CSE 250A. Assignment 1

CSE 250A. Assignment 1

Reading: Russell & Norvig, Chapter 13.
1.1 Conditioning on background evidence [RN 13.16]
It is often useful to consider the impact of specific events in the context of general background evidence,
rather than in the absence of information.
(a) Denoting such evidence by E, prove the conditionalized version of the product rule:
P(X, Y |E) = P(X|Y, E)P(Y |E).
(b) Also, prove the conditionalized version of Bayes rule:
P(X|Y, E) = P(Y |X, E)P(X|E)
P(Y |E)
.
1.2 Conditional independence [RN 13.17]
Show that the following three statements about random variables X, Y , and E are equivalent:
P(X, Y |E) = P(X|E)P(Y |E)
P(X|Y, E) = P(X|E)
P(Y |X, E) = P(Y |E)
You should become fluent with all these ways of expressing that X is conditionally independent of Y
given E.
1.3 Creative writing
Attach events to the binary random variables X, Y , and Z that are consistent with the following patterns of
commonsense reasoning. You may use different events for the different parts of the problem.
(a) Cumulative evidence:
P(X = 1) < P(X = 1|Y = 1) < P(X = 1|Y = 1, Z = 1)
(b) Explaining away:
P(X = 1|Y = 1) P(X = 1),
P(X = 1|Y = 1, Z = 1) < P(X = 1|Y = 1)
(c) Conditional independence:
P(X = 1, Y = 1) 6= P(X = 1)P(Y = 1)
P(X = 1, Y = 1|Z = 1) = P(X = 1|Z = 1)P(Y = 1|Z = 1)
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1.4 Bayes Rule
Suppose that 1% of competitive cyclists use performance-enhancing drugs and that a particular drug test has
a 5% false positive rate and a 10% false negative rate.
(a) Cyclist A tests negative for drug use. What is the probability that Cyclist A is not using drugs?
(b) Cyclist B tests positive for drug use. What is the probability that Cyclist B is using drugs?
1.5 Entropy
(a) Let X be a discrete random variable with P(X = xi) = pi for i ∈ {1, 2, . . . , n}. The entropy H[X]
of the random variable X is a measure of its uncertainty. It is defined as:
H[X] = −
Xn
i=1
pi
log pi
.
Show that the entropy H[X] is maximized when pi =
1
n
for all i. You should do this by computing the
gradient with respect to pi and using Lagrange multipliers to enforce the constraint that P
i
pi = 1.
Later in the course, we will use similar calculations for learning probabilistic models.
(b) The joint entropy of n discrete random variables (X1, X2, . . . , Xn) is defined as:
H(X1, X2, . . . , Xn) = −
X
x1
X
x2
. . .X
xn
P(x1, x2, . . . , xn) log P(x1, x2, . . . , xn),
where the sums range over all possible instantiations of X1, X2, . . . , Xn. Show that if the variables Xi
are independent, then their joint entropy is the sum of their individual entropies: namely,
P(x1, x2, . . . , xn) = Yn
i=1
P(xi) implies H(X1, X2, . . . , Xn) = Xn
i=1
H(Xi).
1.6 Kullback-Leibler distance
Consider two discrete probability distributions, pi and qi
, with P pi =
P qi = 1. The Kullback-Leibler (KL)
distance between these distributions (also known as the relative entropy) is defined as:
KL(p, q) = X
i
pi
log(pi/qi).
(a) Consider the natural logarithm (in base e). By sketching graphs of log(x) and x − 1, verify the
inequality:
log(x) ≤ x − 1,
with equality if and only if x= 1. Confirm this result by differentiation of log(x)−(x − 1).
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(b) Use the previous result to prove that KL(p, q) ≥ 0, with equality if and only if the two distributions
pi and qi are equal.
(c) Using the inequality in (a), as well as the simple equality log x = 2 log √
x, derive the tighter lower
bound:
KL(p, q) ≥
X
i
(
√
pi −
√
qi)
2
.
(d) Provide a counterexample to show that the KL distance is not a symmetric function of its arguments:
KL(p, q) 6= KL(q, p).
Despite this asymmetry, it is still common to refer to KL(p, q) as a measure of distance. Many algorithms for machine learning are based on minimizing KL distances between probability distributions.
1.7 Hangman
Consider the belief network shown below, where the random variable W stores a five-letter word and the
random variable Li ∈ {A, B, . . . , Z} reveals only the word’s ith letter. Also, suppose that these five-letter
words are chosen at random from a large corpus of text according to their frequency:
P(W =w) = COUNT(w)
P
w0 COUNT(w0)
,
where COUNT(w) denotes the number of times that w appears in the corpus and where the denominator is
a sum over all five-letter words. Note that in this model the conditional probability tables for the random
variables Li are particularly simple:
P(Li =`|W =w) = (
1 if ` is the ith letter of w,
0 otherwise.
Now imagine a game in which you are asked to guess the word w one letter at a time. The rules of this game
are as follows: after each letter (A through Z) that you guess, you’ll be told whether the letter appears in
the word and also where it appears. Given the evidence that you have at any stage in this game, the critical
question is what letter to guess next.
L1 L2
W
L3 L4 L5
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Let’s work an example. Suppose that after three guesses—the letters D, I, M—you’ve learned that the letter I does not appear, and that the letters D and M appear as follows:
M D M
Now consider your next guess: call it `. In this game the best guess is the letter ` that maximizes
P
?
L2 =` or L4 =`



