\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{geometry}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage{epsfig}
\usepackage{graphics}
\usepackage{latexsym}
\usepackage{fullpage}
\usepackage{mysymbols}
\usepackage{subfig}
\usepackage{multirow}
\usepackage[parfill]{parskip}
%\newcommand{\argmax}{\operatornamewithlimits{argmax}}
%\theoremstyle{definition}
%\newtheorem{def}[thm]{Definition}
\title{10708 Probabilistic Graphical Models: Final Exam\\
{\small Due Dec 10th by Noon electronically to 10708-instr@cs.cmu.edu
or paper version to Michelle Martin, or by fax to 412-268-3431,}}
\date{}
\begin{document}
\numberwithin{equation}{section}
\maketitle
%\hrule
\vspace{-0.5in} \noindent
Your final must be done individually. You
may not discuss the questions with anyone other than Carlos or the
TAs (you are free to ask us questions by e-mail or in person if you
are having problems with a question). The exam is open book, but not
open-Google, {\it i.e.}, you can use any materials we discussed in
class or linked to from the class website. You are not allowed to
look at other sources. However, you may use a calculator or Matlab
to do numerical computations, if necessary.
Each
question has the name of one of the TAs beside it, to whom you should
direct any inquiries regarding the question. Please submit
your final in two parts, one for each TA. Also, please put the
TA's name on top of the final.
If you hand in your
assignment early, you can get bonus points. \vspace{4pt}
\begin{center}
\begin{minipage}[!h]{0.25\linewidth}
\centering
\begin{tabular}{|c|c|}
\hline
{\sc Handin} & {\sc Bonus} \\
\hline \hline
Monday, Dec 8, 2pm & 3 pts \\
Tuesday, Dec 9, 2pm & 2 pts \\
\hline
\end{tabular}
\end{minipage}
\end{center}
{\large{\textbf{You may {\it NOT} use late days on the final.}}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Short answers {\small [Amr] [6 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[!h]
\begin{center}
{\includegraphics[bb=0 0 917
339,width=\columnwidth]{q1.png}\label{fig:score}} \caption {Graphs
used in question 1}
\end{center}
\end{figure}
\begin{enumerate}
\item For each of the following questions, answer \emph{true} of
\emph{false} justifying your answer in (1-3 sentences).
\begin{enumerate}
\item
\begin{enumerate}
\item $G_1$ is I-equivalent to $G_2$
\item $G_1$ is I-equivalent to $G_3$
\item If $G_1$ is a perfect map for $\cal{P}$, then $G_3$ is an I-map for $\cal{P}$
\end{enumerate}
\item
If $G_1$ is a perfect map for $\cal{P}$, given a infinite samples, $\cal{D}$, from $\cal{P}$:
\begin{enumerate}
\item $\mathrm{Score}_{ML}(G_1;{\cal{D}})= \mathrm{Score}_{ML}(G_2;{\cal{D}})$
\item $\mathrm{Score}_{ML}(G_1;{\cal{D}}) < \mathrm{Score}_{ML}(G_3;{\cal{D}})$
\item $\mathrm{Score}_{BIC}(G_1;{\cal{D}}) = \mathrm{Score}_{BIC}(G_2;{\cal{D}})$
\item $\mathrm{Score}_{BIC}(G_1;{\cal{D}}) < \mathrm{Score}_{BIC}(G_3;{\cal{D}})$
\end{enumerate}
\item Let ${\cal{K}}$ be a cluster graph for a distribution ${\cal{P}}$ (\emph{i.e.}
it satisfies the generalized running intersection and family
preservation properties). If ${\cal{K}}$ happens to be a tree, then
${\cal{K}}$ is a clique tree as well for ${\cal{P}}$. (hint: either
provide a counter example or show that the generalized RIP reduces
to RIP in this case)
\end{enumerate}
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Structure Learning in Undirected Models {\small [Dhruv] [12 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
For this problem, assume that you have i.i.d. data sampled from a
distribution $P(\mathcal{X})$. $P$ is represented by a Markov
Random Field whose graph structure is unknown. However, you do know
that each node has at most $d$ neighbors.
\begin{enumerate}
\item Show why knowing the Markov blanket of each node is sufficient for
determining the graph structure.
\item For any node $X$ and its Markov blanket $\mathrm{MB}(X)$, we know
that \beq P \models (X \perp \mathcal{X} - \{X\} - \mathrm{MB}(X) |
\mathrm{MB}(X)). \eeq Briefly, why might you need \emph{a lot} of
data to test for this conditional independence directly?
\item For disjoint sets of variables $\mathbf{A}$ and $\mathbf{B}$, let
conditional entropy be defined as, \beq \mathrm{H}(\mathbf{A}
|\mathbf{B}) = -\sum_{\mathbf{a}, \mathbf{b}}
P(\mathbf{A}=\mathbf{a}, \mathbf{B}=\mathbf{b}) \log
P(\mathbf{A}=\mathbf{a} | \mathbf{B}=\mathbf{b}) \eeq Prove that for
any node $X$, $\mathrm{H}(X | \mathrm{MB}(X)) = \mathrm{H}(X |
\mathcal{X} - \{X\})$.
