Summer Exam 2009: Math. and Stat. Modelling
[1]
(a) Let A, B and C be three arbitrary events. Write down the missing term represented
by the ‘x’ in the following:
(i) Only A occurs:
A ∩ B̄ ∩ x
(ii) At least one event occurs:
A∪ x ∪C
(iii) At least two events occur:
(A ∩ B ∩ C̄) ∪ (A ∩ B̄ ∩ C) ∪ (Ā ∩ B ∩ C) ∪ x
(iv) One and only one event occurs:
(v) Not more than two events occur:
(A ∩ B̄ ∩ C̄) ∪ x ∪ (Ā ∩ B̄ ∩ C)
x̄
[2 marks each]
(b) Suppose that X1 and X2 are independent normal random variables and that X1 has
mean 0 and variance 2, i.e., X1 ∼ N (0, 2) and that X2 ∼ N (1, 2). Let Y = 2X1 − 2X2 .
(i) What is the distribution of Y ?
(ii) Calculate P(Y ≤ −1).
(iii) Calculate P([Y ≤ −2] ∪ [Y ≥ 3]).
[3,3,4 marks]
1
Solution [1]
(a)
(i) C̄
[2 marks]
(ii) B
[2 marks]
(iii) A ∩ B ∩ C
[2 marks]
(iv) (Ā ∩ B ∩ C̄)
[2 marks]
(v) A ∩ B ∩ C
[2 marks ]
(b)
(i)
E{Y } = 2E{X1 } − 2E{X2 } = 2 · 0 − 2 · 1 = −2.
var{Y } = 4var{X1 } + 4var{X2 } = 4 · 2 + 4 · 2 = 16
Y ∼ N (−2, 16).
[3 marks]
(ii)
P (Y ≤ −1) = P
−1 − (−2)
Y − (−2)
≤
4
4
= P (Z ≤ 1/4) = .5987.
[3 marks]
(iii)
P ([Y ≤ −2] ∪ [Y ≥ 3]) = P (Y ≤ −2) + P (Y ≥ 3)
−2 − (−2)
Y − (−2)
3 − (−2)
Y − (−2)
≤
+P
≥
=P
4
4
4
4
= P (Z ≤ 0) + P (Z ≥ 5/4)
= 0.5 + (1 − P (Z ≤ 5/4))
= 0.5 + (1 − 0.8944) = 0.5 + 0.1056 = 0.6056.
[4 marks]
2
[2]
Consider a random variable X with cumulative distribution function
F (x; θ) = 1 − e−θx − θxe−θx ,
x, θ > 0.
(i) Show that X has the probability density function
f (x; θ) = θ2 xe−θx ,
x, θ > 0.
(ii) The mean of X is given by E{X} = 2/θ. The median of X is that value for which
F (x; θ) = 1/2. Are the mean and median equal?
The likelihood function for N independent random variables X1 , ..., XN , each with the
pdf f (x; θ), is
(θ) = f (x1 ; θ)f (x2 ; θ) · · · f (xN −1 ; θ)f (xN ; θ).
(iii) Explain the idea of ‘maximum likelihood’ estimation. Why can the logarithm of
the likelihood function be used to obtain a maximum likelihood estimate of θ?
(iv) What is the maximum likelihood estimate of θ?
(v) Compare the maximum likelihood estimate with the method of moments estimate
based on the mean.
(vi) If θ = 1, find
P (X < 1 | X < 2).
[4,4,2,4,2,4 marks]
3
Solution [2]
(i) Given
F (x; θ) = 1 − e−θx − θxe−θx ,
x, θ > 0,
the PDF is found by differentiating:
f (x; θ) = θe−θx + θe−θx θx − θe−θx = θ2 xe−θx ,
x, θ > 0.
[4 marks]
(ii) The mean is x = 2/θ, and for this value of x the distribution function is
2
F (x; θ) = 1 − e−θ·(2/θ) − θ · e−θ·(2/θ) = 1 − e−2 − 2e−2 = 0.594,
θ
so it is not also the median for which we would obtain 0.5.
