2021-05-02 19:40:14 +00:00
|
|
|
<!-- vim: set ts=2 sw=2 et tw=80: -->
|
|
|
|
|
|
|
|
---
|
2021-05-08 08:23:29 +00:00
|
|
|
title: Midterm -- Optimization Methods
|
|
|
|
author: Claudio Maggioni
|
2021-05-02 19:40:14 +00:00
|
|
|
header-includes:
|
|
|
|
- \usepackage{amsmath}
|
|
|
|
- \usepackage{hyperref}
|
|
|
|
- \usepackage[utf8]{inputenc}
|
|
|
|
- \usepackage[margin=2.5cm]{geometry}
|
2021-05-08 08:23:29 +00:00
|
|
|
- \usepackage[ruled,vlined]{algorithm2e}
|
|
|
|
- \usepackage{float}
|
|
|
|
- \floatplacement{figure}{H}
|
|
|
|
|
2021-05-02 19:40:14 +00:00
|
|
|
---
|
|
|
|
\maketitle
|
|
|
|
|
|
|
|
# Exercise 1
|
|
|
|
|
|
|
|
## Point 1
|
|
|
|
|
|
|
|
### Question (a)
|
|
|
|
|
|
|
|
As already covered in the course, the gradient of a standard quadratic form at a
|
2021-05-03 09:18:45 +00:00
|
|
|
point $x_0$ is equal to:
|
2021-05-02 19:40:14 +00:00
|
|
|
|
|
|
|
$$ \nabla f(x_0) = A x_0 - b $$
|
|
|
|
|
2021-05-03 09:18:45 +00:00
|
|
|
Plugging in the definition of $x_0$ and knowing that $\nabla f(x_m) = A x_m - b
|
|
|
|
= 0$ (according to the first necessary condition for a minimizer), we obtain:
|
2021-05-02 19:40:14 +00:00
|
|
|
|
|
|
|
$$ \nabla f(x_0) = A (x_m + v) - b = A x_m + A v - b = b + \lambda v - b =
|
|
|
|
\lambda v $$
|
|
|
|
|
|
|
|
### Question (b)
|
|
|
|
|
|
|
|
The steepest descent method takes exactly one iteration to reach the exact
|
|
|
|
minimizer $x_m$ starting from the point $x_0$. This can be proven by first
|
|
|
|
noticing that $x_m$ is a point standing in the line that first descent direction
|
|
|
|
would trace, which is equal to:
|
|
|
|
|
|
|
|
$$g(\alpha) = - \alpha \cdot \nabla f(x_0) = - \alpha \lambda v$$
|
|
|
|
|
|
|
|
For $\alpha = \frac{1}{\lambda}$, and plugging in the definition of $x_0 = x_m +
|
|
|
|
v$, we would reach a new iterate $x_1$ equal to:
|
|
|
|
|
|
|
|
$$x_1 = x_0 - \alpha \lambda v = x_0 - v = x_m + v - v = x_m $$
|
|
|
|
|
|
|
|
The only question that we need to answer now is why the SD algorithm would
|
|
|
|
indeed choose $\alpha = \frac{1}{\lambda}$. To answer this, we recall that the
|
|
|
|
SD algorithm chooses $\alpha$ by solving a linear minimization option along the
|
|
|
|
step direction. Since we know $x_m$ is indeed the minimizer, $f(x_m)$ would be
|
|
|
|
obviously strictly less that any other $f(x_1 = x_0 - \alpha \lambda v)$ with
|
|
|
|
$\alpha \neq \frac{1}{\lambda}$.
|
|
|
|
|
|
|
|
Therefore, since $x_1 = x_m$, we have proven SD
|
|
|
|
converges to the minimizer in one iteration.
|
|
|
|
|
2021-05-03 09:18:45 +00:00
|
|
|
## Point 2
|
2021-05-02 19:40:14 +00:00
|
|
|
|
|
|
|
The right answer is choice (a), since the energy norm of the error indeed always
|
|
|
|
decreases monotonically.
