Lecture 13

The slides for the short talk on stochastic scheduling today are here. 

The method of uniformization that I described in this lecture seems to have been introduced by Jensen (1953), but it was first used extensively in solving queueing problems in the late 1970s (by Grassman (1977) and Keilson (1979)). The Wikipedia article has more about it and summarises the method in these words:

The method involves the constructions of an analogous discrete time Markov chain, where transitions occur according to an exponential distribution with the same parameter in every state.

I have used this idea many dozens of times in my own research. It is usually the best first step to make when  tackling a continuous-time Markov decision problem.

Kalman filter in economics

The Kalman filter was an important tool in space exploration, and is often mentioned in connection with the Apollo XI guidance system. Let me make some remarks about where Kalman filtering ideas are used, in areas adjacent to operations research, such as economics. 

I once asked Lord Eatwell, President of Queens', and a famous economist, "Is the Kalman filter much used in economics". He immediately replied, "Yes, all the time".

Eatwell is one of the compliers of the Palgrave Dictionary of Economics. It is a good place to go if ever you need to find a short article on any economics topic. I searched in Palgrave for the Kalman Filter, and read:

The Kalman filter

The Kalman filter deals with state-space representations where the transition and measurement equations are linear and where the shocks to the system are Gaussian. The procedure was developed by Kalman (1960) to transform (‘filter’) some original observables $y_t$ into Wold innovations at and estimates of the state $x_t$. With the innovations, we can build the likelihood function of the dynamic model. With the estimates of the states, we can forecast and smooth the stochastic process.

The use of unobserved components opens up a new range of possibilities for economic modelling. Furthermore, it provides insights and a unified approach to many other problems. The examples below give a flavour.

The local linear trend model generalizes (1) by the introduction of a stochastic slope, βt, which itself follows a random walk. Thus ...

Econometricians are often interested in building linear models in which some variables are explained by other variables (in some sort of regression model). 

As the values of variables become known over time one wants to update estimates of other variables. The machinery for doing this is provided by the Kalman filter. Notice that the Kalman filter does not have anything to do with the Q assumptions of our LQG model. It is only the Linear Gaussian parts that are relevant.

You might like to try a Google search for "Kalman Filter and finance". Many things turn up. For example, here is a talk, Kalman Filtering in Mathematical Finance. 

Lecture 12

Make sure you have the most recent version of the notes. I have redesigned Chapter 13. Previously, the notes covered only the Gaussian noise case. However, I think it easier to understand how certainty equivalence if I do it without the Gaussian assumptions, as I have now in Theorem 13.1.

When Gaussian assumption are made then $\hat x_t=E[x_t \mid W_t]$ is particularly simple. It is a linear function of $\hat x_0$ and $W_t$ and can be computed recursively using the Kalman filter.

The name "Kalman filter" refers to the estimation equation (13.12) and takes its name from Rudolf Kalman (1930 –), who developed it in the years 1958-64. He also coined the terms controllable and observable, and gave the criteria that we have seen in previous lectures. The fact that a system is controllable iff the matrix $[B\ AB\ \cdots\ A^{n-1}B]$ is of full rank is sometimes called Kalman's criteria. In the IEEE biography of Kalman it is stated

The Kalman filter, and its later extensions to nonlinear problems, represents perhaps the most widely applied by-product of modern control theory. It has been used in space vehicle navigation and control (e.g. the Apollo vehicle), radar tracking algorithms for ABM applications, process control, and socioeconomic systems.

The theory in this lecture is admittedly quite tricky - partly because the notation. As a test of memory, can you say what roles in the theory are taken by each of these?

 $x_t$,  $u_t$, $A$, $B$, $\epsilon_t$, $y_t$, $C$, $\eta_t$, $\hat x_t$, $\Delta_t$, $\xi_t$, $\zeta_t$, $R$, $S$, $Q$, $K_t$, $\Pi_t$, $N$, $L$, $M$, $H_t$,  $V_t$. 

 You will understand the ideas better once you have worked through the details of a scalar example (in which $n=m=p=1$). You do this in Example Sheet 3 Question 2. When you do this question, start by supposing that $\hat x_t=\hat x_{t-1}+u_{t-1}+h_t(y_t-\hat x_{t-1})$, and then find the value of $h_t$ that minimizes the variance of $\hat x_t$. You can start by subtracting $x_t=x_{t-1}+u_{t-1}+3\epsilon_t$ and using $y_t=x_{t-1}+2\eta_t$. You get,

$\hat{x}_{t+1}-{x}_{t+1}=\Delta_{t+1}=\Delta_t-3\epsilon_t-h_t\Delta_t+h_t2\eta_t.$

Then square, take the expected value, and minimize the variance of $\hat x_t$ with respect to $h_t$, to find a formula for $V_{t+1}$ in terms of $V_t$.