[4 marks]
(iii) The likelihood function depends on the unknown parameter(s). The idea is to choose
the parameter value(s) which correspond to the maximum of the likelihood function
(what we obtained being viewed as the most likely occurrence). Since log is a monotonic
transform, the max of the log likelihood function will occur at the max of the likelihood
fn.
[2 marks]
(iv) The likelihood function is
=θ
2N
N
xi e−θxi
i=1
So
log = 2N log θ +
N
log xi − θ
i=1
Then
(d/dθ) log = 2N/θ −
N
N
xi .
i=1
xi = 0 ⇒ θ̂ = 2/x̄.
i=1
[4 marks]
(v) The mean is given by E{X} = 2/θ, so the method of moments estimate is got by
equating x̄ = 2/θ, which gives an identical estimate to maximum likelhood. [2 marks]
(vi)
P (X < 1 | X < 2) =
0.264
F (1)
1 − 2e−1
P (X < 1)
=
=
=
= 0.445.
−2
−2
P (X < 2)
F (2)
1 − e − 2e
0.594
[4 marks]
4
[3]
The temperature (degrees Centigrade) of soil treated with salt was observed over a
period of 9 days. The following sample values xt were recorded,
time, t temp. (o C), xt
1
6.21
2
5.69
3
5.82
4
5.14
5
5.20
6
4.85
7
4.79
8
4.22
9
4.24
Suppose this time series is filtered by the impulse response {gk } where

 1/2, k = 0;
gk = 1/4, k = ±1;

0,
otherwise.
to give the output
yt =
∞
gk Xt−k = g−1 xt+1 + g0 xt + g1 xt−1 .
k=−∞
(i) Plot the series xt against time, for t = 2, . . . , 8 only, and comment on its appearance.
(ii) Calculate the values of yt for t = 2, . . . , 8 and plot them against t = 2, . . . , 8 on the
same plot as in (i).
(iii) Plot the difference series yt − xt against t = 2, . . . , 8 in a new plot.
(iv) Comment on your results and the effect of the linear filtering.
(v) If the input to a linear filter is linear, xt = α + βt, what two properties of the
impulse response {gk } ensure that the output is of the form yt = xt ? [The two
properties hold for the impulse response example above].
[4 marks each]
5
Solution [3]
6
(a)
5.5
5
4.5
4
2
3
4
5
6
7
8
0.2
(b)
0.1
0
−0.1
−0.2
−0.3
2
3
4
5
6
7
8
(i) The series xt against time, for t = 2, . . . , 8 only, is shown in fig (a) as the solid line.
It shows a downward trend with a tendency to ”oscillate” as it goes.
[4 marks]
(ii) The values of yt for t = 2, . . . , 8 are
5.8525, 5.6175, 5.3250, 5.0975, 4.9225, 4.6625, 4.3675
They are plotted against t = 2, . . . , 8 in fig (a) as the dashed line. on the same plot
as in (i).
[4 marks]
(iii) The difference series yt − xt is plotted against t = 2, . . . , 8 in fig (b).
[4 marks]
(iv) The filtered result in (a) is a strongly smoothed version of the xt series. The
difference series shows that the filter has removed the oscillatory behaviour.
[4 marks]
(v) If the input to a linear filter is linear, xt = α + βt, then we want
k
gk xt−k = xt ; i.e.,
k
6
gk [α + β(t − k)] = α + βt.
If we assume that the impulse response sequence is symmetric about k = 0, [this
is condition 1] i.e., g−k = gk , the above is satisfied for all α, β and t if
gk = 1.
k
[This is condition 2]
[4 marks]
7
Summer Exam 2008: Math. and Stat. Modelling
[1]
(a) Given that P(B) = 3/10, P(A ∩ B̄) = 6/10 and P(B ∩ Ā) = 1/10, find
(i) P(A ∩ B).