|
2021-05-03 09:18:45 +00:00
|
|
|
|
|
|
|
To prove that this is true, we first consider a way to express any iterate $x_k$
|
|
|
|
in function of the minimizer $x_s$ and of the missing iterations:
|
|
|
|
|
|
|
|
$$x_k = x_s + \sum_{i=k}^{N} \alpha_i A^i p_0$$
|
|
|
|
|
|
|
|
This formula makes use of the fact that step directions in CG are all
|
|
|
|
A-orthogonal with each other, so the k-th search direction $p_k$ is equal to
|
|
|
|
$A^k p_0$, where $p_0 = -r_0$ and $r_0$ is the first residual.
|
|
|
|
|
|
|
|
Given that definition of iterates, we're able to express the error after
|
|
|
|
iteration $k$ $e_k$ in a similar fashion:
|
|
|
|
|
|
|
|
$$e_k = x_k - x_s = \sum_{i=k}^{N} \alpha_i A^i p_0$$
|
|
|
|
|
|
|
|
We then recall the definition of energy norm $\|e_k\|_A$:
|
|
|
|
|
|
|
|
$$\|e_k\|_A = \sqrt{\langle Ae_k, e_k \rangle}$$
|
|
|
|
|
|
|
|
We then want to show that $\|e_k\|_A = \|x_k - x_s\|_A > \|e_{k+1}\|_A$, which
|
|
|
|
in turn is equivalent to claim that:
|
|
|
|
|
|
|
|
$$\langle Ae_k, e_k \rangle > \langle Ae_{k+1}, e_{k+1} \rangle$$
|
|
|
|
|
|
|
|
Knowing that the dot product is linear w.r.t. either of its arguments, we pull
|
|
|
|
out the sum term related to the k-th step (i.e. the first term in the sum that
|
|
|
|
makes up $e_k$) from both sides of $\langle Ae_k, e_k \rangle$,
|
|
|
|
obtaining the following:
|
|
|
|
|
|
|
|
$$\langle Ae_{k+1}, e_{k+1} \rangle + \langle \alpha_k A^{k+1} p_0, e_k \rangle
|
|
|
|
+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > \langle Ae_{k+1}, e_{k+1}
|
|
|
|
\rangle$$
|
|
|
|
|
|
|
|
which in turn is equivalent to claim that:
|
|
|
|
|
|
|
|
$$\langle \alpha_k A^{k+1} p_0, e_k \rangle
|
|
|
|
+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > 0$$
|
|
|
|
|
|
|
|
From this expression we can collect term $\alpha_k$ thanks to linearity of the
|
|
|
|
dot-product:
|
|
|
|
|
|
|
|
$$\alpha_k (\langle A^{k+1} p_0, e_k \rangle
|
|
|
|
+ \langle Ae_{k+1}, A^k p_0 \rangle) > 0$$
|
|
|
|
|
|
|
|
and we can further "ignore" the $\alpha_k$ term since we know that all
|
|
|
|
$\alpha_i$s are positive by definition:
|
|
|
|
|
|
|
|
$$\langle A^{k+1} p_0, e_k \rangle
|
|
|
|
+ \langle Ae_{k+1}, A^k p_0 \rangle > 0$$
|
|
|
|
|
|
|
|
Then, we convert the dot-products in their equivalent vector to vector product
|
|
|
|
form, and we plug in the definitions of $e_k$ and $e_{k+1}$:
|
|
|
|
|
|
|
|
$$p_0^T (A^{k+1})^T (\sum_{i=k}^{N} \alpha_i A^i p_0) +
|
|
|
|
p_0^T (A^{k})^T (\sum_{i=k+1}^{N} \alpha_i A^i p_0) > 0$$
|
|
|
|
|
|
|
|
We then pull out the sum to cover all terms thanks to associativity of vector
|
|
|
|
products:
|
|
|
|
|
|
|
|
$$\sum_{i=k}^N (p_0^T (A^{k+1})^T A^i p_0) \alpha_i+ \sum_{i=k+1}^N
|
|
|
|
(p_0^T (A^{k})^T A^i p_0) \alpha_i > 0$$
|
|
|
|
|
|
|
|
We then, as before, can "ignore" all $\alpha_i$ terms since we know by
|
|
|
|
definition that
|
|
|
|
they are all strictly positive. We then recalled that we assumed that A is
|
|
|
|
symmetric, so $A^T = A$. In the end we have to show that these two
|
|
|
|
inequalities are true:
|
|
|
|
|
|
|
|
$$p_0^T A^{k+1+i} p_0 > 0 \; \forall i \in [k,N]$$
|
|
|
|
$$p_0^T A^{k+i} p_0 > 0 \; \forall i \in [k+1,N]$$
|
|
|
|
|
|
|
|
To show these inequalities are indeed true, we recall that A is symmetric and
|
|
|
|
positive definite. We then consider that if a matrix A is SPD, then $A^i$ for
|
|
|
|
any positive $i$ is also SPD[^1]. Therefore, both inequalities are trivially
|
|
|
|
true due to the definition of positive definite matrices.
