Method 2 (not requiring starting point)

To solve (5.15), define $ \alpha$ as the element-wise inverse of $ \lambda$. Then, (5.15) can be written $ {\bf B}$$ \alpha$$ ={\bf0}$, which means that $ \alpha$ must be fully in the column space of $ {\bf A}$, or that

$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle = {\bf A} {\bf v},$ (5.17)

for some $ D\times 1$ vector $ {\bf v}$. Therefore, to find the mean of the surrogate density, we solve for the free variable $ {\bf v}$ such that

$\displaystyle {\bf A}^\prime$   $\displaystyle \mbox{\boldmath$\lambda$}$$\displaystyle ({\bf A} {\bf v}) = {\bf z}^*,$ (5.18)


$\displaystyle \mbox{\boldmath$\lambda$}$$\displaystyle ($$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle ) = [1/\alpha_1, 1/\alpha_2, \ldots 1/\alpha_N]^\prime.$ (5.19)

We can solve this by driving the square error to zero:

$\displaystyle \rho({\bf v})=({\bf A}^\prime$   $\displaystyle \mbox{\boldmath$\lambda$}$$\displaystyle ({\bf A} {\bf v}) - {\bf z})^\prime ({\bf A}^\prime$   $\displaystyle \mbox{\boldmath$\lambda$}$$\displaystyle ({\bf A} {\bf v}) - {\bf z}).$ (5.20)

We can easily find derivative of $ \rho({\bf v})$ with respect to $ {\bf v}$ :

$\displaystyle \left[ \frac{\partial \rho}{\partial v_k}\right] = 2 ({\bf A}^\prime$   $\displaystyle \mbox{\boldmath$\lambda$}$$\displaystyle ({\bf A} {\bf v}) - {\bf z})^\prime {\bf A}^\prime {\mbox{\boldmath$\Lambda$}} {\bf A},$ (5.21)

where $ \Lambda$$ ($$ \alpha$$ )$ is the diagonal matrix with diagonal elements

$\displaystyle \Lambda_i = -1/\alpha_i^2.$ (5.22)

We have found that if we use the negative-definite Hessian approximation

$\displaystyle \left[ \frac{\partial^2 \rho}{\partial v_k \partial v_l}\right] \...
...A}^\prime {\mbox{\boldmath$\Lambda$}(\mbox{\boldmath$\alpha$})} {\bf A}\right),$ (5.23)

the resulting Newton-Raphson algorithm has excellent convergence properties when starting with $ \alpha$$ = [1, 1, \ldots 1]^\prime$. Let $ \hat{\mbox{\boldmath $\lambda$}}({\bf z}^*)$ be the value of $ \lambda$ at the solution to (5.18). The algorithm to find $ \hat{\mbox{\boldmath $\lambda$}}({\bf z}^*)$ can be summarized as follows.
  1. Set iteration counter $ n=0$.
  2. To initialize, let $ {\bf v}_0=\left({\bf A}^\prime {\bf A}\right)^{-1}
{\bf A}^\prime {\bf 1},$ where $ {\bf 1}$ is the vector of ones.
  3. $ \alpha$$ _n = {\bf A} {\bf v}_n$. Initially, $ \alpha$ will be the vector of ones.
  4. $ \lambda$$ _n($$ \alpha$$ _n) = [1/\alpha_1, 1/\alpha_2, \ldots 1/\alpha_N]^\prime.$
  5. Compute derivative and Hessian according to (5.21),(5.22) and (5.23), then update $ {\bf v}$:

    $\displaystyle {\bf v}_{n+1} = {\bf v}_n +
\left[ \frac{\partial^2 \rho}{\partial u_k \partial u_l}\right]^{-1}
\left[ \frac{\partial \rho}{\partial u_k}\right].$

  6. Increment $ n$ and go to step 3.
The algorithm is implemnted by software/lam_solve.m with dbound=0. Although this method corresponds to classical methods, it is based on a novel sampling argument and can be extend to other manifolds as we will see.

Baggenstoss 2017-05-19