(ii) P(A).
(iii) P(Ā ∩ B̄).
(iv) P(Ā|B̄).
[8 marks]
(b) A random variable X has E{X 2 } = 15 and var{X} = 6.
(i) Find E{X}.
(ii) Find E{2X − 4}.
(iii) Find var{2 − 2X}.
(iv) Suppose X has a geometric distribution with probability of success p. Then X has
a variance (1 − p)/p2 . Find the value of p.
Assuming that the probability of a flood occurring in any one year is p and is
independent of occurrences in previous years, what is the probability that the first
flood occurs in the 3rd year or subsequently?
[12 marks]
1
Solution [1]
(a)
(i) P(A ∩ B) = P(B) − P(B ∩ Ā) = 0.3 − 0.1 = 2/10 = 0.2.
[2 marks]
(ii) P(A) = P(A ∩ B) + P(A ∩ B̄) = 0.2 + 0.6 = 4/5 = 0.8
[2 marks]
(iii) P(Ā ∩ B̄) = P(Ā) − P(Ā ∩ B) = 0.2 − 0.1 = 1/10 = 0.1.
[2 marks]
(iv) P(Ā|B̄) = P(Ā ∩ B̄)/P(B̄) = 0.1/0.7 = 1/7 = 0.1429.
[2 mark ]
(b)
(i) E{X} =
√
[E{X 2 } − var{X}] = 3.
[3 marks]
(ii) E{2X − 4} = 2E{X} − 4 = 6 − 4 = 2.
[2 marks]
(iii) var{2 − 2X} = (−2)2 var{X} = 4 · 6 = 24.
[2 marks]
(iv) We have that (1 − p)/p2 = 6. Rearranging, 6p2 + p − 1 = 0. This has solutions p =
1/3.−1/2. Only the first can be a probability. Hence p = 1/3. If it is known/realised
the mean of the geometric is 1/p = 3 then again p = 1/3.
[2 marks]
If the flood occurs in the 3rd year or subsequently it has not occurred in the first
two years, an event with probability (1 − p)2 = 4/9.
2
[3 marks]
[2]
Let X denote the proportion of valves produced in a single batch at a manufacturing
plant which have a certain characteristic. X has a probability density function (pdf)
given by
f (x; θ) = θx(θ−1) ,
where 0 < x < 1, and θ > 0.
(i) Find the cumulative distribution function F (x) of X.
(ii) Find the mean of X.
(iii) Find the variance of X and show that it is always positive.
(iv) Explain the idea of ‘method of moments’ parameter estimation.
(v) A sample of size twelve, x1 , . . . , x12 , is observed:
0.94, 0.56, 0.52, 0.82, 0.27, 0.44, 0.73, 0.41, 0.93, 0.49, 0.92, 0.75
Calculate a method of moments estimate of θ.
(vi) Use the method of moments estimate of θ to evaluate
P (X > 0.8 | X > 0.6).
[20 marks]
3
Solution [2]
(i)
x
θy
F (x) =
(θ−1)
dy =
0
x
y θ 0
=
xθ , if 0 < x < 1;
.
0,
otherwise,
[2 marks]
(ii)
1
x · θx(θ−1) dx = θ
E{X} =
0
x(θ+1) 1
θ
=
.
1+θ 0
1+θ
[2 marks]
(iii) The variance is given by E{X 2 } − E 2 {X}. Now
1
E{X } =
x2 · θx(θ−1) dx = θ
2
0
x(θ+2) 1
θ
=
.
0
2+θ
2+θ
So
var{X} =
θ 2
θ
−(
)
2+θ
1+θ
[4 marks]
To show that the variance is positive we need to show that
θ
θ 2
>(
)
2+θ
1+θ
i.e., (1 + 2θ + θ2 )θ > θ2 (2 + θ) which is true since on expanding we get θ + 2θ2 + θ3 >
2θ2 + θ3 which follows since θ > 0.