|
|
|
|
|
|
|
|
[^1]: source: [Wikipedia - Definite Matrix $\to$ Properties $\to$
|
|
|
|
Multiplication](
|
|
|
|
https://en.wikipedia.org/wiki/Definite_matrix#Multiplication)
|
|
|
|
|
|
|
|
Thanks to this we have indeed proven that the delta $\|e_k\|_A - \|e_{k+1}\|_A$
|
|
|
|
is indeed positive and thus as $i$ increases the energy norm of the error
|
|
|
|
monotonically decreases.
|
2021-05-08 08:23:29 +00:00
|
|
|
|
|
|
|
# Question 2
|
|
|
|
|
|
|
|
## Point 1
|
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
### (a) For which kind of minimization problems can the trust region method be
|
|
|
|
used? What are the assumptions on the objective function?
|
|
|
|
|
|
|
|
**TBD**
|
2021-05-08 09:00:45 +00:00
|
|
|
|
|
|
|
### (b) Write down the quadratic model around a current iterate xk and explain the meaning of each term.
|
|
|
|
|
|
|
|
$$m(p) = f + g^T p + \frac12 p^T B p \;\; \text{ s.t. } \|p\| < \Delta$$
|
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
Here's an explaination of the meaning of each term:
|
|
|
|
|
|
|
|
- $\Delta$ is the trust region radius, i.e. an upper bound on the step's norm
|
|
|
|
(length);
|
|
|
|
- $f$ is the energy function value at the current iterate, i.e. $f(x_k)$;
|
|
|
|
- $p$ is the trust region step, the solution of $\arg\min_p m(p)$ with $\|p\| <
|
|
|
|
\Delta$ is the optimal step to take;
|
|
|
|
- $g$ is the gradient at the current iterate $x_k$, i.e. $\nabla f(x_k)$;
|
|
|
|
- $B$ is the hessian at the current iterate $x_k$, i.e. $\nabla^2 f(x_k)$.
|
2021-05-08 09:00:45 +00:00
|
|
|
|
|
|
|
### (c) What is the role of the trust region radius?
|
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
The role of the trust region radius is to put an upper bound on the step length
|
|
|
|
in order to avoid "overly ambitious" steps, i.e. steps where the the step length
|
|
|
|
is considerably long and the quadratic model of the objective is low-quality
|
|
|
|
(i.e. the quadratic model differs by a predetermined approximation threshold
|
|
|
|
from the real objective).
|
|
|
|
|
|
|
|
In layman's terms, the trust region radius makes the method switch more gradient
|
|
|
|
based or more quadratic based steps w.r.t. the confidence in the quadratic
|
|
|
|
approximation.
|
2021-05-08 09:00:45 +00:00
|
|
|
|
|
|
|
### (d) Explain Cauchy point, sufficient decrease and Dogleg method, and the connection between them.
|
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
**TBD**
|
|
|
|
|
|
|
|
**sufficient decrease TBD**
|
|
|
|
|
|
|
|
The Cauchy point provides sufficient decrease, but makes the trust region method
|
|
|
|
essentially like linear method.
|
2021-05-08 09:00:45 +00:00
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
The dogleg method allows for mixing purely linear iterations and purely quadratic
|
|
|
|
ones. The dogleg method picks along its "dog leg shaped" path function made out
|
|
|
|
of a gradient component and a component directed towards a purely Newton step
|
|
|
|
picking the furthest point that is still inside the trust region radius.