L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}
?
.
In other works, pick the letter ` that is most likely to appear in the blank (unguessed) spaces of the word.
For any letter ` we can compute this probability as follows:
P
?
L2 =` or L4 =`



L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}
?
=
X
w
P
?
W =w, L2 =` or L4 =`



L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}
?
, marginalization
=
X
w
P(W =w|L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}
?
P(L2 =` or L4 =`|W =w) product rule & CI
where in the third line we have exploited the conditional independence (CI) of the letters Li given the
word W. Inside this sum there are two terms, and they are both easy to compute. In particular, the second
term is more or less trivial:
P(L2 =` or L4 =`|W =w) = (
1 if ` is the second or fourth letter of w
0 otherwise.
And the first term we obtain from Bayes rule:
P(W =w|L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}
?
=
P(L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}|W =w
?
P(W =w)
P(L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M})
Bayes rule
In the numerator of Bayes rule are two terms; the left term is equal to zero or one (depending on whether
the evidence is compatible with the word w), and the right term is the prior probability P(W = w), as
determined by the empirical word frequencies. The denominator of Bayes rule is given by:
P(L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M})
=
X
w
P(W =w, L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}), marginalization
=
X
w
P(W =w)P(L1 =M, L3 =D, L5 =M, L2 6∈ {D, I, M}, L4 6∈ {D, I, M}|W =w), product rule
where again all the right terms inside the sum are equal to zero or one. Note that the denominator merely
sums the empirical frequencies of words that are compatible with the observed evidence.
Now let’s consider the general problem. Let E denote the evidence at some intermediate round of the
game: in general, some letters will have been guessed correctly and their places revealed in the word, while
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other letters will have been guessed incorrectly and thus revealed to be absent. There are two essential
computations. The first is the posterior probability, obtained from Bayes rule:
P(W =w|E) = P(E|W =w) P(W =w)
P
w0 P(E|W =w0) P(W =w0)
.
The second key computation is the predictive probability, based on the evidence, that the letter ` appears
somewhere in the word:
P
?
Li =` for some i∈ {1, 2, 3, 4, 5}



E
?
=
X
w
P
?
Li =` for some i∈ {1, 2, 3, 4, 5}


W =w
?
P
?
W =w



E
?
.
Note in particular how the first computation feeds into the second. Your assignment in this problem is
implement both of these calculations. You may program in the language of your choice.
(a) Download the file hw1 word counts 05.txt that appears with the homework assignment. The file
contains a list of 5-letter words (including names and proper nouns) and their counts from a large
corpus of Wall Street Journal articles (roughly three million sentences). From the counts in this file
compute the prior probability P(w) = COUNT(w)/COUNTtotal. As a sanity check, print out the
eight most frequent 5-letter words, as well as the eight least frequent 5-letter words. Do your
results make sense?
(b) Consider the following stages of the game. For each of the following, indicate the best next guess—
namely, the letter ` that is most likely (probable) to be among the missing letters. Also report the
probability P(Li = ` for some i ∈ {1, 2, 3, 4, 5}|E) for your guess `. Your answers should fill in the
last two columns of this table.
correctly guessed incorrectly guessed best next guess ` P(Li =` for some i∈ {1, 2, 3, 4, 5}|E)
– – – – – {}
– – – – – {E,O}
D – – I – {}
D – – I – {A}
– U – – – {A,I,E,O,S}
(c) Turn in a hard-copy printout of your source code. Do not forget the source code: it is worth many
points on this assignment.
Checking your work: The demo on Piazza (also under resources) implements this program for words
of length 6-10. You will also find count files for words of these lengths on Piazza, and if you modify
your code to handle these different word lengths, you will be able to check your answers against the
demo. (This is totally optional, though.) Just to be perfectly clear, you are not required in this problem
to implement a user interface or any general functionality for the game of hangman. You will only be
graded on your word lists in (a), the completed table for (b), and your source code in (c).
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