[2 marks]
(iv) The method of moments for parameter estimation equates theoretical and sample
moments so that equations may be solved involving the parameters to be estimated.
[2 marks]
(v) We shall equate E{X} =
θ
1+θ
and x̄ = 0.64833, where x̄ is the mean of the given
sample. Hence θ̂ = x̄/[1 − x̄] = 1.8436.
[4 marks]
(vi)
P (X > 0.8 | X > 0.6) =
1 − F (0.8)
1 − 0.8θ̂
P (X > 0.8)
= 0.5528.
=
=
P (X > 0.6)
1 − F (0.6)
1 − 0.6θ̂
[4 marks]
4
[3]
The temperature (degrees Centigrade) of soil treated with salt was observed over a
period of 9 days. The following sample values were recorded,
time, t temp. (o C), x
1
6.21
2
5.69
3
5.82
4
5.14
5
5.20
6
4.85
7
4.79
8
4.22
9
4.24
(i) Plot the series xt against time, t = 1, . . . , 9 and comment on its appearance.
(ii) Fit a straight line α + βt to the data by the method of medians, clearly showing
your working, and plot the line on the same plot as for (i).
(iii) Calculate the elements yt = xt −(α̂+ β̂t), t = 1, . . . , 9, of the detrended series, where
α̂ and β̂ are your estimates of α and β. Plot the series yt against time, t = 1, . . . , 9.
(iv) Hence calculate the lag-1 sample autocorrelation of the yt using the formula:
ρ̂1 =
1
N −1
N −1
t=1 (yt − ȳ)(yt+1
N
1
2
t=1 (yt − ȳ)
N
− ȳ)
,
where ȳ denotes the average of the y1 , . . . , y9 and N = 9.
(v) Interpret the value of ρ̂1 in terms of your plot of yt .
[20 marks]
5
6.5
6
Solution [3]
5.5
5
4.5
4
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
0.2
0.1
0
−0.1
−0.2
−0.3
(i) The top figure shows the original series which seems to consist of variation about a
decaying trend.
[4 marks]
(ii) We estimate the straight line by medians. The median of 1:9 is 5. Let xl =
median(1 : 4) = 2.5 and xr = median(6 : 9) = 7.5 with yl = median(6.21, 5.69, 5.82, 5.14) =
5.755 and yr = median(4.85, 4.79, 4.22, 4.24) = 4.515.
Then β = (yr − yl)/(xr − xl) = −0.248.
We can get α from α = yr − β ∗ xr = 6.375.
So the fitted line is 6.375 − 0.248t and is shown in the top figure.
[4 marks]
(iii) The residuals yt are
0.083, −0.189, 0.189, −0.243, 0.065, −0.037, 0.151, −0.171, 0.097.
These are plotted in the lower plot.
(iv)
ρ̂1 =
1
N −1
N −1
t=1 (yt − ȳ)(yt+1
N
1
2
t=1 (yt − ȳ)
N
[4 marks]
− ȳ)
=
1
8 × −0.165
1
9 × 0.2041
= −0.9095.
[4 marks]
6
(v) The mean of the residuals is -0.0061, close to zero, and as we see in the lower plot,
the residuals oscillate about zero in a flip/flop way, suggesting a negative correlation,
and quite a pronounced one. The estimate of ρ̂1 = −0.9095 is entirely consistent with
this.
[4 marks]
7
Summer Exam 2007: Math. and Stat. Modelling
[1]
(a) What do we mean by saying A1 , . . . , Ak form a partition of the sample space Ω? Let
A1 , ..., Ak form a partition of a sample space, and B be some event. Use the definition of
conditional probability and the theorem of total probabilities to derive Bayes’ formula
for P(Ai |B).