|
2021-05-08 09:00:45 +00:00
|
|
|
|
|
|
|
Dogleg uses cauchy point if the trust region does not allow for a proper dogleg
|
|
|
|
step since it is too slow.
|
|
|
|
|
|
|
|
Cauchy provides linear convergence and dogleg superlinear.
|
|
|
|
|
|
|
|
### (e) Write down the trust region ratio and explain its meaning.
|
|
|
|
|
|
|
|
$$\rho_k = \frac{f(x_k) - f(x_k + p_k)}{m_k(0) - m_k(p_k)}$$
|
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
The trust region ratio measures the quality of the quadratic model built around
|
|
|
|
the current iterate $x_k$, by measuring the ratio between the energy difference
|
|
|
|
between the old and the new iterate according to the real energy function and
|
|
|
|
according to the quadratic model around $x_k$.
|
|
|
|
|
|
|
|
The ratio is used to test the adequacy of the current trust region radius. For
|
|
|
|
an inaccurate quadratic model, the predicted energy decrease would be
|
|
|
|
considerably higher than the effective one and thus the ratio would be low. When
|
|
|
|
the ratio is lower than a predetermined threshold ($\frac14$ is the one chosen
|
|
|
|
by Nocedal) the trust region radius is divided by 4. Instead, a very accurate
|
|
|
|
quadratic model would result in little difference with the real energy function
|
|
|
|
and thus the ratio would be close to $1$. If the trust region radius is higher
|
|
|
|
than a certain predetermined threshold ($\frac34$ is the one chosen by Nocedal),
|
|
|
|
then the trust region radius is doubled in order to allow for longer steps,
|
|
|
|
since the model quality is good.
|
2021-05-08 09:00:45 +00:00
|
|
|
|
|
|
|
### (f) Does the energy decrease monotonically when Trust Region method is employed? Justify your answer.
|
2021-05-08 08:23:29 +00:00
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
**TBD**
|
|
|
|
|
2021-05-08 08:23:29 +00:00
|
|
|
## Point 2
|
|
|
|
|
|
|
|
The trust region algorithm is the following:
|
|
|
|
|
|
|
|
\begin{algorithm}[H]
|
|
|
|
\SetAlgoLined
|
|
|
|
Given $\hat{\Delta} > 0, \Delta_0 \in (0,\hat{\Delta})$,
|
|
|
|
and $\eta \in [0, \frac14)$\;
|
|
|
|
|
|
|
|
\For{$k = 0, 1, 2, \ldots$}{%
|
|
|
|
Obtain $p_k$ by using Cauchy or Dogleg method\;
|
|
|
|
$\rho_k \gets \frac{f(x_k) - f(x_k + p_k)}{m_k(0) - m_k(p_k)}$\;
|
|
|
|
\uIf{$\rho_k < \frac14$}{%
|
|
|
|
$\Delta_{k+1} \gets \frac14 \Delta_k$\;
|
|
|
|
}\Else{%
|
|
|
|
\uIf{$\rho_k > \frac34$ and $\|\rho_k\| = \Delta_k$}{%
|
|
|
|
$\Delta_{k+1} \gets \min(2\Delta_k, \hat{\Delta})$\;
|
|
|
|
}
|
|
|
|
\Else{%
|
|
|
|
$\Delta_{k+1} \gets \Delta_k$\;
|
|
|
|
}}
|
|
|
|
\uIf{$\rho_k > \eta$}{%
|
|
|
|
$x_{k+1} \gets x_k + p_k$\;
|
|
|