\item For disjoint sets of variables $\mathbf{A}$, $\mathbf{B}$ and
$\mathbf{C}$, we have that \beq \mathrm{H}(\mathbf{A} |\mathbf{B},
\mathbf{C}) \leq \mathrm{H}(\mathbf{A} |\mathbf{B}).\eeq In other
words, information never hurts. Prove that $\mathrm{MB}(X) =
\argmin_{\mathbf{Y}} \mathrm{H}(X|\mathbf{Y})$.
\item Using the intuition developed in the previous parts, describe a
structure learning algorithm for Markov Random Fields, assuming the
constraint that each node has at most $d$ neighbors. Your algorithm
should run in $O(n {n \choose d} c)$ time, where $n$ is the number
of nodes in your model, and $c$ is the complexity of computing the
conditional entropy $\mathrm{H}(X|\mathbf{Y})$, when $|\mathbf{Y}|
\leq d$.
\item If we removed the constraint that each node have at most $d$
neighbors and instead changed our optimization problem to include a
penalty term, $\mathrm{MB}(X) = \argmin_{\mathbf{Y}}
\{\mathrm{H}(X|\mathbf{Y}) + |Y| \}$, how would the time complexity
of the algorithm change?
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ Trading Inference Accuracy vs. Efficiency {\small [Amr 16 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\emph{\textbf{Note:}} Nearly half of this problem is marked as extra
credit.
As we know from class, compact representation
does not translate into efficient inference. For trees, the cost of
inference is linear in the size of the factors. For loopy graphs, we
can either convert it to a clique tree and then use belief
propagation, or just run the belief propagation algorithm on the
loopy graph(which is equivalent to first constructing a bipartite
factor graph as we discussed in class, and then running BP on it.)
In the former case, inference is exact but its cost is exponential
in the largest clique size. In the later case inference is
approximate, but its cost is linear in the size of the largest
factor. The above two solutions comprises two extremes in the space
of efficiency vs. exactness. In both cases inference is carried via
BP over a given graph: in clique trees, the graph vertices
correspond to maximal cliques, while in factor graphs, the graph
vertices correspond to factors and singleton variables in the
original network. Region graphs generalize the above two solutions
by filling the gap between them, and allowing for trading efficiency
vs. exactness. In this question we will explore this connection in
more details.
\textbf{Region Graphs (K\&F 10.3):} Recall that a region graph is a
directed acyclic graph with vertices correspond to regions which are
subsets of the variables. If an edge exists from region $r$ to
region $r'$ then scope[$r'$] $ \subset$ scope[$r$]. Each factor
$\phi$ in the original network is assigned to a top-most region $r$
such that Scope[$\phi$]$ \subset$ scope[$r$]. The region graph is
calibrated via a message-passing algorithm similar to BP. The exact
message passing rules are given in 10.37 in (K\&F). Upon
calibration, the belief of each region is computed using 10.36 in
(K\&F).
\begin{figure}[!h]
\begin{center}
\subfloat[${\cal{T}}_1$]
{\includegraphics[width=0.4\linewidth]{ct2.pdf}\label{fig:ct}}\hspace{0.05\linewidth}
\subfloat[The split operator: $AB \perp EF
|CD$]{\includegraphics[width=0.5\linewidth]{split.pdf}\label{fig:split}}
\hspace{0.05\linewidth} \end{center} \caption{}
\end{figure}
\subsection{Clique trees and factor graphs as region graphs}
\begin{enumerate}
\item Represent the clique tree ${\cal{T}}_1$ in Figure \ref{fig:ct}
as a region graph ${\cal{R}}_1$? (Hint: the region graph has two
levels)
\item ${\cal{T}}_1$ can be calibrated using BP as follows. First pick a
root, then pass messages upward from leaves to root using BP (i.e.
sum-product), and finally pass messages downward from root to leaves
using BP. Similarly ${\cal{R}}_1$ can be calibrated using message
passing as in 10.37 in K\&F. If we initialize all messages in
${\cal{R}}_1$ to 1, show that there exists a schedule of sending
messages that 1) will converge in only one pass, and 2) will achieve
the same final beliefs over all cliques and sepsets as in
${\cal{T}}_1$. (Hint: it is easier to find a schedule of sending
messages over ${\cal{R}}_1$ that parallel the messages sent during
calibrating the clique tree)
\item Given a clique tree ${\cal{T}}$ with cliques ${\cal{\textbf{C}}}$ and sepsets
${\cal{\textbf{S}}}$, sketch a scheme of representing it as an
equivalent region graph ${\cal{R}}$. (hint: generalize your solution
to part 1). If ${\cal{T}}$ has $N$ cliques, how many regions and
edges does ${\cal{R}}$ contain?
\item Given a factor graph, show how to represent it as a region graph. (hint: this is
easy) We can also show that loopy belief propagation is equivalent
to running the region graph's message-passing algorithm over the
above representation (You do NOT have to show that).
