Recent content by jimbodonut

i figured out a) and b). It is a chi-square distribution since the ||x|| = sqrt(x1^2 + x2^2 + ... + xn^2) thus ||x||^2 = x1^2 + x2^2 + ... + xn^2. Since xi~N(0,1), it is chi-square. and the expectation of a chi-square distribution is its degrees of freedom... in this case... E(||x||^2) = n...

Homework Statement Suppose x = (x1, x2, ..., xn)T ∈ Rn is a random vector drawn from the n-dimensional standard Gaussian distribution N(0, I), where 0 = (0, 0, ..., 0)^T (0 vector transpose) and I is the identity matrix. (a) What distribution does ||x||^2 follow? Justify your answer. (b) On...