|
}
|
|
|
|
\Else{
|
|
|
|
$x_{k+1} \gets x_k$\;
|
|
|
|
}
|
|
|
|
}
|
|
|
|
\caption{Trust region method}
|
|
|
|
\end{algorithm}
|
|
|
|
|
|
|
|
The Cauchy point algorithm is the following:
|
|
|
|
|
|
|
|
\begin{algorithm}[H]
|
|
|
|
\SetAlgoLined
|
|
|
|
Input $B$ (quadratic term), $g$ (linear term), $\Delta_k$\;
|
|
|
|
\uIf{$g^T B g \geq 0$}{%
|
|
|
|
$\tau \gets 1$\;
|
|
|
|
}\Else{%
|
|
|
|
$\tau \gets \min(\frac{\|g\|^3}{\Delta_k \cdot g^T B g}, 1)$\;
|
|
|
|
}
|
|
|
|
|
|
|
|
$p_k \gets -\tau \cdot \frac{\Delta_k}{\|g\|^2 \cdot g}$\;
|
|
|
|
\Return{$p_k$}
|
|
|
|
\caption{Cauchy point}
|
|
|
|
\end{algorithm}
|
|
|
|
|
|
|
|
Finally, the Dogleg method algorithm is the following:
|
|
|
|
|
|
|
|
\begin{algorithm}[H]
|
|
|
|
\SetAlgoLined
|
|
|
|
Input $B$ (quadratic term), $g$ (linear term), $\Delta_k$\;
|
|
|
|
$p_N \gets - B^{-1} g$\;
|
|
|
|
|
|
|
|
\uIf{$\|p_N\| < \Delta_k$}{%
|
|
|
|
$p_k \gets p_N$\;
|
|
|
|
}\Else{%
|
|
|
|
$p_u = - \frac{g^T g}{g^T B g} g$\;
|
|
|
|
|
|
|
|
\uIf{$\|p_u\| > \Delta_k$}{%
|
|
|
|
compute $p_k$ with Cauchy point algorithm\;
|
|
|
|
}\Else{%
|
|
|
|
solve for $\tau$ the equality $\|p_u + \tau * (p_N - p_u)\|^2 =
|
|
|
|
\Delta_k^2$\;
|
|
|
|
$p_k \gets p_u + \tau \cdot (p_N - p_u)$\;
|
|
|
|
}
|
|
|
|
}
|
|
|
|
\caption{Dogleg method}
|
|
|
|
\end{algorithm}
|
|
|
|
|
|
|
|
## Point 3
|
|
|
|
|
|
|
|
The trust region, dogleg and Cauchy point algorithms were implemented
|
|
|
|
respectively in the files `trust_region.m`, `dogleg.m`, and `cauchy.m`.
|
|
|
|
|
|
|
|
## Point 4
|
|
|
|
|
|
|
|
### Taylor expansion
|
|
|
|
|
|
|
|
The Taylor expansion up the second order of the function is the following:
|
|
|
|
|
|
|
|
$$f(x_0, w) = f(x_0) + \langle\begin{bmatrix}48x^3 - 16xy + 2x - 2\\2y - 8x^2
|
|
|
|
\end{bmatrix}, w\rangle + \frac12 \langle\begin{bmatrix}144x^2 -16y + 2 - 16 &
|
|
|
|
-16 \\ -16 & 2 \end{bmatrix}w, w\rangle$$
|
|
|
|
|
|
|
|
### Minimization
|
|
|
|
|
|
|
|
The code used to minimize the function can be found in the MATLAB script
|
|
|
|
`main.m` under section 2.4. The resulting minimizer (found in 10 iterations) is:
|
|
|
|
|
|
|
|
$$x_m = \begin{bmatrix}1\\4\end{bmatrix}$$
|
|
|
|
|
|
|
|
### Energy landscape
|
|
|
|
|
|
|
|
The following figure shows a `surf` plot of the objective function overlayed
|
|
|
|
with the iterates used to reach the minimizer:
|
|
|
|
|
|
|
|
![Energy landscape of the function overlayed with iterates and steps (the white
|
|
|
|
dot is $x_0$ while the black dot is $x_m$)](./2-4-energy.png)
|
|
|
|
|
|
|
|
The code used to generate such plot can be found in the MATLAB script `main.m`
|
|
|
|
under section 2.4c.
