We are learning matrix calculus in class. This is really new for me and I am trying to get extra practice by going through old assignments. One of the questions asks to find the derivative of the function below with respect to β and if possible to find the value of the elements of β that maximize the function. After taking logs to simplify the original function, I took the partial with respect to β and set it equal to zero. From that point I tried and tried with every linear algebra operation I could think of, but I wasn't able to solve for β.
I posted this question in another forum and a member from there was nice enough to verify that my derivative was correct but pointed out that the original function doesn't have a global maximum. He did say that if x1 and x2 are assumed to be linearly independent, then as a result of cancelling out common factors I would end up with one linear equation for x1'β and one for x2'β, which can then be solved for, and then knowing the scalar products of β with two linearly independent vectors, I can reconstruct β.
I know that linear independence in this case would mean that c1x1 + c2x2 = [0,0]' and thus c1=c2=0, but I still don't see how that helps - I thought it might mean that the coefficients of x1 and x2 (the first 2 terms in my derivative) would be zero but how can they be if their numerators are 1? Nor do I know how to reconstruct β as suggested.
I'm just about close to banging my head against the wall over this but would someone be kind enough to provide me with some step-by-step help? Thanks!
\(\displaystyle
x_{1} =
\begin{pmatrix}
a\\
b\\
\end{pmatrix}\quad
x_{2} =
\begin{pmatrix}
c\\
d\\
\end{pmatrix}\quad
\beta =
\begin{pmatrix}
\beta_{1}\\
\beta_{2}\\
\end{pmatrix}\quad
Y =
\begin{pmatrix}
y_{1}\\
y_{2}\\
\end{pmatrix}\quad
\)
\(\displaystyle
f(x_{1},x_{2},y_{1},y_{2},\beta) =
\frac{1}{x_{1}'\beta}exp\{\frac{-y_{1}}{x_{1}'\beta}\}\frac{1}{x_{2}'\beta}exp\{\frac{-y_{2}}{x_{2}'\beta}\}
\)
My derivative set equal to zero:
\(\displaystyle
\frac{\partial lnf}{\partial\beta}=-\frac{1}{x_{1}'\beta}x_{1}-\frac{1}{x_{2}'\beta}x_{2}+\frac{x_{1}y_{1}}{x_{1}'\beta.x_{1}'\beta}+\frac{x_{2}y_{2}}{x_{2}'\beta.x_{2}'\beta}=0
\)
I posted this question in another forum and a member from there was nice enough to verify that my derivative was correct but pointed out that the original function doesn't have a global maximum. He did say that if x1 and x2 are assumed to be linearly independent, then as a result of cancelling out common factors I would end up with one linear equation for x1'β and one for x2'β, which can then be solved for, and then knowing the scalar products of β with two linearly independent vectors, I can reconstruct β.
I know that linear independence in this case would mean that c1x1 + c2x2 = [0,0]' and thus c1=c2=0, but I still don't see how that helps - I thought it might mean that the coefficients of x1 and x2 (the first 2 terms in my derivative) would be zero but how can they be if their numerators are 1? Nor do I know how to reconstruct β as suggested.
I'm just about close to banging my head against the wall over this but would someone be kind enough to provide me with some step-by-step help? Thanks!
\(\displaystyle
x_{1} =
\begin{pmatrix}
a\\
b\\
\end{pmatrix}\quad
x_{2} =
\begin{pmatrix}
c\\
d\\
\end{pmatrix}\quad
\beta =
\begin{pmatrix}
\beta_{1}\\
\beta_{2}\\
\end{pmatrix}\quad
Y =
\begin{pmatrix}
y_{1}\\
y_{2}\\
\end{pmatrix}\quad
\)
\(\displaystyle
f(x_{1},x_{2},y_{1},y_{2},\beta) =
\frac{1}{x_{1}'\beta}exp\{\frac{-y_{1}}{x_{1}'\beta}\}\frac{1}{x_{2}'\beta}exp\{\frac{-y_{2}}{x_{2}'\beta}\}
\)
My derivative set equal to zero:
\(\displaystyle
\frac{\partial lnf}{\partial\beta}=-\frac{1}{x_{1}'\beta}x_{1}-\frac{1}{x_{2}'\beta}x_{2}+\frac{x_{1}y_{1}}{x_{1}'\beta.x_{1}'\beta}+\frac{x_{2}y_{2}}{x_{2}'\beta.x_{2}'\beta}=0
\)