\end{enumerate}
\subsection{Building region graphs top-down}
The cost of sending messages over the region graph is exponential in
the largest region size \footnote{Assuming a naive implementation of
region graphs. In practice, for each region, one should only store
its factors and then perform variable elimination when sending each
message, therefore the effective width is the largest clique size
that results during this inference step. As we know from class, we
can always estimate this size once by inspecting the factors used in
the elimination process and the elimination order}. Therefore, by
constructing a region graph with \emph{bounded} region size, one can
trade efficiency vs. exactness. We will define \emph{the width of a
region graph} to be the size of its largest region.
\textbf{Definition 1: Region graph invariance:} There are operations
that when applied on a region graph ${\cal{R}}$, result in an
equivalent region graph ${\cal{R}}'$ such that after calibration,
both ${\cal{R}}$ and ${\cal{R}}'$ agrees on the same marginals. One
of these operations is \emph{split} illustrated in Figure
\ref{fig:split} and defined as follows. Consider an outer region
(that is a region in the top-most level) $C_r$. If we can find a
partitioning of the variables in scope[$C_r$] as
$(C_{r1},C_{r2},C_{r3})$ such that $C_{r1} \perp C_{r3} |C_{r2}$,
then we can replace $C_r$ with three regions: $C_{r12}, C_{r23}$ and
$C_{r2}$ whose scope is given by $C_{r1} \cup C_{r2}, C_{r2} \cup
C_{r3}$ and $C_{r2}$ respectively. Moreover, $C_{r2}$ will become a
child of both $C_{r12},C_{r23}$, and all edges $C_r \rightarrow
C_j$ will emanate from one of the new regions with the smallest
scope that contains scope[$C_j$] (see Figure \ref{fig:split} for an
example).
\begin{enumerate}
\item Given a clique tree ${\cal{T}}$ with cliques ${\cal{\textbf{C}}}$ and sepsets
$\cal{\textbf{S}}$, let $R_T$ be its equivalent region graph as you
derived in 3.1.3. Let $R$ be a region graph that consists of a
single region $r$ such that scope[$r$]= ${\cal{X}}$ (i.e. all the
variables). Show that using a sequence of split operations, $R$ can
be morphed into the region graph ${\cal{R}}_T$.
\item Briefly conclude that inference using ${\cal{R}}_T$ is thus
exact.
\item\textbf{[Extra credit 4pts]} Given ${\cal{R}}_T$,one can not reduce its width further using
the split operator. However, one can define a \emph{soft\_split}
operator that behaves exactly like \emph{split}, but only requires
that $C_{r1}$ is almost independent from $C_{r3} |C_{r2}$. We define
almost independent such that $I(C_{r1};C_{r3} |C_{r2}) < $
threshold. Clearly the resulting region graph after the soft\_split
operation will result in approximate inference. Using this new
operator, show how to build a region graph with width $W$. (Hint:
there is a subtle point here that you should think about. What
happened if based on the soft\_split criteria a region $r=ABCD$ that
has two children $r_1=BCD$ and $r_3=ABC$ is splitted into $AB$, $B$
and $BCD$?)
\item\textbf{[Extra credit 2pts]} Why is it hard to evaluate the operator \emph{soft\_split} in
practice? (hint: this is an instance of a chicken and egg problem)
\end{enumerate}
\subsection{Building region graphs bottom-up\textbf{ [Extra credit]}}
One interpretation of the operation of calibrating a region graph is
that it is equivalent to optimizing a free energy functional called
Kikuchi free energy. In this view, the fixed-point regions' beliefs
are maxima of this functional. Lets assume that for a calibrated
region graph ${\cal{R}}$ that $F({\cal{R}})$ is the value of the
free energy achieved. We will use $F({\cal{R}})$ as a proxy for the
approximation quality when ${\cal{R}}$ is used for inference. More
specifically, ${\cal{R}}_1$ gives a better approximation than
${\cal{R}}_2$ if $F({\cal{R}}_1) > F({\cal{R}}_2)$. Starting from a
base region graph, we would like to find the best region graph with
bounded width $W$ by greedily adding a bigger outer region at each
iteration\footnote{An outer region is a region with no parent.}. We
will use a factor graph as the base region graph\footnote{in this
case, the Kikuchi free energy reduces to the Bethe free energy
discussed in class}.
\begin{enumerate}
\item \textbf{ [5 points]} Assume there exists an oracle named Candidates(${\cal{R}}$)
that given a region graph, it returns candidate regions that are NOT
strictly contained in any current region in ${\cal{R}}$(but
definitely overlap with some regions in ${\cal{R}}$,why?). Starting
from a base factor graph, give an algorithm that iteratively builds
the best (in the sense defined above) region graph of bounded width
$w$.
\item\textbf{ [2 points]} How would you implement the Candidates(${\cal{R}}$) oracle? (your
answer should be brief and \emph{efficient}, there is no need for a
theoretical justification [this is an open problem], you can just
give the intuition behind your reasoning)
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{KL Projection in Assumed Density Filtering {\small [Dhruv] [12 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In class, we discussed the Boyen-Koller algorithm, an instance of
assumed density filtering, where the belief state is represented by
a clique tree. At each time step, the belief state becomes more
complex, and we project it into a simpler clique tree by doing a
simple marginalization. This approach may seem like a hack, but, in
this question, you will show that this marginalization is a
well-defined, KL-minimizing projection.