|
|
|
|
|
|
|
|
## Point 5
|
|
|
|
|
|
|
|
### Minimization
|
|
|
|
|
|
|
|
The code used to minimize the function can be found in the MATLAB script
|
|
|
|
`main.m` under section 2.5. The resulting minimizer (found in 25 iterations) is:
|
|
|
|
|
|
|
|
$$x_m = \begin{bmatrix}1\\5\end{bmatrix}$$
|
|
|
|
|
|
|
|
### Energy landscape
|
|
|
|
|
|
|
|
The following figure shows a `surf` plot of the objective function overlayed
|
|
|
|
with the iterates used to reach the minimizer:
|
|
|
|
|
|
|
|
![Energy landscape of the Rosenbrock function overlayed with iterates and steps
|
|
|
|
(the white dot is $x_0$ while the black dot is $x_m$)](./2-5-energy.png)
|
|
|
|
|
|
|
|
The code used to generate such plot can be found in the MATLAB script `main.m`
|
|
|
|
under section 2.5b.
|
|
|
|
|
|
|
|
### Gradient norms
|
|
|
|
|
|
|
|
The following figure shows the logarithm of the norm of the gradient w.r.t.
|
|
|
|
iterations:
|
|
|
|
|
|
|
|
![Gradient norms (y-axis, log-scale) w.r.t. iteration number
|
|
|
|
(x-axis)](./2-5-gnorms.png)
|
|
|
|
|
|
|
|
The code used to generate such plot can be found in the MATLAB script `main.m`
|
|
|
|
under section 2.5c.
|
|
|
|
|
|
|
|
Comparing the behaviour shown above with the figures obtained in the previous
|
|
|
|
assignment for the Newton method with backtracking and the gradient descent with
|
|
|
|
backtracking, we notice that the trust-region method really behaves like a
|
|
|
|
compromise between the two methods. First of all, we notice that TR converges in
|
|
|
|
25 iterations, almost double of the number of iterations of regular NM +
|
|
|
|
backtracking. The actual behaviour of the curve is somewhat similar to the
|
|
|
|
Netwon gradient norms curve w.r.t. to the presence of spikes, which however are
|
|
|
|
less evident in the Trust region curve (probably due to Trust region method
|
|
|
|
alternating quadratic steps with linear or almost linear steps while iterating).
|
|
|
|
Finally, we notice that TR is the only method to have neighbouring iterations
|
|
|
|
having the exact same norm: this is probably due to some proposed iterations
|
|
|
|
steps not being validated by the acceptance criteria, which makes the method mot
|
|
|
|
move for some iterations.
|
2021-05-08 12:19:37 +00:00
|
|
|
|
|
|
|
# Exercise 3
|
|
|
|
|
2021-05-10 12:53:57 +00:00
|
|
|
We first show that the lemma holds for $\tau \in [0,1]$. Since
|
|
|
|
|
|
|
|
$$\|\tilde{p}(\tau)\| = \|\tau p^U\| = \tau \|p^U\| \text{ for } \tau \in [0,1]$$
|
|
|
|
|
|
|
|
Then the norm of the step $\tilde{p}$ clearly increases monotonically as $\tau$
|
|
|
|
increases. For the second criterion, we compute the quadratic model for a
|
|
|
|
generic $\tau \in [0,1]$:
|
|
|
|
|
|
|
|
$$m(\tilde{p}(\tau)) = f - \frac{\tau^2 \|g\|^2}{g^TBg} - \frac12 \frac{\tau^2
|
|
|
|
\|g\|^2}{(g^TBg)^2} g^TBg = f - \frac12 \frac{\tau^2 \|g\|^2}{g^TBg}$$
|
|
|
|
|
|
|
|
$g^T B g > 0$ since we assume $B$ is positive definite, therefore the entire
|
|
|
|
term in function
|
|
|
|
of $\tau^2$ is negative and thus the model for an increasing $\tau \in [0,1]$
|
|
|
|
decreases monotonically (to be precise quadratically).
