Description
The purpose of this assignment is to give you experience with Bayesian decision theory. The first
four problems are of the pen-and-paper type, while problem 5 is a computer problem. Some of the
problems simply require the straightforward application of the principles covered in class. These are
intended to give you some practice on the Bayesian manipulations that are usually involved in decision
theory. Others involve somewhat deeper thinking and extend in some way what we have seen in class.
The problems are not ordered by difficulty.
1. In this problem we will consider the traditional probability scenario of coin tossing. However, we
will consider two variations. First, the coin is not fair. Denoting by S the outcome of the coin toss we
have
PS (heads) = α, α ∈ [0, 1].
Second, you do not observe the coin directly but have to rely on a friend that reports the outcome of
the toss. Unfortunately your friend is unreliable, he will sometimes report heads when the outcome was
tails and vice-versa. Denoting the report by R we have
PR|S(tails|heads) = θ1 (1)
PR|S(heads|tails) = θ2 (2)
where θ1, θ2 ∈ [0, 1]. Your job is to, given the report from your friend, guess the outcome of the toss.
a) Given that your friend reports heads, what is the optimal decision function in the minimum probability of error sense. That is, when should you guess heads, and when should you guess tails?
b) Consider the case θ1 = θ2. Can you give an intuitive interpretation to the rule derived in a)?
c) You figured out that if you ask your friend to report the outcome of the toss various times, he will
produce reports that are statistically independent. You then decide to ask him to report the outcome
n times, in the hope that this will reduce the uncertainty. (Note: there is still only one coin toss, but
the outcome gets reported n times). What is the new minimum probability of error decision rule?
d) Consider the case θ1 = θ2 and assume that the report sequence is all heads. Can you give an intuitive
interpretation to the rule derived in c)?
2. Problem 2.2.2 in DHS.
3. Problem 2.5.23 in DHS (feel free to use MATLAB here to compute eigenvalues, matrix vector
products, and so forth. The goal is not evaluate your mastery of the mechanics of linear algebra, but
to give you an intuitive understanding of what whitening is).
4. Problem 2.9.43 in DHS
1 ECE 271A
5. (computer) This is the first in a series of computer problems where we will get a feel for the
difficulty of practical pattern recognition problems, such as computer vision. The goal is to segment
the “cheetah” image (shown below in the left) into its two components, cheetah (foreground) and grass
(background).
To formulate this as a pattern recognition problem, we need to decide on an observation space. Here
we will be using the space of 8 × 8 image blocks, i.e. we view each image as a collection of 8 × 8 blocks.
For each block we compute the discrete cosine transform (function dct2 on MATLAB) and obtain
an array of 8 × 8 frequency coefficients. We do this because the cheetah and the grass have different
textures, with different frequency decompositions and the two classes should be better separated in the
frequency domain. We then convert each 8 × 8 array into a 64 dimensional vector because it is easier
to work with vectors than with arrays. The file Zig-Zag Pattern.txt contains the position (in the 1D
vector) of each coefficient in the 8 × 8 array. The file TrainingSamplesDCT 8.mat contains a training
set of vectors obtained from a similar image (stored as a matrix, each row is a training vector) for each
of the classes. There are two matrices, TrainsampleDCT BG and TrainsampleDCT FG for foreground
and background samples respectively.
To make the task of estimating the class conditional densities easier, we are going to reduce each
vector to a scalar. For this, for each vector, we compute the index (position within the vector) of the
coefficient that has the 2nd largest energy value (absolute value). This is our observation or feature X.
(The reason we do not use the coefficient with the largest energy is that it is always the so-called “DC”
coefficient, which contains the mean of the block). By building an histogram of these indexes we obtain
the class-conditionals for the two classes PX|Y (x|cheetah) and PX|Y (x|grass). The priors PY (cheetah)
and PY (grass) should also be estimated from the training set.
a) using the training data in TrainingSamplesDCT 8.mat, what are reasonable estimates for the prior
probabilities?
b) using the training data in TrainingSamplesDCT 8.mat, compute and plot the index histograms
PX|Y (x|cheetah) and PX|Y (x|grass).
c) for each block in the image cheetah.bmp, compute the feature X (index of the DCT coefficient with
2nd greatest energy). Compute the state variable Y using the minimum probability of error rule based
on the probabilities obtained in a) and b). Store the state in an array A. Using the commands imagesc
and colormap(gray(255)) create a picture of that array.
d) The array A contains a mask that indicates which blocks contain grass and which contain the
cheetah. Compare it with the ground truth provided in image cheetah mask.bmp (shown below on the
right) and compute the probability of error of your algorithm.
Notes:
• in TrainingSamplesDCT 8.mat each matrix row contains the “zig-zag scanned” vector of coefficients. So, for a) and b) you do not need to compute the DCT, etc.
• the training data that we provide was created by representing images as doubles in the range [0, 1].
You need to represent the image in the same way, in order to have consistency between the train
and test sets. This can be done by using the commands im2double(img) or double(img)/255
on MATLAB (where img is the array that contains the image as read from cheetah.bmp).
• in Zig-Zag Pattern.txt the zig-zag pattern goes from 0 to 63, not 1 to 64. Please remember
2 ECE 271A
this as MATLAB starts indexes from 1. (You can just add 1 to the numbers in the file to get to
the MATLAB coordinates).