\\
Consider the clique tree $T_1$ (corresponding to the complex belief
state in Boyen-Koller): \beq ABC - BCD - CDE \eeq \noindent Let the
cliques be calibrated, so that we have all the clique marginal
probabilities.
\\
Consider also the clique tree $T_2$ (corresponding to the simpler
belief state in BK): \beq AB - BC - CD - DE \eeq \noindent A KL
projection of a distribution $P$ over a set of distributions $S$ is
given by, \beq P_{S} = \text{arg min}_{Q \in S}KL(P||Q) \eeq
\noindent Let $P$ denote the distribution given by the calibrated
clique tree $T_1$; and let $S$ denote the set of distributions
represented by clique tree $T_2$ (i.e., for which $T_2$ is an
I-map). Show that the KL Projection of $P$ over $S$ is given by
setting the clique probabilities in $T_2$ to be the marginals of the
corresponding clique probabilities in $T_1$. That is, $P_{T_2}(AB) =
\sum_{C}P_{T_1}(ABC)$, and so on.
\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Gaussian Graphical Models {\small [Dhruv] [13 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Warming up}
Lets first get acquainted with the canonical parameterization of
Gaussian distributions. (Notation: Throughout this question, we will
refer to Gaussians in standard form as $N(\cdot; \mu, \Sigma)$ and
Gaussians in canonical form as $N_c(\cdot; \eta, \Lambda)$) You are
given the joint distribution over $X$ and $Y$ in standard form:
\beq P(X, Y) = N\big(X, Y; \mu = \left[ \begin{array}{c} 3 \\
6 \end{array} \right], \Sigma = \left[ \begin{array}{cc} 1 & 0.5 \\
0.5 & 0.75
\end{array} \right]\big)\eeq
\begin{enumerate}
\item Write down $P(X, Y)$ in canonical form.
\item Write down $P(Y)$ in canonical form.
\item Using your answers from parts (1) and (2), write down $P(X | Y)$ in canonical form.
(Hint 1: To multiply two Gaussians in canonical form, you simply add
the parameters, and to divide two Gaussians in canonical form, you
subtract the parameters, filling in with zeros as necessary. Refer
to the Kalman filter slides for more details.)
(Hint 2: Your answer to part (c) in canonical form will be
represented as a multivariate distribution over $X$ and $Y$, not a
univariate distribution over $X$ as would be the case in standard
form.) \end{enumerate}
\begin{figure}[!h]
\begin{center}
\subfloat[Gaussian Graphical Model]
{\includegraphics[width=0.3\linewidth]{gaussjt.pdf}\label{fig:ggm}}\hspace{0.05\linewidth}
\subfloat[Junction
Tree]{\includegraphics[width=0.4\linewidth]{jt.pdf}\label{fig:jt}}
\hspace{0.05\linewidth} \end{center}
\caption{}
\end{figure}
\subsection{Message passing in Gaussian Graphical Models}
In Figure~\ref{fig:ggm} we give you a Gaussian graphical model with
the following conditional probability distributions (all given in
canonical form):
\begin{eqnarray*}
P(A) & = & N_c\big(A; \eta = 9, \Lambda = 1 \big) \\
P(B) & = & N_c\big(B; \eta = 1, \Lambda = 0.6 \big) \\
P(C|A) & = & N_c\big(C, A; \eta = \left[
\begin{array}{c} 0 \\ 0
\end{array} \right], \Lambda = \left[ \begin{array}{cc}
1 & -1 \\ -1 & 1
\end{array} \right]\big) \\
P(D|B) & = & N_c\big(D, B; \eta = \left[
\begin{array}{c} 0.5 \\ -0.5
\end{array} \right], \Lambda = \left[ \begin{array}{cc}
0.25 & -0.25 \\ -0.25 & 0.25
\end{array} \right]\big) \\
P(E|C, D) & = & N_c\big(E, C, D; \eta = \left[
\begin{array}{c} 0 \\ 0 \\ 0
\end{array} \right], \Lambda = \left[ \begin{array}{ccc}
1 & -9 & 0.5 \\ -9 & 81 & -4.5 \\ 0.5 & -4.5 & 0.25
\end{array} \right]\big) \\
P(F|E) & = & N_c\big(F, E; \eta = \left[
\begin{array}{c} 2 \\ 2
\end{array} \right], \Lambda = \left[ \begin{array}{cc}
2 & 2 \\ 2 & 2
\end{array} \right]\big)
\end{eqnarray*}
In this problem, you will use the Shafer-Shenoy message passing
scheme for Gaussians to perform exact inference on this model. The
Shafer-Shenoy algorithm for Gaussian graphical models is analogous
to the algorithm presented in class for discrete models. However,
you will need to take into account how to multiply potentials
together and how to marginalize out variables in the canonical
Gaussian setting as discussed in 5.1. In your answers to the
questions below, your Gaussian distribution should be written in
\emph{canonical} form.
\begin{enumerate}
\item Give an elimination ordering for the Bayesian network in
Figure~\ref{fig:ggm} that would result in the junction tree in
Figure~\ref{fig:jt}.