[6 marks]
(b) Valves manufactured at a certain plant have to pass a visual inspection. The inspections are spread over 10 days. On any given day the inspector is one of A, B and
C. Inspector A inspects on 5 days, inspector B on 3 days and inspector C on 2 days.
It is known that 12% of the valves fail the inspection when the inspector is A, 10%
fail if the inspector is B and 8% if the inspector is C. We observe one of the 10 days.
Suppose D is the event that exactly one of the three valves examined on that day fails
the inspection.
(i) Show that the conditional probability of D given that A was the inspector is approximately 0.279.
[Recall that if an event has a probability p of occurring in each of m independent trials,
the probability of j occurrences is (m!/[j!(m − j)!])pj (1 − p)m−j ,
j = 0, ..., m.]
(ii) Find P(D).
(iii) What is the probability that C was the inspector given that D occurred?
[14 marks]
1
Solution [1]
(a) Ω =
k
i=1
Ai and Ai ∩ Aj = ∅, ∀i = j.
[2 marks]
Given A1 , . . . , Ak form a partition of a sample space, and P(Ai ) > 0 for i = 1, . . . , k.
Let B be any event, with P(B) > 0. The Theorem of Total Probability says that
k
k
[2 marks]
P(B) = j=1 P(B ∩ Aj ) = j=1 P(B|Aj )P(Aj ).
P(Ai |B) =
P(B|Ai )P(Ai )
P(B|Ai )P(Ai )
= k
.
P(B)
j=1 P(B|Aj )P(Aj )
[2 marks]
(b)
(i) Let X be the number of valves that fail on the selected day. Conditional on A
being the inspector, X has the binomial distribution with parameters m = 3 and
p = 0.12. So
3
P(D|A) = P(X = 1|A) =
p(1 − p)2 = 3 · 0.12 · 0.882 ≈ 0.279.
1
[4 marks]
(ii) By the law of total probability
P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)
= P(D|A)
But
3
2
5
+ P(D|B) + P(D|C)
10
10
10
3
P(D|B) = P(X = 1|B) =
p(1 − p)2 = 3 · 0.10 · 0.92 ≈ 0.243
1
3
P(D|C) = P(X = 1|C) =
p(1 − p)2 = 3 · 0.08 · 0.922 ≈ 0.203
1
Thus P(D) ≈ 0.253.
[6 marks]
(iii)
P(C|D) =
P(D|C)P(C)
0.203 · 0.2
P(C ∩ D)
=
≈
= 0.16.
P(D)
P(D)
0.253
[4 marks]
2
[2]
(a) Suppose that X1 and X2 are independent random variables and that X1 has mean
10 and variance 9, i.e., X1 ∼ N (10, 9) and that X2 ∼ N (10, 16). Let Y = X1 − X2 .
(i) What is the distribution of Y ?
(ii) Calculate P(Y > 2).
(iii) Calculate P(−2 ≤ Y ≤ 2).
[10 marks]
(b) Two machines, A and B, are used to manufacture water pipes. A random sample
of pipe diameters (cms) from each machine was taken.
The sample mean and sample standard deviation for the 15 measurements of the
pipes from A are: x̄A = 5.017 and sA = 0.0805. The sample mean and sample standard
deviation for the 9 measurements of the pipes from B are: x̄B = 5.102 and sB = 0.1312.
Assume the true unknown variances are equal.
(i) Write down in terms of the null, H0 , and alternative hypothesis, H1 , the hypothesis
test to determine if machine B produces pipes with mean diameter exceeding that
from machine A by more than 1mm?
(ii) Show that the pooled estimate of the common variance is approximately 0.0104.
(iii) Carry out the hypothesis test at the 5% level.
[10 marks]
3
Solution [2]
(a)
(i)
Y ∼ N (10 − 10, 9 + 16) = N (0, 25).
[3 marks]
(ii)
P(Y > 2) = 1 − P(Y ≤ 2) = 1 − P(
= 1 − P(Z ≤
2−0
Y −0
≤
)
5
5
2
) = 1 − Φ(2/5)
5
= 1 − 0.6554 = 0.3446.