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50 100 150 2Homework Set One
ECE 271A
Department of Computer and Electrical Engineering
University of California, San Diego
Nuno Vasconcelos
The purpose of this assignment is to give you experience with Bayesian decision theory. The first
four problems are of the pen-and-paper type, while problem 5 is a computer problem. Some of the
problems simply require the straightforward application of the principles covered in class. These are
intended to give you some practice on the Bayesian manipulations that are usually involved in decision
theory. Others involve somewhat deeper thinking and extend in some way what we have seen in class.
The problems are not ordered by difficulty.
1. In this problem we will consider the traditional probability scenario of coin tossing. However, we
will consider two variations. First, the coin is not fair. Denoting by S the outcome of the coin toss we
have
PS (heads) = α, α ∈ [0, 1].
Second, you do not observe the coin directly but have to rely on a friend that reports the outcome of
the toss. Unfortunately your friend is unreliable, he will sometimes report heads when the outcome was
tails and vice-versa. Denoting the report by R we have
PR|S(tails|heads) = θ1 (1)
PR|S(heads|tails) = θ2 (2)
where θ1, θ2 ∈ [0, 1]. Your job is to, given the report from your friend, guess the outcome of the toss.
a) Given that your friend reports heads, what is the optimal decision function in the minimum probability of error sense. That is, when should you guess heads, and when should you guess tails?
b) Consider the case θ1 = θ2. Can you give an intuitive interpretation to the rule derived in a)?
c) You figured out that if you ask your friend to report the outcome of the toss various times, he will
produce reports that are statistically independent. You then decide to ask him to report the outcome
n times, in the hope that this will reduce the uncertainty. (Note: there is still only one coin toss, but
the outcome gets reported n times). What is the new minimum probability of error decision rule?
d) Consider the case θ1 = θ2 and assume that the report sequence is all heads. Can you give an intuitive
interpretation to the rule derived in c)?
2. Problem 2.2.2 in DHS.
3. Problem 2.5.23 in DHS (feel free to use MATLAB here to compute eigenvalues, matrix vector
products, and so forth. The goal is not evaluate your mastery of the mechanics of linear algebra, but
to give you an intuitive understanding of what whitening is).
4. Problem 2.9.43 in DHS
1
5. (computer) This is the first in a series of computer problems where we will get a feel for the
difficulty of practical pattern recognition problems, such as computer vision. The goal is to segment
the “cheetah” image (shown below in the left) into its two components, cheetah (foreground) and grass
(background).
To formulate this as a pattern recognition problem, we need to decide on an observation space. Here
we will be using the space of 8 × 8 image blocks, i.e. we view each image as a collection of 8 × 8 blocks.
For each block we compute the discrete cosine transform (function dct2 on MATLAB) and obtain
an array of 8 × 8 frequency coefficients. We do this because the cheetah and the grass have different
textures, with different frequency decompositions and the two classes should be better separated in the
frequency domain. We then convert each 8 × 8 array into a 64 dimensional vector because it is easier
to work with vectors than with arrays. The file Zig-Zag Pattern.txt contains the position (in the 1D
vector) of each coefficient in the 8 × 8 array. The file TrainingSamplesDCT 8.mat contains a training
set of vectors obtained from a similar image (stored as a matrix, each row is a training vector) for each
of the classes. There are two matrices, TrainsampleDCT BG and TrainsampleDCT FG for foreground
and background samples respectively.
To make the task of estimating the class conditional densities easier, we are going to reduce each
vector to a scalar. For this, for each vector, we compute the index (position within the vector) of the
coefficient that has the 2nd largest energy value (absolute value). This is our observation or feature X.
(The reason we do not use the coefficient with the largest energy is that it is always the so-called “DC”
coefficient, which contains the mean of the block). By building an histogram of these indexes we obtain
the class-conditionals for the two classes PX|Y (x|cheetah) and PX|Y (x|grass). The priors PY (cheetah)
and PY (grass) should also be estimated from the training set.
a) using the training data in TrainingSamplesDCT 8.mat, what are reasonable estimates for the prior
probabilities?
b) using the training data in TrainingSamplesDCT 8.mat, compute and plot the index histograms
PX|Y (x|cheetah) and PX|Y (x|grass).
c) for each block in the image cheetah.bmp, compute the feature X (index of the DCT coefficient with
2nd greatest energy). Compute the state variable Y using the minimum probability of error rule based
on the probabilities obtained in a) and b). Store the state in an array A. Using the commands imagesc
and colormap(gray(255)) create a picture of that array.
d) The array A contains a mask that indicates which blocks contain grass and which contain the
cheetah. Compare it with the ground truth provided in image cheetah mask.bmp (shown below on the
right) and compute the probability of error of your algorithm.
Notes:
• in TrainingSamplesDCT 8.mat each matrix row contains the “zig-zag scanned” vector of coefficients. So, for a) and b) you do not need to compute the DCT, etc.
• the training data that we provide was created by representing images as doubles in the range [0, 1].
You need to represent the image in the same way, in order to have consistency between the train
and test sets. This can be done by using the commands im2double(img) or double(img)/255
on MATLAB (where img is the array that contains the image as read from cheetah.bmp).
• in Zig-Zag Pattern.txt the zig-zag pattern goes from 0 to 63, not 1 to 64. Please remember
2
this as MATLAB starts indexes from 1. (You can just add 1 to the numbers in the file to get to
the MATLAB coordinates).
50 100 150 200 250
50
100
150
200
250
50 100 150 200 250
50
100
150
200
250
300 250
50
100
150
200
250
3 ECE 271A
