© 2020 Imperial College London Page 1
MATH97085
BSc, MSci and MSc EXAMINATIONS (MATHEMATICS)
May-June 2020
This paper is also taken for the relevant examination for the
Associateship of the Royal College of Science
Stochastic Simulation
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1. (a) The congruential generator to generate pseudo-standard uniform variates, Ui, defines
Ui =
Xi
m
,
where,
Xi = (aXi−1 + b) mod m, i = 1, 2, 3, . . .
with integer m and X0, a, b ∈ {0, 1, . . . ,m− 1}.
(i) What is the value of b for a multiplicative generator? (1 mark)
(ii) Define the period, k, of such a generator. What is the largest value of k when b 6= 0?
(3 marks)
(iii) Consider the following R code:
cg = function(n, m, a, b, x0)
{
x = c(x0, rep(0,(n-1)))
for(i in 2:n)
x[i] = (a * x[i - 1]+b) %% m
data.frame(x = x/m)
}
x1 = cg(500, 2^(40), 5^3, 0, 1)
x2 = cg(500, 2^(45), 11, 0, 1)
x3 = cg(500, 2^(12), 2^6, 1, 2^2)
Find the period for each of the generators which produce x1, x2, and x3. (4 marks)
The figures below show lag-one scatter plots of the data produced by the R code above.
Figure A Figure B Figure C
To which sequence (out of x1, x2, and x3) do each of the Figures (out of A, B and C)
correspond? (2 marks)
(b) Given a sequence of n variates which are claimed to be from a U(0, 1) distribution, describe
how you would perform a chi-squared goodness-of-fit test on the numbers (between 0 and 99)
formed by taking the first two decimal places of the generated sequence. (5 marks)
(c) Describe two methods to investigate the dependence structure of a sequence of generated
U(0, 1) variates. (5 marks)
(Total: 20 marks)
MATH96054/MATH97085 Stochastic Simulation (2020) Page 2
2. (a) Describe in detail the rejection method for generating samples from a probability density function
(pdf) fX(·) using a rejection envelope gY (·). (3 marks)
(b) Describe in detail the Metropolis-Hastings (M-H) scheme to generate samples from a pdf fX(·)
using a transition kernel q(· | ·). (4 marks)
(c) Describe in detail the inversion algorithm to generate samples from a pdf fX(·). (2 marks)
(d) Consider schemes to simulate samples from a pdf fX(x) ∝ f ∗X(x), where,
f ∗X(x) =

exp(−|x|), −1 < x < 1;
0, otherwise.
(i) Determine the acceptance probability of a rejection scheme to generate samples from fX(x)
using a uniform envelope, i.e.
gY (y) =

0.5, −1 < y < 1;
0, otherwise.
(2 marks)
(ii) Consider the following R code to implement a M-H scheme to simulate from fX(·),
f = function(x){
c=2*(1-exp(-1))
ifelse(x>-1 & x<1, exp(-abs(x))/c, 0)
}
qpdf = function(y, sigma){
rnorm(1, mean=y, sd=sigma)
}
MH = function(n){
x=vector("double",n); x[1]=0; u=runif(n); sigma=0.1
for(i in 2:n){
y=qpdf(x[i-1],sigma)
x[i] = ifelse(u[i]}
return(x)
}
(I) What transition kernel is being used in this implementation? (1 mark)
(II) What R code should replace **** for this code to work? (1 mark)
(III) Describe the effect of changing the variable sigma. (2 marks)
(e) Why is it beneficial for a simulation scheme to generate independent samples? (2 marks)
(f) With justification, which approach, out of rejection, inversion and M-H would you choose to
apply in order to generate samples from the pdf outlined in part (d). (3 marks)
(Total: 20 marks)
MATH96054/MATH97085 Stochastic Simulation (2020) Page 3
3. Consider the following integral,
θ =
∫
h(x) dx =
∫
φ(x)fX(x) dx.
(a) Given X1, . . . , Xn ∼ fX(·).
(i) Write down the Monte-Carlo estimator θˆ for θ in terms of the Xi, i = 1, . . . , n. (1 mark)
(ii) Prove that θˆ is unbiased for θ. (2 marks)
(iii) Find an expression for the variance of θˆ in terms of φ(·), fX(·) and θ. (2 marks)
(b) Consider estimating the integral,
θ =
∫ 1
0
exp(−x3) dx.
