Solutions for BU7527
Mike Peardon mjp@maths.tcd.ie
Section A
1.
(a) The outcomes (for winnings, not including the stake) are:
Dice 1
Dice 2
1
2
3
4
5
6
1
2
3
4
5
6
120
60
40
30
24
20
240
120
80
60
48
40
360
180
120
90
72
60
480
240
160
120
96
80
600
300
200
150
120
100
720
360
240
180
144
120
i. 12 of the 36 outcomes result in winnings greater than e175, so the probability of
being in profit is 1/3.
ii. The expected profit can be computed from averaging over the 36 outcomes. E[W ] =
−e3.5 (so on average, the player loses).
(b) If event A is defined as “P makes a profit” and event B is “Q makes a profit” then we
have P (A|B) = 0.8, P (A|B c ) = 0.08 and P (B) = 0.7 (so P (B c ) = 0.3). Then
P (A) = P (A|B)P (B) + P (A|B c )P (B c ) = 0.8 × 0.7 + 0.08 × 0.3 = 58.4%.
2.
(a) With fX (x) =
1
(x
32
+ 1)(x + 2)(3 − x), the cdf is given by
FX (x) =
Z x
−1
fX (x̄)dx̄ =
and so P (X > 0) = 1 − FX (x) = 1 −
11
128
3
7
1 4
y + y2 −
y
16
64
128
=
117
128
(b)
E[X] =
Z 3
xf (x) dx =
−1
and
Z 3
E[X 2 ] =
x2 f (x) dx =
−1
So the variance is E[X 2 ] − E[X]2 =
3.
7
3
−
19
15
19
15
2
=
7
3
164
225
(a) The binomial experiment is a series of random trials, with the following properties;
•
•
•
•
(b)
Each trial asks a question with a true-false answer.
All trials are independent
The number of trials is fixed a priori
The probability of a true answer is fixed (usually labelled p).
i. P0 =15 C0 p0 q 1 5 ≈ 3.5%
ii. P (k > 3) = 1 − P0 − P1 − P2 − P3 ≈ 35.2%
1
(c) Since the Poisson process describes memoryless events, the result is just the probability there are m events in an interval of length T , ie.
P (N = n + m in 2T |N = n in T ) =
4.
(µT )m e−µT
m!
(a) For X to be a martingale need E[Xk ] = E[Xk−1 ]. Now
26p 1
1−p
E[Xk ] = 3pE[Xk−1 ] +
E[Xk−1 ] =
+
E[Xk−1 ]
9
9
9
so that for a martingale,
4
26p 1
+ =1→p= .
9
9
13
(b) Possible values of X in this game are 1,3,9,27,81,243,729. The game stops when the
player has e1 or e729. The optional stopping theorem for stopping time T gives
E[XT ] = 81 = 729P (XT = 729) + 1 × P (XT = 1)
which gives P (XT = 1) =
81
91
= 89%.
Section B
1.
(a) The central limit theorem enables the probablity a sample mean lies in some range close
to the true mean to be computed reliably, regardless of the underlying distribution that
is being sampled provided the mean and variance of the underlying distribution are
known. Reliable predictions regarding the deviation from the true mean can then be
made and confidence intervals can be computed. It is usually necessary to estimate
the variance of the underlying distribution by measuring the sample variance; under
these circumstances, the Student t-distribution is the required one, but this converges
towards the normal distribution for fairly modest sample sizes and by sample sizes of
about 50, the distributions are indistinguishable.
(b) The Student t-distribution is needed when the sample is used both to estimate the
expected value of the underlying distribution and also its variance. For this example,
we have
1 2
X̄ − X̄ 2
s2 =
n−1
From the data, X̄ = 13.94, X̄ 2 = 195.0 and so s = 0.412. Using the information
provided in the question, the width of the 95% confidence interval is then w = 2.78s =
1.14. A 95% confidence interval is then [12.8, 15.1].
(c) The χ2 -statistic is used to estimate how likely a model is true given some sampled
data. If a model is compatible with (ie not ruled out with high probability by) some
data, the χ2 statistic should be of similar size to the number of degrees of freedom
(the number of experiments - the number of parameters determined from the data).
2
k
1
2
3
4
x
0.2
0.5
1.3
2.6
Y
1.51
2.22
3.61
6.18
σ
0.1
0.1
0.03
0.04
Model
(M-Y)^2/sigma
1.4
1.21
2
4.84
3.6
0.111111111
6.2
0.25
chi^2= 6.411111111
chi^2/N_df= 1.6027778
The resulting value, χ2 = 6.41 is large and makes the model unlikely. More data
needed to be sure!
2.
(a) X and Y are not independent, since we can not factorise the joint distribution function
into a product of a function of X alone and one of Y alone. The marginal distribution
for X is
Z 1
10 6
− x
fX (x) =
fXY (x, y) dy =
7
7
0
(b)
E[X] =
Z 1
0
E[XY ] =
Z 1
3
3
x̄fX (x̄)dx̄ = , E[Y ] =
7
7
dx̄
Z 1
0
0
dȳ x̄ȳfXY (x̄, ȳ) =
so the covariance, E[XY ] − E[X]E[Y ] =
5
28
− 37 ×
3
7
5
28
1
= − 196
(c) Theorem of conditional probability gives
P (X < 1/2|Y < 1/2) =
P (X < 1/2 ∩ Y < 1/2)
P (Y < 1/2)
and with
P (X < 1/2 ∩ Y < 1/2) =
1
2
Z
dy
0
and
P (Y < 1/2) =
1
2
Z
dy
0
1
2
Z
0
Z 1
dx
0
dx
6
5
2 − x − y2 =
7
14
6
17
2 − x − y2 =
7
28
we get
P (X < 1/2|Y < 1/2) =
10
17
(d)
P (X > Y ) =
Z 1
dx
0
3.
Z x
0
dy
6
1
2 − x − y2 =
7
2
(a) A Markov process is a series of stochastic transitions, where the probability of a system
moving from one state to another depends only on those two states and not on the
history or future of the system. This process can be described this way as the definition
of the transition probabilities given in the question obeys this relation.
3
(b) In steady state, the probabilities must obey the balance equation
1
1
πk−1 × = πk × 1 −
k
k+1
and the result follows.
(c) Following from this, a recursion can be built
πk = α
(k + 1)!
k+1
π0 = α
π0
k! k!
k!
with α a constant (that will normalise the probabilities). Finding α involves solving
P
for the constraint k πk = 1, which gives α = 1/2e and the steady state follows.
The state with the highest probability is found by examining the first few steps in the
recursion, which are
4
3
π1 = 2π0 , π2 = π1 , π3 = π2 . . .
2
9
and we see π2 is the biggest probability as later terms then fall off monotonically.
4