You will not be asked to reproduce the proof of Theorem 13.1 or 13.3 in examinations. You should simply know that $\hat{x}_t$ is computed from $\hat{x}_{t-1}$ and $y_t$ in the linear manner specified by (13.12), and that the covariance matrix $V_t$ satisfies a Riccati equation. You are not expected to memorize Riccati equations.

Notice that the Riccati equation for $V_t$, i.e. $V_t = g\, V_{t-1}$ runs in the opposite time direction to the one we had for $\Pi_t$ in lecture 10, where $\Pi_{t-1} = f\, \Pi_t$. We are given $V_0$ and $\Pi_h$.

Lecture 11

Controllability and observabliity stand in a dual relationship to one another. This is clear in the necessary and sufficient conditions: that $[B\ AB\ \cdots A^{n-1}B]$ must be rank $n$ (for controllability), and 

$\begin{bmatrix}C\\\ CA\\ \vdots\\ CA^{n-1}\end{bmatrix}$ must be of rank $n$  (for observability). 

This duality is also nicely exhibited in the starred question on Example Sheet 3, Question 11. You should now be able to do this question.

Notice that a system can be stabilizable without being controllable. E.g. the scalar system with $x_{t+1}=(1/2)x_t$is trivially stabilizable, but it is not controllable.

Lecture 10

There is a nice video of what optimal control can do when it is coupled to today's fast microprocessors and feedback controllers. Watch this:

Quadrocopter pole acrobatics

The idea of controllability is straightforward and we can understand it using ideas of linear algebra.

Consider again the broom balancing problem. I mentioned in lectures that it is not possible to stabilize two brooms at equilibrium simultaneously if they are the same lengths, but it is possible if the lengths are different. This is because of the forms:

$A=\begin{pmatrix}0 & 1 & 0 & 0\\ \alpha & 0 & 0& 0\\ 0 & 0 & 0& 1\\ 0 & 0& \beta & 0\end{pmatrix}\quad B=\begin{pmatrix}0 \\-\alpha\\0\\-\beta\end{pmatrix}\quad M_4=\begin{pmatrix}0 & -\alpha & 0 & -\alpha^2\\ -\alpha & 0 & -\alpha^2& 0\\ 0 & -\beta & 0& -\beta^2\\ -\beta & 0& -\beta^2 & 0\end{pmatrix}$

where $\alpha = g/L_1$, $\beta=g/L_2$.

So $M_4$ is of rank 4 iff $\alpha$ and $\beta$ are different. However, this assumes that as you move your hand you must keep it at constant height from the ground. Things might be different if you can move your hand up and down as well.

In 1996, paper 2, question 10 there was this tripos question about controllability:

A fantasy kingdom has populations of x vampires and y humans. At the start of each of the four yearly seasons the king takes a census and then intervenes to admit or expel some humans. Suppose that the population dynamics of the kingdom are governed by the plant equations:

$x_{t+1} = x_t + (y_t – Y)$,
$y_{t+1} = y_t – (x_t – X) + u_t$,

where $x_t$ and $y_t$ are integers representing the respective populations of vampires and humans in the $t$-th season, $t=0,1,2,3,\dotsc$; $X$ and $Y$ represent equilibrium values in the population without regal intervention; ...

[ some more description]

Can the king regain equilibrium by only expelling humans?
Suppose, the census is taken during the day, so the king can only count humans, and thus $u_t$ is allowed only to depend on $y_t$. Show that ...

The last part of the question is about observability (see Lecture 12). The full question is in this file of past questions

Lecture 9

The linear-quadratic regulator (LGR) that we met in Lecture 9 is one part of the solution to the linear-quadratic-Gaussian control problem (LQG). This problem is the subject of Lectures 9-12. The LQG problem is perhaps the most important problem in control theory. It has a full solution: given in terms of the Kalman Filter (a linear-quadratic estimator) and the linear-quadratic regulator.

Following today's lecture you can do Questions 8–10 on Example Sheet 2. I mostly did question 10 in today's lecture. Here is an important hint. Do not try to solve these problems by plugging in values to the general Riccati equation solution in equations. It is always better to figure out the solution from scratch, by the method of backwards induction. Conjecture for yourself that the optimal value function, $F(x,t)$, is of a form that is quadratic in the state variable, plus some $\gamma_t$ (if it is a problem with white noise). Then find the explicit form of the recurrences, working backwards inductively from a terminal time $h$, 

In general, solutions to a Riccati equation can be computed numerically, but not algebraically. However, one can find an full solution in the special case that $x_t$ and $u_t$ are both one-dimensional (Question 9). There is also a fully-solvable special case in which $x_t$ is two-dimensional and $u_t$ is one-dimensional (Question 10). A useful trick in some problems is to observe that $\Pi_t^{-1}$ satisfies a recurrence relation (Questions 9 and 10). 