(i) Describe the hit-or-miss method for estimating θ.
Consider the following R code:
a=0;b=1;c=1;n=10000
u=runif(n, min=a,max=b)
v=runif(n, min=0,max=c)
thetahat = ****
What R code should replace **** in the code above in order for thetahat to be the
hit-or-miss Monte-Carlo estimate of θ? (4 marks)
(ii) Describe how you would construct the crude Monte-Carlo estimator for θ given a sequence
of independent variates U1, . . . , Un ∼ U(0, 1). (2 marks)
(iii) Using inversion, one can generate from the following pdf,
fX(x) =

k cos(kx)√
1− e−2 , 0 < x < 1;
0, otherwise,
where k = arccos(e−1).
Consider the following R implementation of an inversion scheme to generate from fX(·),
n=10000;k=acos(exp(-1));k1=sqrt(1-exp(-2))
u=runif(n)
x= ****
What R code should replace **** in the section of code above so that x contains variates
with pdf fX(·)? (4 marks)
(iv) Determine an estimator of θ given a sequence of independent variates X1, . . . , Xn ∼ fX(·),
described in part (iii). (3 marks)
(v) Without performing any calculations, but giving your reasoning, which estimator out of
those calculated in parts (i), (ii) and (iv), would you expect to have the lowest variance?
(2 marks)
(Total: 20 marks)
MATH96054/MATH97085 Stochastic Simulation (2020) Page 4
4. (a) Given random variables X and Y with joint probability density (pdf) function fX,Y (x, y), and
conditional probability density functions fY |X(y|x) and fX|Y (x|y) , outline the two-stage Gibbs
sampler to generate (Xt, Yt), t = 1, 2, . . . from the joint pdf. (3 marks)
(b) It is supposed that X and Y have a joint distribution for which the conditional pdfs are
X|Y = y ∼ Exp(y) and Y |X = x ∼ Exp(x).
(i) Show that these conditional distributions would be consistent with a joint distribution whose
pdf is fX,Y (x, y) ∝ exp(−xy), x, y > 0.
Explain why this function fX,Y (x, y) is not actually a valid pdf. (3 marks)
(ii) What issue with Gibbs sampling does the lack of validity of fX,Y (x, y) highlight?
(1 mark)
(iii) Now suppose that the X and Y have a joint distribution with pdf fX,Y (x, y) ∝ exp(−xy)
for x ∈ (0, 1), y ∈ (0, 1) and fX,Y (x, y) = 0 otherwise.
In this case, outline the Gibbs sampler to generate from fX,Y (x, y). (4 marks)
(c) Let the joint pdf of X and U be given by
fX,U(x, u) =

1, 0 < u < 8
pi
√
x(1− x), 0 < x < 1;
0, otherwise.
(i) Verify that
fX(x) =

8
pi
√
x(1− x), 0 < x < 1;
0, otherwise.
(1 mark)
(ii) Prove that U |X = x ∼ U(0, a) and X|U = u ∼ U(b, c) where you should specify a in
terms of x, and b and c in terms of u. (5 marks)
(iii) An implementation of the Gibbs sampler to simulate from the joint pdf is given below.
N=5000
X=c(runif(1));U=c(runif(1))
for(t in 1:N){
a=**1**
U[t+1]=runif(1,0,a)
b=**2**
c=**3**
X[t+1]=runif(1,b,c)
}
What R code should replace **1**, **2** and **3** in order for X to contain variates
with pdf fX(·)? (3 marks)
(Total: 20 marks)
MATH96054/MATH97085 Stochastic Simulation (2020) Page 5
Module: MATH96054/MATH97085
Setter: McCoy
Checker: Hallsworth
Editor: Hallsworth
External: Jennison
Date: May 18, 2020
Version: Draft version for checking
BSc, MSci and MSc EXAMINATIONS (MATHEMATICS)
May – June 2020
MATH96054/MATH97085 Stochastic Simulation (Solutions)
The following information must be completed:
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of a sound knowledge of a good part of the material and the solution of straightforward
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BSc, MSc and MSci EXAMINATIONS (MATHEMATICS)
May – June 2020
This paper is also taken for the relevant examination for the Associateship of the
Royal College of Science.