\item
Using the Family Preserving Property, assign the given CPDs to
appropriate cliques in the junction tree, then compute the initial
clique potentials $\Pi_1^{(0)}$, $\Pi_2^{(0)}$, $\Pi_3^{(0)}$, and
$\Pi_4^{(0)}$.
\item Compute $P(C, D, E)$ using Shafer-Shenoy message passing. Write
down the three messages that were needed to compute this
probability. Both your final answer and your messages should be
represented as Gaussian distributions in canonical form.
\item Given the messages computed for part 3, what additional message would
you need if you wanted to compute $P(A|C)$? Write down this
message. (Note: you do not need to compute $P(A|C)$.)
\item Given that a minimal junction tree for a Bayesian network with $n$
nodes can have at most $n$ cliques, what is the time complexity of
Shafer-Shenoy on a Gaussian graphical model with $n$ nodes and
induced tree width $w$? Briefly justify your answer in one or two
sentences. (Hint: Time complexity of matrix inversion for a $k
\times k$ matrix is $O(k^3)$.) What is the running time of a
discrete model over the same BN structure, where each variable takes
on at most $c$ values?
\item \textbf{{[Extra Credit] {\small [3 pts]}}} If we wanted to compute $P(C, D,
E)$, an alternative to message passing would have been to multiply
together all the CPDs, form a single matrix for the distribution
$P(A,B,C,D,E,F)$ and directly marginalize out all the other
variables. What is the time complexity of this operation? Briefly
justify your answer in one or two sentences.
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Variational Free Energy {\small [Amr] [18 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Once upon a time there were three bears, a mother bear, a father
bear, and a baby bear. The mother was a frequentist; the father a
pragmatic Bayesian. However, the baby bear thought that his parents
were terribly silly -- his mother prone to knitting sweaters that
were far too snug, that overfit; his father far too stern in his
insistence that he choose a single sweater to wear each day. The
baby bear preferred to think of himself as wearing a distribution
over sweaters, making him the only proper Bayesian bear in the whole
forest.
Knowing the bears' fondness for parameter estimation, a little
blonde girl comes along to steal their work. Finding no one home,
she looks at the desks of each bear and finds,
\begin{defn}[Maximum Likelihood]
\label{def:em-ml}
If $y$ are the observed variables and $\theta$ the model parameters
then the maximum likelihood criterion is
$$\hat{\theta}_{ML} = \argmax_{\theta}\log{p(y|\theta)}$$
\end{defn}
\begin{defn}[Maximum a Posteriori]
\label{defn:em-map}
If $y$ are the observed variables and $\theta$ the model parameters
then the maximum a posteriori criterion is
$$\hat{\theta}_{MAP} = \argmax_{\theta}\log{p(\theta|y)}$$
\end{defn}
\begin{defn}[Fully Bayesian]
\label{defn:em-bayes}
If $y$ are the observed variables and $\theta$ the model parameters
then the fully Bayesian criterion demands the full posterior
$$p(\theta|y) = \frac{p(\theta)p(y|\theta)}{p(y)}$$
\end{defn}
\subsection{EM-ML}
It is well known to any frequentist bear that, denoting the latent
variables $z$ and introducing any distribution over latent variables,
$q(z)$,
\begin{align*}
\log{p(y|\theta)}
&\geq \sum_zq(z)\log{\frac{p(y,z|\theta)}{q(z)}} \\
&= E_{q(z)}[\log{p(y,z|\theta)}] + H[q(z)]
\equiv F(q,\theta) .
\end{align*}
The EM algorithm consists of maximizing a lower bound on $p(y|\theta)$ by
iterating the following steps over time $t$:
\begin{align*}
\textrm{\bf E-step: }\, & q^{(t+1)} = \argmax_{q}F(q,\theta^{(t)}) \\
\textrm{\bf M-step: }\, & \theta^{(t+1)} = \argmax_{\theta}F(q^{(t+1)},\theta)
\end{align*}
\begin{enumerate}
\item In the E-step, prove that $q^{(t+1)} = p(z|y,\theta^{(t)})$.
\item In class it was explained that the M-step updated parameters
using expected counts. Show that in the E-step expected counts are
computed using
\begin{align*}
E_{q(z)}[\textrm{Count}({\bf A}_O = {\bf a}_O, {\bf A}_H = {\bf a}_H)] &=
\sum_{j = 1}^{R}{\bf 1}({\bf A}^{(j)}_O = {\bf a}_O)P({\bf A}_H = {\bf a}_H|O^{(j)},\theta^{(t)})
\end{align*}
where $O^{(j)}$ is the $j^{th}$ record in the data set, ${\bf A}_O$
are the observed variables, ${\bf A}_H$ the unobserved (latent)
variables, and $\theta^{(t)}$ the estimate of the Bayesian network
parameters at step $t$ of the EM algorithm. ${\bf 1}({\bf A}_O^{(j)} =
{\bf a}_O)$ is an indicator function for whether variables ${\bf A}_O$
take on value ${\bf a}_O$ is record $j$. \label{ques:em-ml:b}
\item You are given two inference routines, one for variable
elimination and another for junction trees. Which routine is more
appropriate for the E-step in part~\ref{ques:em-ml:b} ? Briefly
explain you answer.