I do not expect you to memorize the general form of the Riccati equation for the exams. The important thing is to remember enough of what it is for, how it looks, and how it is derived, so that you could reconstruct its derivation in the context of a particular problem, as you do in Question 9.

Lecture 8

Hints for Examples Sheet 2. In Question 2 you will want to mimic the calculation done in the proof of Bruss's odds algorithm. You cannot solve this by simply applying the algorithm directly.

You will find that the policy improvement algorithm is useful in answering Questions 5, 6 and 7. In Question 5, you will find $\lambda$ and $\phi$ for a certain policy ("On seeing a filling station, stop and fill the tank"), and then look for a condition that the policy improvement algorithm will not improve it (or, equivalently, that this $\lambda$ and $\phi$ satisfy the optimality equation.

In Question 6 you will use policy improvement idea in the context of a discounted-cost problem. You find $F(x)$ for a simple policy (in which the engineer allocates his time randomly), and then improve it by a step of the policy improvement algorithm. This leads to a policy in which the engineer puts all his maintenance effort into the machine with greatest value of $c_i(x_i+q_i)/(1-\beta b_i)$. This policy is better, but may not yet be optimal.

There is another interesting way to motivate the optimality equation in the average cost case. This can be made rigorous and helps us understand the relative value function $\phi(x)$.

Let $F(x)$ be the infinite-horizon value function when the discount factor is $\beta$. Then we know this satisfies the optimality equation

$F(x) = \min_u \{ c(x,u) +\beta E[ F(x_1) \mid x_0=x,u_0=u ] \}$ .

Pick some state, say state 0. By subtracting $\beta F(0)$ from both sides of the above, we obtain

$F(x) – F(0) + (1–\beta)F(0) = \min_u \{ c(x,u) + β E[ F(x_1) – F(0) \mid x_0=x,u_0=u ] \}$

One can show that as  $\beta\to 1$ we have have $F(x) – F(0) \to \phi(x)$ and $(1–\beta)F(0) \to\lambda$ (the average-cost). Thus we obtain

$\phi(x) + \lambda = \min_u \{ c(x,u) + E[ \phi(x_1) \mid x_0=x,u_0=u ] \}$ 

and this is our average-cost optimality equation. If you would like to understand why $(1–\beta)F(0) \to\lambda$  see this small note about the connection between the average-cost problem and the discounted-cost problem with $\beta$ near 1.

It is also interesting to think about the following (which I mentioned briefly in lectures today). Suppose we have a deterministic stationary Markov policy, say π, with u=f(x). Suppose we have λ and φ such that

φ(x) + λ = c(x,f(x)) + Σ_y φ(y) P(y | x )     (1)

where P(y | x) is the transition matrix under π.
Suppose π induces an ergodic Markov chain (i.e. a Markov chain that is irreducible and positive recurrent) and this has invariant distribution μ. We know that

μ_y = Σ_x μ_x P( y | x)     (2)

Multiplying (1) through by μ_x and then summing on x, we get

Σ_x μ_x φ(x) + λ Σ_x μ_x

= Σ_x μ_x c(x,f(x)) + Σ_y Σ_x μ_x φ(y) P( y | x)

which, using (2) gives

Σ_x μ_x φ(x) + λ Σ_x μ_x = Σ_x μ_x c(x,f(x)) + Σ_y μ_y φ(y).

Then using Σ_x μ_x = 1, and cancelling the terms that are equal on both sides, we have

λ = Σ_x μ_x c(x,f(x))

and so we see that λ is the average-cost of policy π.

Lecture 7

The Gittins index theorem is one of the most beautiful results in the field of Markov decision processes. Its discovery and proof in 1974 is due to John Gittins. The proof I have given in today's lecture is very different to Gittins's original proof. It was first presented in On the Gittins index for multiarmed bandits. Ann. Appl. Prob. 2, 1024-33, 1992. Proof is pages 1026-27.

For today's lecture I presented some slides on the Gittins Index. You may enjoy looking through the entire talk. It will look better if you download the .pdf file to your computer and read the presentation with a .pdf viewer such as acrobat, rather than try to read within your browser.

The theory of the MABP and proof of the Gittins index theorem is non-examinable. However, I thought you would enjoy seeing this beautiful result and how it can be proved. Lectures 1-6 have covered everything you need to know in order to understand the Gittins index. Today's lecture has also been an opportunity to revise ideas that we already met in problems on job scheduling and pharmaceutical trials. On Examples sheet 2, #3 you are asked to calculate the Gittins indices in a simple example - which should help to test your understanding of today's lecture.

