology is slightly suboptimal, since we will never expect a dice to come up 3.5.) 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 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.

bunderbunder 8y ago
Also, the "E[foo]" notation is something you'd pick up in an introductory statistics course. Which, IMO, means it's perfectly appropriate to use it without further explanation in this sort of context. It's not really reasonable to expect technical subjects like this to always be presented in a way that's easily digestible to people who lack any background in the subject area. This article is clearly aimed at people who are studying machine learning, and anyone who is studying machine learning should already have a good command of basic statistics in linear algebra.
johndough 8y ago
Slightly more formated: http://mathb.in/28658 http://mathb.in/28658

em500 8y ago

Rewind to the start of the paragraph: "Let \mathbf{x} be a random vector with N dimensions." You're assumed to already know what a "random vector with N dimensions" is. It's very resonable then to also assume you know what expectations and covariances of random vectors are, and some of their basic properties, such as linearity of expectations and quadratic forms of covariance matrices, since all of these are typically taught in the same course.