\end{enumerate}
\subsection{EM-MAP}
The father bear has scrawled down the following equation:
\begin{equation}
\log{p(\theta|y)} \geq
E_{q(z)}[\log{p(y,z|\theta)}] + H[q(z)] + \log{p(\theta)}
\equiv F(q,\theta)
\label{eqn:var-map}
\end{equation}
where $q(z)$ is some distribution over unobserved variables $z$, which
is typically called the variational free distribution, and $p(\theta)$
is a parameter prior. The EM algorithm consists of maximizing a lower
bound on $p(\theta|y)$.
\begin{align*}
\textrm{\bf E-step: }\, & q^{(t+1)} = \argmax_{q}F(q,\theta^{(t)}) \\
\textrm{\bf M-step: }\, & \theta^{(t+1)} = \argmax_{\theta}F(q^{(t+1)},\theta)
\end{align*}
\begin{enumerate}
\item Prove equation~\ref{eqn:var-map}.
\end{enumerate}
\subsection{Fully Bayesian}
The bears come home to find the little blonde girl rummaging through
their desks. Regular bears would simply maul her. However, since these
are not regular bears they chain the girl up and give her choice:
answer the following questions or be eaten alive.
\begin{enumerate}
\item Briefly explain why computing $p(y)$ exactly is difficult ?
\item Assuming some free distribution that factors over latent variables and
parameters, $q(z,\theta) = q(z)q(\theta)$, prove that
\begin{equation*}
\log{p(y)} \geq
\int q(\theta)\ln{\frac{p(\theta)}{q(\theta)}}\,d\theta + \int q(\theta)\sum_{z}q(z)\ln{\frac{p(y,z|\theta)}{q(z)}}\,d\theta
\equiv F(q(z),q(\theta))
\end{equation*}
\item Prove that maximizing $F(q(z),q(\theta))$ corresponds to
minimizing the KL-divergence $D(q(z,\theta)||p(z,\theta|y))$.
\end{enumerate}
\noindent We can estimate the marginal likelihood $p(y)$ by maximizing
the lower bound $F(q(z),q(\theta))$ with the following EM-style
algorithm (which you do not need to prove):
\begin{align*}
\textrm{\bf E-step: }\, & q^{(t+1)}(z) \propto \exp\int{q^{(t)}(\theta)\ln{p(y,z|\theta)}}\,d\theta \\
\textrm{\bf M-step: }\, & q^{(t+1)}(\theta) \propto p(\theta)\exp\int\ln{p(y,z|\theta)q^{(t+1)}(z)}\,dz
\end{align*}
When asked to derive this, the little blonde girl decided that she'd
rather be eaten alive. The end.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{MPE inference in MRFs with Graph-Cuts {\small [Dhruv] [23 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Recall that an MPE (Most Probable Explanation) query attempts to
find the most probable assignment to all the non-evidence variables.
More precisely, if $\mathcal{X}$ is a set of random variables, $E$
is the set of evidence variables, and $W = \mathcal{X} - E$ is the
set of query variables:
\begin{align}
\text{MPE} \, (W \mid e) = \argmax_{w} P (W=w \mid E=e).
\end{align}
Also recall that an MRF can be parametrized with energy functions:
\begin{align}
P(\mathcal{X}) & = \frac{1}{\mathcal{Z}}\, \prod \Phi_C(X_C) \\
& = \frac{1}{\mathcal{Z}} \, \prod \exp (-\mathcal{E}_C(X_C)),
% & = \sum \log (\mathcal{E}_{\mathcal{C}}) - \log \mathcal{Z}
%\mathcal{E} (\mathcal{X}) = - \log(P(\mathcal{X})).
\end{align}
where $\mathcal{E}_C(X_C) = -\log \Phi_C(X_C)$ is the energy of the clique $C$.
With this parametrization, the MPE problem can be rewritten as:
\begin{align}
\argmax_{\mathcal{X}} P( \mathcal{X}) & = \argmax_{\mathcal{X}} \log P( \mathcal{X}) \\
& = \argmin_{\mathcal{X}} \left( \sum \mathcal{E}_{\mathcal{C}} - \log \mathcal{Z} \right) \\
& = \argmin_{\mathcal{X}} \sum \mathcal{E}_{\mathcal{C}},
\end{align}
where we can ignore $\mathcal{Z}$ in the last step because the partition function does not depend on the assignment
of the variables. Thus we have reparametrized the MPE problem into that of energy minimization.
In this question, you will show that for a certain class of energy
functions, MPE problem in binary-valued MRFs can be solved optimally
using a very simple graph-cut algorithm. The most surprising part of
this reduction is that the algorithm is guaranteed to return the
optimal solution in polynomial time, regardless of the structural
complexity of the underlying graph, which is in contrast to most
inference methods we have seen so far, where polynomial-time
solutions were obtainable only for graphs with low tree widths.
