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Stat 141/Bioeng 141
Homework 2 - SOLUTIONS
PROBLEM 1
We want to find the value of N that maximizes the function N e−N L/G . Strange, but true:
If you take the log of the function, the logged function achieves its maximum at the same
point the original function does. This is a neat trick, often used in statistics, that makes the
calculus of this a little easier:
NL
log N e−N L/G = log N −
G
d
1
L
log N e−N L/G =
−
dN
N
G
1
L
0=
−
N
G
G
N̂ =
L
Taking the first derivative and setting it equal to 0 will give you the critical values for
the function. To confirm that this is indeed a maximum, take the second derivative, and
evaluate it at the critical value:
d2
1
log N e−N L/G = − 2
2
dN
N
L2
=− 2
G
This is clearly a negative number, so the second derivative test tells us that this critical
value is a maximum.
PROBLEM 2
To find the fraction covered by at least one read, we used 1 − P(X = 0). To find the
fraction covered by at least three reads, we extend this to 1 − P(X = 0) − P(X = 1) − P(X =
2). The solution can be worked on a hand calculator, or run quickly in R using
ppois(2, 4.6, lower.tail = F)
(4.6)0 e−4.6 (4.6)1 e−4.6 (4.6)2 e−4.6
−
−
0!
1!
2!
= 0.8373
1 − P(X = 0) − P(X = 1) − P(X = 2) = 1 −
PROBLEM 3
1
P(at least one overlap) = 1 − P(no overlaps)
= 1 − (P(one configuration is not an overlap))2
= 1 − (P(at least one letter doesn’t match))2
= 1 − (1 − P(all three letters match))2
3 !2
1
=1− 1−
4
3 6 !
1
1
=1− 1−2
+
4
4
3 6
1
1
=2
−
4
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PROBLEM 4
First, let’s deal with the two hints:
What is the fraction of the original long sequence covered by at least one read?
(6.9)0 e−6.9
≈ 0.999
0!
What is the mean number of contigs?
1−
N e−6.9 = 0.001N
From this, we can sort out that the number of letters covered is about 0.999G, and the
mean number of contigs is 0.001N . Doing a bit of arithmetic, we find:
total length covered
mean number of contigs
0.999G
=
0.001N
G L
= 999 ×
N
L
999
=
L
6.9
= 144.8L
mean size of contig ≈
PROBLEM 5
A finite, aperiodic, irreducible Markov chain has transition probability matrix P and
stationary distribution ψ. Let k be some constant, 0 < k < 1. We wish to show that a chain
with transition probability P F = kP + (1 − k)I also has stationary distribution ψ.
2




p11 . . . p1n
1 ... 0

..  + (1 − k)  .. . . .. 
P F = k  ... . . .
 .
. . 
. 
pn1 . . . pnn
0 ... 1


kp11 + 1 − k . . .
kp1n


..
..
...
PF = 

.
.
kpn1
. . . kpnn + 1 − k
Let’s see what happens when we pre-multiply this matrix by ψ. Without loss of generality,
we can look only at the first entry of the resulting vector:
ψP F (first entry only) = ψ1 × (kp11 + 1 − k) + · · · + ψn × kpn1
= ψ1 kp11 + ψ1 − ψ1 k + · · · + ψn kpn1
= ψ1 − ψ1 k + k (ψ1 p11 + · · · + ψn pn1 )
= ψ1 − ψ1 k + kψ1
= ψ1
Note that, because ψ is the stationary distribution of the Markov chain with transition
matrix P , ψ1 p11 + · · · + ψn pn1 = ψ1 by definition. The algebra for all other entries will be
similar. We conclude that P F has the same stationary distribution as P .
Over the long run, the proportion of time spent in each state will be about the same for
both P and P F . Note that, in P F , all of the off-diagonal elements are smaller than in P ,
while the main diagonal is larger. You’ll be more likely to see a single state repeat itself
several times in P F than in P .
PROBLEM 6
Although there are 27 possible arrangements of the three states, the initial probability
vector and transition probability matrix both contain several zeroes. In the end, only two
hidden state sequences are actually possible, greatly reducing the scope of this problem.
(a)
Hidden states
S1 S2 S1
S1 S3 S2
Probability
1 × .5 × .5 × 0 × 1 × 0 = 0
1 × .5 × .5 × .5 × 1 × .5 = 0.0625
Here, P(O|λ) = 0.0625.
(b)
Hidden states
S1 S2 S1
S1 S3 S2
Probability
1 × .5 × .5 × .5 × 1 × .5 = 0.0625
1 × .5 × .5 × .5 × 1 × .5 = 0.0625
Here, P(O|λ) = 0.125.
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PROBLEM 7
Consider an HMM with the following parameters:
1. Three states: A, B, C
2. The alphabet is {1, 2, 3}
3. Transition probability matrix


0.5 0 0.5
P =  0 0.5 0.5 
0.5 0.5 0
4. Initial probability vector π = [1/3 1/3 1/3]
5. Emission probability matrix
eA (1) = .98 eA (2) = .01 eA (3) = .01
eB (1) = .01 eB (2) = .98 eB (3) = .01
eC (1) = .01 eC (2) = .01 eC (3) = .98
Now consider the observed sequence O = 1, 2. There are nine hidden state sequences, G,
that could give O:
G
P(O|G)
G
P(O|G)
G
P(O|G)
AA .98 × .01 BA .01 × .01 CA .01 × .01
AB .98 × .98 BB .01 × .98 CB .01 × .98
AC .98 × .01 BC .01 × .01 CC .01 × .01
It is clear that the hidden sequence AB maximizes P(O|G). However, looking at the
transition matrix P , P(G = AB) = 0.
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