Stochastic Simulation (Solutions)
Date: Tuesday, 26th May 2020
Time: AM
Time Allowed: 1.5 Hours for MATH96 paper; 2 Hours for MATH97 papers
This paper has 3 Questions (MATH96 version); 4 Questions (MATH97 versions).
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c© 2020 Imperial College London MATH96054/MATH97085 Page 1 of 9
seen ⇓
1. (a) (i) b = 0 for a multiplicative generator.
1 A
(ii) The period k is the smallest positive integer such that
Xi+k = Xi, ∀i
The sequence {Xi, . . . , Xi+k−1} repeats.
2 A
If b 6= 0 the largest value of k is m (as i mod m ∈ {0, 1, . . . ,m− 1}, i ∈ N). 1 A
meth seen ⇓(iii) A multiplicative generator with m = 2β has period m/4 if a mod 8 ≡ 3 or 5 and X0
is odd.
x1: m = 240, a = 53, b = 0, x0 = 1, so this is multiplicative and a mod 8 = 5 so
period = 238.
x2: m = 245, a = 11, b = 0, x0 = 1, so this is multiplicative and a mod 8 = 3 so
period = 243.
x3: m = 212, a = 26, b = 1, x0 = 22, so,
x1 = (2622 + 1) mod 212 = 28 + 1,
x2 = (26(28 + 1) + 1) mod 212 = 26 + 1,
x3 = (26(26 + 1) + 1) mod 212 = 26 + 1,
so xi+1 = xi, i > 1 and hence the period =1.
4 D
unseen ⇓Figure A is the plot for sequence x3 as it eventually gets stuck at the value (26+1)/212.
Sequence x2 has a = 11 and hence is associated with Figure C (all points on lines
with gradient 11)
Finally, x1 is associated with Figure B. 2 D
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 2
sim. seen ⇓
(b) Let vi, i = 1, . . . , n be the number represented by the first two decimal places of ui, i.e.
vi = b100uic, alternatively, let d1i and d2i represent the first two decimal places of the
observed ui, i = 1, . . . , n respectively and let vi = 10d1i + d2i. Let,
Oj =
n∑
i=1
I[vi=j], j = 0, . . . , 99; where I[vi=j] =

1 if vi = j;
0 otherwise.
If the sequence follows a standard uniform distribution then we would expect each of the
numbers 0, . . . , 99 to occur with equal probability so Ei = n/100.
Form the test statistics
χ2 =
99∑
i=0
(Oi − Ei)2
Ei
.
Under the null hypothesis that the sequence is uniformly distributed the test statistic will
approximately follow a Chi-Squared distribution with 100−1 = 99 degrees of freedom. The
value of the test statistic is then compared to the percentage points of the χ299 distribution
to see if there is evidence to reject the null hypothesis.
5 C
seen ⇓(c) The dependence structure of a sequence, U1, . . . , Un of generated U(0, 1) variates can be
tested by the following:
∗ Correlogram: Plot of the sample autocorrelation sequence
ρˆk =
1
n
∑n−k
i=1 (Ui − U¯)(Ui+k − U¯)
1
n
∑n
i=1(Ui − U¯)2
:
against k. An approximate 95% confidence interval (CI) for the estimated
autocorrelation sequence is
− 1
n
± 1.96√
n
;
Values of ρˆk that lie outside the CI give evidence to reject the hypothesis that the
sequence is independent.
∗ Serial test for digits: Let njk be the number of times the digit j is followed by the
digit k. In this case we have
O0 = n00, O1 = n01, . . . , O9 = n09, O10 = n10, . . . , O99 = n99
E0 = n100 , E1 =
n
100 , . . . , E99 =
n
100 .
This can be tested with a Chi-Squared test.
∗ Gap Test: Choose a digit, say 3, then record the length of the subsequence lying
between occurrences of the digit 3.
X1, X2, X3, 3, X5, X6, . . . , Xk+4︸ ︷︷ ︸
gap length= k
, 3, . . .