[3 marks]
(iii)
P(−2 ≤ Y ≤ 2) = 1 − P(Y ≤ −2) − P(Y > 2) = 1 − 2P(Y > 2)
= 1 − 2 · 0.3446 = 0.3108.
[4 marks]
(b)
(i) H0 : µA − µB = −0.1 versus H1 : µA − µB < −0.1. Also α = 0.05.
[3 marks]
(ii) a pooled estimate of the common variance:
s2p =
(nA − 1)s2A + (nB − 1)s2B
(nA + nB − 2)
14 · 0.08052 + 8 · 0.13122
=
= (0.0907 + 0.1377)/22 = 0.0104
22
[3 marks]
(iii)
Then the evaluated statistic is
(x̄A − x̄B ) − (−0.1)
5.017 − 5.102 + 0.1
= 0.1019√0.1778 = 0.349.
1
1
sp
nA + nB
The test is single (lower) tailed. t22;0.05 = −1.717 and since 0.349 is much larger we do
not reject the null hypothesis.
[4 marks]
4
[3]
(a)
(i) Describe how lag-k scatterplots can be used to extract information from a time
series.
(ii) A plot of the estimated autocorrelation sequence is also often used; give one advantage and one disadvantage of this method compared to lag-k scatterplots.
[8 marks]
(b) The attached figure shows two panels, labelled ‘a’ and ‘b,’ of scatterplots.
(i) Each scatterplot in panel ‘a’ is a plot of the values of a real hydrological time series,
xt against lagged values xt+k ; the three cases are for k = 1, 6 and 12, as shown.
The series observations are at monthly intervals. What can you deduce about the
time series from these scatterplots? What might the time series look like? What
might the autocorrelation sequence look like?
(ii) Each scatterplot in panel ‘b’ is a plot of the values of a real scientific time series, xt
against lagged values xt+k ; the three cases here are for k = 1, 2 and 3, as shown.
The series observations are at daily intervals. What can you deduce about the time
series from these scatterplots? What might the time series look like? What might
the autocorrelation sequence look like?
[12 marks]
5
Solution [3]
(a)
(i) Lag-k scatterplots show any obvious relationship between time series values xt and
their lagged values xt+k .
[2 marks]
In particular, they can detect a linear relationship, showing for example a positive or
negative correlation.
[2 marks]
They can indicate the strength of the correlation by the degree of clustering of the
points.
[1 mark]
They can also indicate non-linear relationships (e.g., curvature, circularity)
[1 mark]
(ii) Autocorrelation sequences neatly summarize many scatterplots.
[1 mark]
But they assume a linear relationship between xt and xt+k .
[1 mark]
(b)
(i) xt is positively correlated with xt+1 (positive-gradient straight line would be fitted).
xt is negatively correlated with xt+6 and positively correlated with xt+12 .
[3 marks]
As the series is monthly this indicates an annual seasonal effect (cyclical) with
period 12, so that peaks/troughs every 12 months are followed by troughs/peaks 6
months later.
[2 marks]
The autocorrelation is 1 at lag 0, high positive at lag 1, and oscillates with a trough
at lag 6, peak at lag 12 etc,
[1 mark]
(ii) xt is negatively correlated with xt+1 , but for xt and xt+2 , and xt and xt+3 no
correlation can be perceived.
[3 marks]
The series is likely of a flip-flop or alternating positive/negative nature. [2 marks]
The autocorrelation is 1 at lag 0, significantly negative at lag 1, and then would be
close to zero at lags 2 and 3.
[1 mark]
6
Summer Exam 2006: Math. and Stat. Modelling
[1]
(a) Given that P(A) = 7/10, P(A|B) = 1/2 and P(B|A) = 1/7, find
(i) P(A ∩ B).
(ii) P(B).
(iii) P(Ā ∩ B̄).
(iv) P(Ā|B̄).
