MAST20004 Probability
Assignment Three [Due 4:00 pm Monday 07/10]
There are 5 problems in total, of which 3 randomly chosen ones will be marked.
You are expected to submit answers to all questions, otherwise a mark penalty will
apply. Calculations and reasoning must be given in order to obtain full credit.
Problem 1. Choose a number X at random from the set {1, 2, 3, 4, 5}, then
choose a number Y at random from the subset {1, · · · , X}.
(i) Find the joint pmf of X and Y .
(ii) Find the conditional pmf of X given Y = 3.
(iii) Are X and Y independent?
(iv) Compute the expected value of X
Y
.
Problem 2. Let (X, Y ) be a bivariate random variable whose joint pdf is given
by
fX,Y (x, y) =
{
Cy
x3
, 0 < x < 1 and 0 < y < x2,
0, otherwise.
(i) Compute the constant C, and the marginal pdf’s of X and Y respectively.
(ii) Compute fY |X(y|x) and deduce that
E[Y |X] = 2
3
X2.
(iii) Compute fX|Y (x|y) and E[X|Y ].
Problem 3. Let (X, Y ) be a general bivariate normal random variable.
(i) If Cov(X, Y ) = 0, show that X, Y are independent.
(ii) If Var[X] = Var[Y ], show that X + Y and X − Y are independent.
(iii) Assume that µX = 0, σ2X = 1, µY = −1, σ2Y = 4, ρ = 1/2. Compute
P(X + Y > 0) and P(X + Y > 0|2X − Y = 0).
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Problem 4. Suppose that X, Y are independent random variables, both uni-
formly distributed over [0, 1].
(i) Find the pdf of R = |X − Y | and E[R].
(ii) Find the joint pdf of U = X + Y and V = X/Y .
[Hint: The trickier part is to find out what the transformed region T (D) is, where
D is the unit square in the x-y plane. There is not a general way of doing so,
and one needs to argue example by example. For this problem, we can first figure
out the range of u, and then for each u within this range we further figure out
the range of v. It is typical that this range of v, denoted as Ru and visualized in
the (u, v)-plane as the vertical segment Lu = {(u, v) : v ∈ Ru}, will depend on
u which falls between the graphs of two functions of u. The region T (D) is then
determined by figuring out what Lu swipes out as u runs over its own range.
(iii) Show that the bivariate random variable (U, V ) defined by
U =
√
−2 logX cos(2piY ), V =
√
−2 logX sin(2piY )
is a standard bivariate normal random variable with parameter ρ = 0.
[Remark: By taking the U-component, this gives a simple way of generating a
standard normal random variable. To some extent, this method is better than the
one we discussed in lecture using U = Φ−1(X) where Φ is the Cdf of N(0, 1), since
Φ−1 is very hard to obtain. This is a nice illustration of the philosophy that: when
we are working on a one dimensional problem, sometimes it could be substantially
easier if we look at the problem from a multi-dimensional perspective. Another
example of such kind is the computation of
∫∞
−∞ e
−x2/2dx we did in the lecture.
This idea is further developed and appreciated in the subject of complex analysis.]
(iv) By using a given pair of independent uniform random variables (X, Y ) over
[0, 1], find a way to construct a bivariate random variable (Z,W ) satisfying
E[Z] = E[W ] = 0, Var[Z] = Var[W ] = 5, Cov(Z,W ) = 4.
[Remark. Essentially, this method allows us to generate a general bivariate normal
random variable from a pair of independent uniform random variables over [0, 1]]
Problem 5. [Hard] For this problem, you might need to use the following so-
called inclusion-exclusion principle without proof. Let A1, A2, · · · , An be n events.
Then
P(A1 ∪ · · · ∪ An) =
n∑
i=1
P(Ai)−
∑
16iP(Ai ∩ Aj)
+
∑
16iP(Ai ∩ Aj ∩ Ak)− · · ·+ (−1)n−1P(A1 ∩ · · · ∩ An).
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A little girl is painting on a blank paper. Suppose that there is a total number
of N available colors. At each time she selects one color randomly and paints on
the paper. It is possible that she picks a color that she has already used before.
Different selections are assumed to be independent.
(1) Suppose that the littile girl makes n selections.
(1-i) If red and blue are among the available colors, let R (respectively, B) be the
event that her painting contains color red (respectively, blue). What is P(R) and
P(R ∪B)?
(1-ii) Suppose that she is about to make the (n + 1)-th selection. What is the
probability that she will obtain a new color in this selection?
[Hint: discuss according to the specific color in her (n+ 1)-th selection.]
(1-iii) Suppose that n = N . For 1 6 i 6 N, let Ei be the probability that her
painting does not contain color i. By using the inclusion-exclusion principle to
∪Ni=1Ei, show that
N ! =
N∑
k=0
(−1)k
(
N
k
)
(N − k)N .
(1-iv) Let D be the number of different colors she obtain among her n selections.
By writing N − D as a sum of Bernoulli random variables, compute E[D] and
Var[D].
(2) Let S be the number of selections needed until every available color has been
selected by the little girl.
(2-i) Find the pmf of S.
[Hint: consider {S > n} and use the inclusion-exclusion principle to compute this
probability.]
(2-ii) For 0 6 i 6 N − 1, let Xi be the random variable that after obtaining i
different colors, the number of extra selections needed until further obtaining a
new color. By understanding the distributions of these Xi’s and their relationship
with S, show that
E[S] = N ×
(
1 +
1
2
+ · · ·+ 1
N
)
.
Since the harmonic series H(N) = 1 + 1
2
+ · · · + 1
N
has logarithmic growth (i.e.
H(N)
logN
→ 1 as N → ∞), this result shows that when N is large, on average the
little girl needs to make about N logN selections before obtaining all different
colors.
(3) Let T be the number of selections until the little girl picks a color that she
has obtained before.
(3-i) Find the pmf and expected value of T .
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(3-ii) Consider E[T ] as a function of N . What is the growth rate of E[T ] as
N →∞? You don’t need to solve this problem mathematically. Simply make an
educated guess.
[Hint: firstly, use the formula
E[T ] =
∞∑
k=0
P(T > k)
given in Tutorial 6, Problem 4 to simplify the expression of the mean. Secondly,
relate E[T ] with either some standard Taylor series to guess the growth rate or
relate it with Poisson random variables and the central limit theorem. The central
limit theorem (which we will learn soon) says, if X1, X2, · · · are independent and
identically distributed with finite variance, then
lim
n→∞
P
(
Sn − E[Sn]√
Var[Sn]
6 x
)
= Φ(x), for all x ∈ R,
where Sn = X1+· · ·+Xn and Φ(x) is the Cdf of the standard normal distribution.]
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