Exam 2ST065 20151021
43/15
Patrik Andersson
EXAM Probability theory and statistical inference I , 2ST065 (7.5 hp).
Wednesday 21/10 2015, 8.00 - 12.00.
Examiner: Patrik Andersson.
Allowed tools.
– Formulae for the course Probability Theory and Statistical Inference
– Math Handout (by Lars Forsberg)
– Pocket calculator.
– Physical dictionary (or word-list).
Notes in the permitted aids are not allowed.
If you feel that something in the wording of the problem is unclear, write under what
assumptions you are solving the problem.
After turning in your test, you may keep the test-pages with the question-statements.
1. A chess game can be won, drawn or lost and a player plays with the white or black
pieces. A player plays with the white pieces in 50% of the games. If he plays against
a higher rated player he wins 60% of the games and games that was won against
a higher rated player was played with the black pieces in 30% of the games. That
the player plays with the white pieces is independent of the opponent having higher
rating.
What is the probability that a game is won when playing with the black pieces
against a higher rated player?
2. The Pareto distribution has density function
fX (x) =
↵x↵m
, x > xm ,
x↵+1
where xm > 0 and ↵ > 0 are parameters. Let us assume that we know that xm = 1.
The following observations on X are made
Obs.
X
1
1.7908
2
7.4022
3
1.0706
4
1.0386
5
1.6070
Find the maximum likelihood estimate of P (X > 10).
3. Consider the the bivariate density fX,Y (x, y) = 2x(x + y 2 ), 0 < x < 1, 0 < y < 1.
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Exam 2ST065 20151021
43/15
Patrik Andersson
(a) Calculate P (X  0.5|Y > 0.5).
(b) Calculate P (X  0.5|Y = 0.5).
4. To evaluate the efficacy of a new drug it was given to 100 patients and the e↵ect
was measured. For comparison, an already existing drug was given to 50 patients.
The result of the study is given below, lower values are better.
Drug
New
Old
Sample size
100
50
Mean
9.8705
10.9407
Std. dev.
1.8840
2.0396
(a) Make a 95% confidence interval for the di↵erence in e↵ect between the new
and old drug. State clearly what assumptions are necessary.
(b) We wish to test if the new drug is better than the old. State appropriate
hypotheses and calculate the p-value.
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Exam 2ST065 20151021
44/15
Patrik Andersson
Solutions Probability theory and statistical inference I , 2ST065 (7.5 hp).
Wednesday 21/10 2015
Examiner: Patrik Andersson.
1. Set the notation A =white pieces, B =win, C =higher rated player. The information
given is P (A) = 0.5, P (B|C) = 0.6, P (Ā|B \ C) = 0.3.
We calculate
P (Ā \ B \ C)
P (Ā|B \ C)P (B|C)P (C)
P (Ā|B \ C)P (B|C)
=
=
1 P (A)
P (Ā \ C)
P (Ā)P (C)
0.3 · 0.6
=
= 0.36.
0.5
P (B|Ā \ C) =
2. We first calculate,
P (X > 10) =
Z
1
10
↵
x↵+1
dx = . . . =
✓
1
10
◆↵
.
Since the quantity we wish to estimate is a function of ↵, we proceed to find the
MLE of ↵. Denoting our observations x1 , . . . , xb , the likelihood function is,
n
Y
↵
↵+1 ,
x
i=1 i
L(↵) =
and the log-likelihood is thus,
n
X
l(↵) = n log(↵)
(↵ + 1) log(xi ).
i=1
The MLE is the solution to
l0 (↵) =
n
↵
That is
n
X
log(xi ) = 0.
i=1
n
.
i=1 log(xi )
↵
ˆ = Pn
Plugging in our observations we get ↵
ˆ = 1.5798. Using the invariance property of
the MLE, the estimate of the above probability is
✓ ◆↵ˆ
1
\
P (X > 10) =
= 0.0263.
10
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Exam 2ST065 20151021
44/15
Patrik Andersson
3. (a) By the definition of conditional probabilities,
P (X  0.5|Y > 0.5) =
P (X  0.5, Y > 0.5)
.
P (Y > 0.5)
We find the probabilites by integrating,
Z 0.5 Z 1
Z 0.5 Z 1
P (X  0.5, Y > 0.5) =
fX,Y (x, y)dydx =
2x(x + y 2 )dydx
0
0.5
0
0.5
◆
Z 0.5 ✓
Z 0.5
Z 0.5 
3 1
3
y
x 1 0.5
7x
=
2x xy +
dx =
2x
+
dx =
(x2 + )dx
3 0.5
2
3
12
0
0
0
3
2
0.5
7 · 0.5
11
=
+
= ,
3
24Z Z 96
Z 1
1
1
7x
1
7
5
P (Y > 0.5) =
fX,Y (x, y)dydx =
(x2 + )dx = +
= .
12
3 24
8
0
0.5
0
Thus,
P (X  0.5|Y > 0.5) =
11/96
11
=
5/8
60
(b) Here, we begin by finding the conditional density.
fX|Y (x, 0.5) =
fX,Y (x, y)
= Cx(x + 0.25).
fY (0.5)
We find the constant C by,
Z 1
1 0.25
11
(x2 + 0.25x)dx = +
= ,
3
2
24
0
so that C = 24/11 and
fX|Y (x, 0.5) =
So,
Z
11 2
(x + 0.25x).
24
Z
24 0.5 2
P (X  0.5|Y = 0.5) =
fX|Y (x, 0.5)dx =
(x + 0.25x)dx
11 0
0
✓ 3
◆
✓
◆
24 0.5
0.25 · 0.52
24 1
1
1
24
7
=
+
=
+
=
+
= .
11
3
2
11 24 32
11 352
44
0.5
4. (a) We have two large samples so the two-sided confidence intervall is
s
2
ˆnew
ˆ2
µ̂new µ̂old ± z0.025
+ old
nnew nold
r
1.88402 2.03962
= 9.8705 10.9407 ± 1.96
+
= [ 1.7455, 0.3949]
100
50
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Exam 2ST065 20151021
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Patrik Andersson
(b) We will test H0 : µnew µold = 0 against Ha : µnew
a large sample so the test-statistic is
µ̂new
Z=q
2
ˆnew
nnew
µ̂old
+
2
ˆold
nold
9.8705
=q
1.88402
100
10.9407
+
2.03962
50
µold < 0. Again, this is
=
3.11.
Looking in the table for the standard normal distribution we find the p-value
to be 1 0.9991 = 9 · 10 4 .
3