If the sequence is uniform and independent, the distribution of gap lengths, K, should
be Geometric( 110), which can be tested with a Chi-Squared test.
5 A
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 3
seen ⇓
2. (a) GeneratingX with a known pdf fX(·) using rejection employs the use of a rejection envelope
gY (·) (which we can generate from). The pdf gY (·) must have the following properties:
∗ the support of gY encompasses that of fX , i.e. fX(x) > 0⇒ gY (x) > 0,
∗ there exists M > 0 such that ∀x s.t. fX(x) > 0,
fX(x)
gY (x)
≤M <∞.
Rejection Sampling Algorithm
1. Generate Y = y ∼ gY (.).
2. Generate U = u ∼ U(0, 1).
3. If u ≤ fX(y)
MgY (y) set X = y.
4. Otherwise GOTO 1.
3 A
(b) GeneratingX from known pdf fX(·) using Metropolis-Hastings (M-H), employs a transition
kernel q(y|x) ≥ 0, which satisfies ∫
q(y|x)dy = 1.
q(y | x) and should be chosen to ensure irreducibility (the conditional pdf should be non-
zero over the range of X).
Metropolis-Hastings Procedure (continuous state space):
1. Initialise the chain: start from an arbitrary X0, possibly sampled from an initial
pdf/prior. Set n = 1.
2. Given Xn−1 = x, generate a candidate value Y = y from the proposal density q(y|x).
3. Set Xn = y with probability α(x, y), where
α(x, y) = min
{
fX(y)q(x|y)
fX(x)q(y|x) , 1
}
,
otherwise, set Xn = x.
4. Replace n by n+ 1 and return to Step 2.
Note that as X0 is not generated directly from fX the algorithm needs a burn-in period.
Also, MH does not produce independent samples.
4 A
meth seen ⇓
(c) Generating X from known pdf fX(·) using inversion requires calculating the inverse cdf
F−1X (x).
Inversion Algorithm:
1. Generate U = u ∼ U(0, 1)
2. Set X = x = F−1X (u)
2 A
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 4
(d) (i)
M = sup
x
f ∗X(x)
gY (x)
= sup
−12e−|x| = 2.
Acceptance probability, θ, is given by
θ =
∫ 1
−1 f
∗
X(x) dx
M
= 2
∫ 2
0 e−x dx
2
=
[
−e−x
]1
0
= 1− e−1.
2 B
(ii) (I) The transition kernel is a Normal density centred at the current value with variance
0.01,
q(y|x) = 10√
2pi
e−50(y−x)2 , y ∈ R. 1 B
(II) **** should be replaced with f(y)/f(x[i-1]). Note that q(y|x) is symmetric,
so does not need to be included as the terms cancel.
1 B
(III) sigma controls how close the distribution of the proposal around the current
value. Reducing sigma means that the proposed values are accepted with higher
probability, will be more correlated and slowly mixing with less opportunity to
explore the space. If sigma is set too high, then the acceptance probability may
be too low and be inefficient.
2 A
(e) Many statistical procedures rely on the independence assumption either directly or through
the Central Limit Theorem.
2 B
(f) Answer should justify with a discussion of computational complexity and timing
considerations. Accept any well reasoned answer based on e.g. avoiding complex
calculations, computational time or a discussion of excluding M-H because of lack of
independence. 3 C
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 5
seen ⇓
3. (a) (i)
θ = E (φ(x))) , θˆ = 1
n
n∑
i=1
φ(Xi). 1 A
(ii)
E(θˆ) = 1
n
n∑
i=1
E(φ(Xi)) =
1
n
n∑
i=1
θ = θ.
So θˆ is unbiased for θ.
2 A
(iii)
var(θˆ) = 1
n2
n∑
i=1
var(φ(Xi)) =
1
n
(
E(φ2(X))− θ2
)
= 1
n
[∫
φ2(x)fX(x) dx− θ2
]
2 A
part seen ⇓
(b) (i) Given θ = ∫ ba h(x) dx with 0 ≤ h ≤ c. We generate points uniformly in the rectangle
(a, b)× (0, c) and estimate the area from the proportion that lie under h(x).