(v) P(B̄|Ā).
(b) A random variable X has mean E{X} = 8 and variance var{X} = 4.
(i) Find E{X 2 }.
(ii) Find E{3X + 4}.
(iii) Find var{1 − 8X}.
(iv) Suppose X has a binomial distribution with probability of success p and the number
of trials is m. Then X has a mean of mp and variance mp(1 − p). Find the values
of m and p.
1
Solution [1]
(a)
(i) P(A ∩ B) = P(B|A)P(A) = (1/7)(7/10) = 1/10 = 0.1.
[2 marks]
(ii) P(B) = P(A ∩ B)/P(A|B) = (1/10)/(1/2) = 1/5 = 0.2
[2 marks]
(iii) P(Ā ∩ B̄) = P(Ā) − P(Ā ∩ B) = P(Ā) − P(Ā|B)P(B) = (3/10) − (1/2)(1/5) =
2/10 = 0.2.
[4 marks]
(iv) P(Ā|B̄) = P(Ā ∩ B̄)/P(B̄) = (2/10)/(4/5) = 1/4 = 0.25.
(v) P(B̄|Ā) = P(Ā ∩ B̄)/P(Ā) = (2/10)/(3/10) = 2/3 = 0.667.
[1 mark]
[1 mark ]
(b)
(i) E{X 2 } = var{X} + E 2 {X} = 4 + 64 = 68.
[3 marks]
(ii) E{3X + 4} = 3E{X} + 4 = 24 + 4 = 28.
[2 marks]
(iii) var{1 − 8X} = (−8)2 var{X} = 64 · 4 = 256.
[2 marks]
(iv) We have that 8 = mp and 4 = mp(1 − p) so 1 − p = 0.5 and p = 0.5. Then m = 16.
[3 marks]
2
[2]
Let X denote the proportion of valves produced in a single batch at a manufacturing
plant which have a certain characteristic. X has a probability density function (pdf)
given by
f (x; θ) =
1 ( 1 −1)
,
x θ
θ
where 0 < x < 1, and θ > 0.
(i) Find the cumulative distribution function F (x) of X.
(ii) Find the mean of X.
(iii) The median and mean of a random variable are both measures of the location of its
probability density function. The median of X is that value x for which F (x) = 1/2.
Find the median of X and show that it is equal to the mean if θ = 1.
The likelihood function for N independent random variables X1 , ..., XN , each with the
pdf f (x; θ), is
(θ) = f (x1 ; θ)f (x2 ; θ) · · · f (xN −1 ; θ)f (xN ; θ).
(iv) Explain the idea of ‘maximum likelihood’ estimation. Why can the logarithm of
the likelihood function be used to obtain a maximum likelihood estimate of θ?
(v) The proportions with the characteristic observed in 12 independent batches are
given by
0.94, 0.56, 0.52, 0.82, 0.27, 0.44, 0.73, 0.41, 0.93, 0.49, 0.92, 0.75
What is the maximum likelihood estimate of θ?
(vi) Use the maximum likelihood estimate of θ to evaluate
P (X > 0.5 | X > 0.1).
3
Solution [2]
(i)
F (x) =
0
x
x
1 ( 1 −1)
dy = y 1/θ 0 =
y θ
θ
x1/θ , if 0 < x < 1;
.
0,
otherwise,
[4 marks]
(ii)
E{X} =
0
1
1 1
x( θ +1) 1
1
x · x( θ −1) dx =
=
.
θ
1+θ 0
1+θ
1
[4 marks]
(iii) The median is the solution of F (x) = 0.5. From (i) this is given by x = (1/2)θ .
Further, when θ = 1 both the mean and median are 1/2.
[2 marks]
(iv) The likelihood function depends on the unknown parameter(s). The idea is to choose
the parameter value(s) which correspond to the maximum of the likelihood function
(what we obtained being viewed as the most likely occurrence). Since log is a monotonic
transform, the max of the log likelihood function will occur at the max of the likelihood
fn.