|
|
|
|
|
|
|
|
Now we show that the monotonicity claims hold also for $\tau \in [1,2]$. We
|
|
|
|
define a function $h(\alpha)$ (where $\alpha = \tau - 1$) with same monotonicity
|
|
|
|
as $\|\tilde{p}(\tau)\|$ and we show that this function monotonically increases:
|
|
|
|
|
|
|
|
$$h(\alpha) = \frac12 \|\tilde{p}(1 - \alpha)\|^2 = \frac12 \|p^U + \alpha(p^B -
|
|
|
|
p^U)\|^2 = \frac12 \|p^U\|^2 + \frac12 \alpha^2 \|p^B - p^U\|^2 + \alpha (p^U)^T
|
|
|
|
(p^B - p^U)$$
|
|
|
|
|
|
|
|
We now take the derivative of $h(\alpha)$ and we show it is always positive,
|
|
|
|
i.e. that $h(\alpha)$ has always positive gradient and thus that is it
|
|
|
|
monotonically increasing w.r.t. $\alpha$:
|
|
|
|
|
|
|
|
$$h'(\alpha) = \alpha \|p^B - p^U\|^2 + (p^U)^T (p^B - p^U) \geq (p^U)^T (p^B - p^U)
|
|
|
|
= \frac{g^Tg}{g^TBg}g^T\left(- \frac{g^Tg}{g^TBg}g + B^{-1}g\right) =$$$$= \|g\|^2
|
|
|
|
\frac{g^TB^{-1}g}{g^TBg}\left(1 - \frac{\|g\|^2}{(g^TBg)(g^TB^{-1}g)}\right) $$
|
|
|
|
|
|
|
|
Since we know $B$ is symmetric and positive definite, then $B^{-1}$ is as well.
|
|
|
|
Therefore, we know that the term outside of the parenthesis is always positive
|
|
|
|
or 0. Therefore, we now only need to show that:
|
|
|
|
|
|
|
|
$$\frac{\|g\|^2}{(g^TBg)(g^TB^{-1}g)} \leq 1 \Leftrightarrow \|g\|^2 \leq
|
|
|
|
(g^TBg)(g^TB^{-1}g)$$
|
|
|
|
|
|
|
|
since both factors in the denominator are positive for what we shown before.
|
|
|
|
|
|
|
|
We now define a inner product space $\forall a, b \in R^N, \; {\langle a,
|
|
|
|
b\rangle}_B = a^T B b$. We now prove that this is indeed a linear product space
|
|
|
|
by proving all properties of such space:
|
|
|
|
|
|
|
|
- **Linearity w.r.t. the first argument:**
|
|
|
|
|
|
|
|
${\langle x, y \rangle}_B + {\langle z,
|
|
|
|
y \rangle}_B = x^TBy + z^TBy = (x + z)^TBy = {\langle (x + z), y \rangle}_B$;
|
|
|
|
|
|
|
|
- **Symmetry:**
|
|
|
|
|
|
|
|
${\langle x, y \rangle}_B = x^T B y = (x^T B y)^T = y^TB^Tx$, and
|
|
|
|
since $B$ is symmetric, $y^TB^Tx = y^TBx = {\langle y,x \rangle}_B$;
|
|
|
|
|
|
|
|
- **Positive definiteness:**
|
|
|
|
|
|
|
|
${\langle x, x \rangle_B} = x^T B x > 0$ is true since B is positive definite.
|
|
|
|
|
|
|
|
Since ${\langle x, y \rangle}_B$ is indeed a linear product space, then:
|
|
|
|
|
|
|
|
$${\langle g, B^{-1} g \rangle}_B \leq {\langle g, g \rangle}_B, {\langle B^{-1}
|
|
|
|
g, B^{-1} g \rangle}_B$$
|
|
|
|
|
|
|
|
holds according to the Cauchy-Schwartz inequality. Now, if we expand each inner
|
|
|
|
product we obtain:
|
|
|
|
|
|
|
|
$$g^T B B^{-1} g \leq (g^T B g) (g^T (B^{-1})^T B B^{-1} g)$$
|
|
|
|
|
|
|
|
Which, since $B$ is symmetric, in turn is equivalent to writing:
|
|
|
|
|
|
|
|
$$g^T g \leq (g^TBg) (g^T B^{-1} g)$$
|
|
|
|
|
|
|
|
which is what we needed to show to prove that the first monotonicity constraint
|
|
|
|
holds for $\tau \in [1,2]$.
|
|
|
|
|
2021-05-08 12:19:37 +00:00
|
|
|
**TBD**
|