Hit-or-Miss Algorithm:
1. Sample U = ui ∼ U(a, b), V = vi ∼ U(0, c), i = 1, . . . , n
2. Define
θˆ = c(b− a)︸ ︷︷ ︸
area of
rectangle
1
n
n∑
i=1
I(vi ≤ h(ui))︸ ︷︷ ︸
frequency of “dart”
under h
We have h(x) = exp(−x3), x ∈ (0, 1).
So a = 0, b = 1 and c = supx∈(0,1) exp(−x3) = 1 and **** should be replaced with
the R code
c*(b-a)*mean(v <= exp(-u^3)).
Alternatively, c*(b-a)*length(v[v<=exp(-u^3)])/n,
or c*(b-a)*sum(v <= exp(-u^3))/n.
4 C
meth seen ⇓(ii) We have
θ =
∫ 1
0
exp(−x3) dx =
∫ b
a
φ(x)fX(x) dx,
where fX(x) = 1, x ∈ (0, 1) and φ(x) = exp(−x3), x ∈ (0, 1). The crude MC
estimator is given by,
θˆ = 1
n
n∑
i=1
exp(−U3i ). 2 B
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 6
(iii) We have, for x ∈ (0, 1),
FX(x) =
∫ x
0
k cos(ky)√
1− e−2 dy =
1√
1− e−2 [sin(ky)]
x
0 ,
= sin(kx)√
1− e−2 .
To find F−1X (·), x ∈ (0, 1),
U = sin(kx)√
1− e−2 ⇒ X = F
−1
X (U) =
1
k
arcsin(U
√
1− e−2).
To generate from fX(·) using inversion:
1. Generate U = u ∼ U(0, 1).
2. Set X = x = arcsin(u
√
1− e−2)/k.
Hence ***** should be replaced with the commands x=asin(u*k1)/k. 4 D
(iv) We have
θ =
∫ 1
0
exp(−x3) dx =
∫ b
a
φ(x)fX(x) dx,
where
fX(x) =
k cos(ky)√
1− e−2 , x ∈ (0, 1), and hence φ(x) =
exp(−x3)√1− e−2
k cos(kx) , x ∈ (0, 1).
The MC estimator is given by
θˆ = 1
n
n∑
i=1
exp(−X3i )
√
1− e−2
k cos(kXi)
.
3 B
(v) We would expect the estimator outlined in (iv) to have the lowest variance. We know
that hit-or-miss always has higher variance than crude MC. The density outlined in (iv)
mimics the shape of exp(−x3) better than the uniform (cf Importance sampling).
2 B
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 7
4. (a) Two-stage Gibbs sampler given the conditional pdfs for fX,Y (x|y) and fY |X(y|x).
Set X0 = x0. Set t = 1.
1. Generate Yt = yt ∼ fY |X(·|xt−1);
2. Generate Xt = xt ∼ fX|Y (·|yt);
3. Set t = t+ 1 and goto 1. 3
(b) We have fX|Y (x|y) = y exp(−xy), x > 0 and fY |X(y|x) = x exp(−xy), y > 0.
(i)
fX,Y (x, y) ∝ exp(−xy)⇒ fX(x) ∝
∫ ∞
0
exp(−xy) dy =
[
exp(−xy)
−x
]∞
0
= 1
x
, x > 0,
fY |X(y|x) = fX,Y (x, y)
fX(x)
∝ x exp(−xy), x > 0.
So Y |X = x ∼ Exp(x). Similarly X|Y = y ∼ Exp(y) as required for consistency
with joint distribution.
2
This is not a valid pdf as ∫∞0 1/y dy is not finite. 1
(ii) This can be a danger as you can implement the Gibbs sampler with conditional
distributions without having to verify that that they correspond to a valid joint
distribution.
1
(iii) If the conditional pdfs are truncated to (0, 1), then we have
fY |X(y|x) = x exp(−xy)∫ 1
0 x exp(−xy) dy
= x exp(−xy)1− e−x
Similarly for fX|Y . To generate from fY |X (and fX|Y ) we can use inversion. For
0 < y < 1, we have
FY |X(y|x) =
∫ y
0
x exp(−xt)
1− e−x dt =
1− e−xy
1− e−x .