For simplicity, we will assume that the evidence variables have
been instantiated and absorbed into the node potentials,
and allow only pairwise cliques. Thus, if the MRF has graph structure
$\mathcal{G} = (V, E)$, our energies are of the form:
\begin{align}
\mathcal{E} (\mathcal{X}) = \sum_{i \in V} \mathcal{E}_i (X_i) +
\sum_{(i,j) \in E} \mathcal{E}_{ij} (X_i,X_j)
\end{align}
In addition, we will assume that the energy function is
\emph{submodular}, which means:
\begin{align}
\mathcal{E}_{ij} (0,0) + \mathcal{E}_{ij} (1,1) \le
\mathcal{E}_{ij}(1,0) + \mathcal{E}_{ij}(0,1) \qquad \forall (i,j)
\end{align}
\paragraph{Review of min-cut.} Before we proceed, here is a quick review of the min-cut problem.
The $s$-$t$ min-cut problem is defined for networks, which are directed graphs with a vertex set $V,$ plus two special nodes
called the source $s$, and the sink $t$. We have a set of directed edges $E$ over $V \cup \{s,t\}$, where each edge is associated
with a non-negative cost $cost(v_i,v_j)$. A \emph{cut} is a disjoint partition of $V \cup \{s,t\}$ into $V_{s} \cup V_{t}$ such that $s \in V_s$,
and $t \in V_t$. The cost of the cut is given by the sum of the cost on the edges originating in $V_s$ and ending in $V_t$.
\begin{align}
cost(V_s,V_t) = \sum_{v_i \in V_s, v_j \in V_t} cost(v_i, v_j)
\end{align}
The \emph{min-cut} is the partition $V_s, V_t$ that achieves the minimum cost. These edges $(v_i,v_j)$ which originate in $V_s$
and end in $V_t$ are called cut-edges. Polynomial time algorithms for finding this min-cut
exist which are based on max-flow computations.
So how do we reduce the MPE problem to one of computing min-cut on a
graph? Intuitively, we need to design our graph so that a cut
corresponds to an assignment of $\mathcal{X}$, and the cost of a cut corresponds to the
energy of the corresponding assignment (plus perhaps a constant).
We will construct a directed
graph $\mathcal{G}' = (V', E')$, such every variable $X_i \in
\mathcal{X}$ corresponds to a node $v_i$.
%and every undirected edge
%$(X_i,X_j)$ in the MRF corresponds to two edges $(v_i, v_j)$ and
%$(v_j, v_i)$ in $E'$.
In addition there are two special nodes, a
source $(s)$ and a sink $(t)$. The source connects \emph{to} all
$v_i$ $(\ie, (s,v_i) \in E', \forall i)$, and the sink has an edge
\emph{from} all $v_i$ $(\ie, (v_i,t) \in E', \forall i)$.
\textbf{Construction 1: Unary Energies Only.} Assume for this part that there are no
pairwise energies ($\ie,\, \mathcal{E}_{ij} (X_i, X_j) = 0, \; \forall (X_i, X_j)$). Consider unary energy
$\mathcal{E}_i$ for a variable $X_i$. If $\mathcal{E}_i(0) <
\mathcal{E}_i (1)$, then we put weight $\mathcal{E}_i (1) -
\mathcal{E}_i (0)$ on edge $(s,v_i)$, and weight 0 on edge $(v_i,
t)$. If $\mathcal{E}_i(0) > \mathcal{E}_i (1)$, then we put weight
$\mathcal{E}_i (0) - \mathcal{E}_i (1)$ on edge $(v_i,t)$, and
weight 0 on edge $(s, v_i)$. Otherwise, if $\mathcal{E}_i(0) =
\mathcal{E}_i (1)$, we put weight 0 on both edges.
Figure~\ref{fig:gc_unary} shows this visually (zero weight edges have not been drawn).
%\begin{figure}[t]
%\centering \subfloat[Construction 2]
%{\includegraphics[width=0.5\linewidth]{graph_cuts_unary.pdf}
%\label{fig:gc_unary}} \subfloat[Construction 3]
%{\includegraphics[width=0.25\linewidth]{graph_cuts_pairwise2.pdf}
%\label{fig:gc_pairwise}}
%\end{figure}
\begin{figure}[t]
\centering \subfloat[Construction 1]
{\includegraphics[width=0.5\linewidth]{graph_cuts_unary.pdf}
\label{fig:gc_unary}} \subfloat[Construction 2]
{\includegraphics[width=0.25\linewidth]{graph_cuts_pairwise2.pdf}
\label{fig:gc_pairwise}}
\end{figure}
\begin{enumerate}
\item Describe how a cut in this constructed graph would look like (structurally)? Specifically, can a node be connected to two cut-edges?
Can a node be connected to zero cut-edges?
\item Using the intuition for what a cut looks like, describe how a cut in this graph ($\mathcal{G}'$) would correspond to an
assignment of $\mathcal{X}$.
(\emph{Hint}: construct a mapping that would take a cut as input and
produce an assignment of $\mathcal{X}$. Prove that this mapping is a
bijection.)