[2 marks]
(v) The likelihood function is
=θ
−N
N
( 1 −1)
xi θ
i=1
So
Then
N
1
( − 1) log xi .
log = −N log θ +
θ
i=1
N
N
1
log xi = 0 ⇒ θ̂ = −(1/N )
log xi .
(d/dθ) log = −N/θ −
θ2
i=1
i=1
[4 marks]
(vi)
P (X > 0.5 | X > 0.1) =
P (X > 0.5)
1 − F (0.5)
1 − 0.51/θ̂
= 0.758.
=
=
P (X > 0.1)
1 − F (0.1)
1 − 0.11/θ̂
[4 marks]
4
[3]
A model for a time series {Xt } containing a seasonal component is Xt = St + Yt , where
{St } is the seasonal component and {Yt } is a stationary time series with zero mean.
(i) Suppose we wish to use the methods of analysis of stationary stochastic processes
to analyse {Xt }. Explain why it is then necessary to remove the seasonal term
from the time series before carrying out the analysis.
If the seasonal component has period s then the seasonal effect can be removed by
(s)
creating the sth difference, Xt
= Xt − Xt−s . The following N = 12 sample values {xt }
of such a river flow time series were recorded:
0.1856, −0.9653, −0.1383, 0.8554, −0.1149, −1.0615, 0.0535, 1.2669, −0.2662,
− 0.7202, −0.1145, 0.9307.
The observations were taken every 3months (ie., quarterly).
(ii) Plot the series against time, t = 1, ..., 12. What is the apparent seasonal period s
and is this consistent with the times of the observations?
(s)
(iii) Remove the seasonal effect by calculating the sample sth difference series, {xt }
of length N − s.
(s)
(iv) From {xt } calculate the sample cross-correlation for lags k = 1, 2 and 3 using
N −s−k
ρ̂k =
t=1
(s)
(s)
(xt − x̄(s) )(xt+k − x̄(s) )
,
N −s (s)
(s) )2
t=1 (xt − x̄
(s)
where x̄(s) is the sample mean of {xt }.
(v) Based on your results in (iv) is there any strong evidence to reject the hypothesis
that the stationary time series {Yt − Yt−s } is an uncorrelated series? Fully explain
your reasoning.
5
Solution [3]
(i) A seasonal term in the time seriesviolates the condition that the mean of {Xt } is
fixed, a requirement for a stationary stochastic process.
[2 marks]
(ii) Plot of the series is shown here (left plot).
original data
deseasonalized data
1.5
0.5
0.4
1
0.3
0.2
0.5
0.1
0
0
−0.5
−0.1
−0.2
−1
−0.3
−1.5
0
5
10
−0.4
0
15
2
time
4
time
6
8
The apparent seasonal period is s = 4, which is consistent with the times of the observations (quarterly, so that four observations cover a year, and we would expect annual
cycles.)
[4 marks]
(s)
(iii) The 8 deseasonalized observations {xt } are
−0.3005, −0.0962, 0.1918, 0.4115, −0.1513, 0.3414, −0.1680, −0.3362.
[4 marks]
(iv) The sample mean of the deseasonalized data is x̄(s) = −0.0134.
Using
N −s−k
ρ̂k =
t=1
(s)
(s)
(xt − x̄(s) )(xt+k − x̄(s) )
,
N −s (s)
(s) )2
(x
−
x̄
t
t=1
we obtain
ρ̂1 = −0.0316, ρ̂2 = −0.1108, ρ̂3 = −0.1007.
[6 marks]
6
(s)
(v) Now Xt
= Xt − Xt−s = Yt − Yt−s , so that the deseasonalized data can be thought
of as a sample from {Yt − Yt−s }. The computed autocorrelations are all small in
size, and thus there is no evidence from these that {Yt − Yt−s } is anything other
than an uncorrelated process.
[4 marks]
7