The prospecting problem in 6.5 and Weiztman's Pandora's boxes problem in 7.5 are really the same problem, and a simple example to which the Gittins index theorem provides the answer.

Burglar problem

I think it will be helpful if I say something about Examples Sheet 1, #8.

Since this is a case of positive programming and $F(x)=x$ satisfies the optimality equation in the region $x\geq x^*=(1-q)/q\lambda$, it  follows from Theorem 4.2 that it is optimal for the burglar to retire if he has accumulated loot of at least $x^*$.

The difficulty in this problem lies in the question of whether or not  an optimal policy exists when $x<x^*$. We cannot use Theorem 5.2 to  claim that an optimal policy exists because we are not in a D or N programming case.

Let $x$ be his accumulated loot so far. We know finite horizon problems are easier to analyse, so we might look at a problem in which the burglar may burgle at most $s$ further days, with optimality equation
$$F_s(x) = \max\left[ x, (1-q)\int_0^\infty F_{s-1}(x+y) \lambda e^{-\lambda y}   dy\right],$$ with $F_0(x)=x$. For this problem a OSLA rule is optimal.  It is easily proved by induction that it is optimal to follow policy $\pi^*$, defined as "stop iff $x\geq x^*=(1-q)/q\lambda$".

One way we could proceed would be by actually calculating $F(\pi^*,x)$ for $x<x^*$ and showing it satisfies the DP equation. This is possible but difficult.

An alternative is to show that $F(x)\leq F(\pi^*,x)$.  To do that, let's consider a problem with artificially greater rewards. Suppose that $x_0=x<x^*$, and on the $(s+1)$th day of burglary the burglar receives (if he is not caught) his usual exponentially distributed increment, plus an additional increment of $x^*$. Having received all of this, he will certainly want to stop since he will now have more loot than $x^*$. But in this problem, with greater rewards, the maximal reward the burglar can obtain is certainly bounded above by
$$F_s(x)+(1-q)^s(x+(s+1)\tfrac{1}{\lambda} + x^*),$$ which is the maximum amount he could obtain by stopping after no more than $s$ days, plus the expected reward he will obtain if he stops after the $(s+1)$th day. The second term accounts for the possibility he is caught before $s+1$ (through the multiplying factor of $(1-q)^s$), but ignores the fact that he cannot actually get both this and the reward from stopping somewhere amongst the first $s$ days.  The important thing is that the second term tends to $0$ as $s\to\infty$. This implies that the maximal amount he could get in the original problem, $F(x)$, satisfies $F(x)\leq \lim_{s\to\infty}F_s(x)=\lim_{s\to\infty}F_s(\pi^*,x)=F(\pi^*,x)$.  So in fact $\pi^*$ is an optimal policy.

Lecture 6

Today we looked at some stopping problems. Many problems are like this. They are present in financial mathematics in the theory of option pricing (Black-Scholes). The holder of an American option is allowed to exercise the right to buy (or sell) the underlying asset at a predetermined price at any time before or at the expiry date. The valuation of American options is essentially an optimal stopping problem.

You can read more about Bruss's odds algorithm in his paper: "Sum the odds to one and stop" Annals of Probability, Vol. 28, 1384–1391, (2000). I showed you  how to use this to address the Secretary Problem. A very interesting related problem is the "last arrivals problem". Consider a shopkeeper who would like to go home early just after he has served the last potential customer of the day.

I will say more about the mine prospecting problem of Section 6.5 next time.

Let me finish with some hints about questions on Examples Sheets 1 and 2. Question 6 on Sheet 1 can be solved by realizing that there are only two stationary Markov policies. You simple need to decide which is best. Question 7 is a simple stopping problem that can be solved using a OSLA rule. Question 8 is tricky. You should use Theorem 4.2 to prove that $F(x)=x$ when $x\geq x^*=(1-q)/(q\lambda)$. But the difficult part of this question is to convince yourself that an optimal policy exists when $x<x^*$. 

Question 9 on Sheet 1 is important because it should convince you that it is possible to solve stochastic dynamic programming equations using linear programming (at least if both state space and action space are finite.)  It there are $n$ states and $m$ actions (controls) available in each state then we will have an LP in $n$ variables, with $nm$ linear inequality constraints. 

When on sheet 2 you do Question 1 you might think about using Bruss's Odds Algorithm. Although you cannot apply the algorithm directly, you should be able to solve this problem using the same scheme of proof as we used to prove the optimality of the odds algorithm in this lecture.