Setting
U = 1− e
−xY
1− e−x ⇒ U(1− e
−x) = 1− e−xY ⇒ −xY = log(1− U(1− e−x))
⇒ Y = F−1Y |X(U |x) = −
1
x
log(1− U(1− e−x))
Gibbs sampler:
1. Set t = 1, X0 = x0.
2. Generate U = u ∼ U(0, 1), set Yt = yt = − 1xt−1 log(1 − u(1 − e−xt−1)), then
Yt ∼ Exp(0,1)(xt−1), where Exp(0,1)(·) is an exponential distribution truncated to
(0, 1).
3. Generate U = u ∼ U(0, 1), set Xt = xt = − 1yt log(1 − u(1 − e−yt)), then
Xt ∼ Exp(0,1)(yt).
4. Set t = t+ 1 and goto 1. 4
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 8
(c) (i)
fX(x) =
∫
fX,U(x, u) du =
∫ 8√x(1−x)/pi
0
1 du = 8
pi
√
x(1− x), 0 < x < 1,
as required. 1
(ii)
fU |X(u|x) = fU,X(u, x)
fX(x)
= pi
8
√
x(1− x)
, 0 < u < 8
pi
√
x(1− x).
i.e U |X = x ∼ U(0, a) where a = 8
pi
√
x(1− x). Now,
fU(u) =
∫
0pi
√
x(1−x)
1 dx
u <
8
pi
√
x(1− x)⇒ piu8 <
√
x(1− x)⇒ pi
2u2
64 < x(1− x)⇒ x
2 − x+ pi
2u2
64 < 0
⇒
x− 1−
√
1− 4pi2u264
2
x− 1 +
√
1− 4pi2u264
2
 < 0
⇒ b = 12 −
1
8
√
16− pi2u2 < x < 12 +
1
8
√
16− pi2u2 = c.
So
fU(u) =
∫ c
b
1 dx = 14
√
16− pi2u2, 0 < u < 4
pi
,
and
fX|U(u|x) = 4√16− pi2u2 ,
1
2 −
1
8
√
16− pi2u2 < x < 12 +
1
8
√
16− pi2u2
i.e. X|U = u ∼ U(b, c) where b = 12 − 18
√
16− pi2u2 and c = 12 + 18
√
16− pi2u2. 5
(iii) From part (ii):
**1** = 8*sqrt(X[t]*(1-X[t]))/pi
**2** =0.5-(1/8)*sqrt(16-U[t+1]^2*pi^2)
**3** = 0.5+(1/8)*sqrt(16-U[t+1]^2*pi^2)
3
Commentary
Q1 Mainly a straightforward question, though students may find a(iii) challenging, for (b) they have
seen the test for a single digit.
Q2 The first two parts are bookwork, though we concentrated more on the discrete case, so the MH
scheme may be a little more challenging.
Q3 Mainly application of standard results, but the inversion is a little fiddly and the R code sections
may be harder for them as this has not appeared in previous exams.
Q4 Based on Chapter 7 of “Introducing Monte Carlo Methods with R, Use R” by Robert and Casella.
They have only seen the basic Gibbs sampler in lectures.
MATH96054/MATH97085 Stochastic Simulation (Solutions) (2020) Page 9
ExamModuleCode QuestionNComments for Students
MATH97085 1
This question was generally well done. The only significant comment is that many candidates 
failed to distinguish between methods designed to test for uniformity of output and those 
designed to test for serial dependence.
MATH97085 2
This question was found straightforward. Rather few answers provided a careful argument about 
why we would want independent samples, and too many candidates asserted (incorrectly) that 
inversion was not an option with the density in (d).
MATH97085 3
Question was generally done well.  There were some mistakes in the inversion technique for the 
trigonometric pdf and the calculation of the full details of the hit‐or‐miss algorithm was not given 
in all cases
MATH97085 4
This questions proved challenging for many students, particularly part ( c) where many students 
were unable to derive the values for b and c.
If your module is taught across multiple year levels, you might have received this form for each level of the module. You are only required to fill this out once for each question.
Please record below, some brief but non‐trivial comments for students about how well (or otherwise) the questions were answered. For example, you may wish to comment on common errors and 
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