L(β;σ^2) = -T/2*(log(2π))-T/2*(log(σ^2))-(1/2σ^2)*(y-Xβ)'(y-Xβ)
Can someone clarify that ∂^2 L/∂β∂σ^2 = (1/σ^4)*(X'Xβ-X'y) ??
My tutor has it as -(My answer) but I can't work out why. I'll assume this is a mistake unless this can be proved otherwise? I have sent an e-mail to him but he normally takes a couple of days!
Thanks
Differentiate L wrt $\beta$ gives $X'(y-X\beta)/\sigma^2$. (There are two minus signs which give the +). This is easy if $\beta$ is just a number, but also works if it is a vector. Differentiate wrt to $\sigma^2$ gives minus the quoted result as $\partial (1/x) /\partial x = -1/x^2$