\item Prove that the cost of a cut in
the above constructed graph is equal to the energy of the corresponding assignment
(plus possibly a constant that does not depend on the
assignment of variables).
(\emph{Hint}: Use the fact that these energies can be shifted by a constant without affecting the MPE outcome. Thus you can assume
that either $\mathcal{E}_i(0)$ or $\mathcal{E}_i(1)$ is 0, by subtracting the smaller of the two from both.)
\end{enumerate}
\textbf{Construction 2: Pairwise Energies.} Consider a pairwise
energy term $\mathcal{E}_{ij}$ for a pair of variables $(X_i, X_j)$.
We now show how the four parameters describing this pairwise energy (eqn~\ref{eqn:pairwise})
will be incorporated onto weights on edges in $\mathcal{G'}$.
We notice that this pairwise energy can be broken into four terms as
shown in equation ~\ref{eqn:split}. Note that the first term is a constant, the second term is a function of only $X_i$, the third term
is a function of only $X_j$, and the fourth term is a function of both $X_i$ and $X_j$.
\begin{align}
\begin{tabular}{c c |c|c| c |c|c|}
\cline{3-4}\cline{6-7}
\multirow{2}{*}{$\mathcal{E}_{ij}$} & \multirow{2}{*}{=} & $\mathcal{E}_{ij} (0,0)$ & $\mathcal{E}_{ij} (0,1)$ & \multirow{2}{*}{=} & $A$ & $B$ \\
\cline{3-4}\cline{6-7}
& & $\mathcal{E}_{ij} (1,0)$ & $\mathcal{E}_{ij} (1,1)$ & & $C$ & $D$\\
\cline{3-4}\cline{6-7}
\end{tabular}
\label{eqn:pairwise}
\end{align}
\begin{align}
\begin{tabular}{|c|c| c c c |c|c| c |c|c| c |c|c|}
\cline{1-2}\cline{6-7}\cline{9-10}\cline{12-13} $A$ & $B$ &
\multirow{2}{*}{=} & \multirow{2}{*}{$A$} & \multirow{2}{*}{+} & 0 &
0 & \multirow{2}{*}{+} & 0 & $D-C$
& \multirow{2}{*}{+} & 0 & $B+C-A-D$\\
\cline{1-2}\cline{6-7}\cline{9-10}\cline{12-13}
$C$ & $D$ & & & & $C-A$ & $C-A$ & & 0 & $D-C$ & & 0 & 0\\
\cline{1-2}\cline{6-7}\cline{9-10}\cline{12-13}
\end{tabular}
\label{eqn:split}
\end{align}
Corresponding to each of the terms, we will add weights
on edges in the graph.
\begin{itemize}
\item For the fourth term, we set the weight of
edge $(v_i,v_j)$ as $(B+C-A-D)$, and that of $(v_j,v_i)$ to 0.
\item For the second and third terms, we
follow same rules as construction 1 by treating them as unary energies. If $C-A>0$,
we add a weight of $C-A$ on edge $(s,v_i)$, otherwise we add a weight of $A-C$ on edge
$(v_i,t)$. If $D-C>0$,
we add a weight of $D-C$ on edge $(s,v_j)$, otherwise we add a weight of $C-D$ on edge
$(v_j,t)$. For example,
figure~\ref{fig:gc_pairwise} shows the weights added on edges when
$C>A$ and $C>D$.
\item The first term ($A$) is ignored.
\end{itemize}
It should be noted that this contribution of
pairwise energies would be added to any contribution from the unary
energies. Thus in the current example, the final weight on $(s,v_i)$
would be $\mathcal{E}_i(1) - \mathcal{E}_i(0) + C - A$.
Finally, here are your questions:
\begin{enumerate}
\setcounter{enumi}{3}
\item Extend your previous solution to include pairwise energies. Specifically, show that the cost of a cut on the graph
constructed by the above construction is equal to the energy of the
corresponding assignment (plus possibly a constant that does not depend on the
assignment of variables).
\item In construction 2 why did we not add any edges for the first term in equation~\ref{eqn:split} ($\ie$, A)?
\item Using the intuition developed by the past couple of questions, provide a polynomial-time algorithm for performing MPE
inference queries in binary MRFs with submodular energies. (You can use a min-cut algorithm as a black-box).
\item We know from class that the MPE problem in general graphs is NP-complete. Yet you were able to provide a polytime
algorithm in the above part. Why?
\item \textbf{Extra Credit:} Can this method be extended to non-binary MRFs? If each variable $X_i$ were a $K$-ary variable, can we still
construct a graph where a cut corresponds to an assignment of $\mathcal{X}$? Provide either an algorithm for MPE inference in
$K$-ary MRFs, or comment why finding such an algorithm is hard.
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Feedback {\small [0 pts]}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The following are questions that we use to calibrate the exam in future years. Your answers are appreciated.
\begin{enumerate}
\item How many hours did it take to complete the exam ?
\item Which two questions did you find hardest ?
\item Which two questions did you find easiest ?
\end{enumerate}
\end{document}