Math 302, Assignment 4
You are required to explain the mathematical reasoning behind your answers.
Due Friday, Oct 14th, by 11:59 PM on Canvas.
(1) Let X take values {1, 2, 3, 4, 5}. Its probability mass function is given by
Table 1. The p.m.f. of X
k
P(X = k)
1
2
1/7 1/14
3
4
3/14 2/7
5
2/7
(a) Calculate E [X].
(b) Calculate E[|X − 2|].
(2) Consider the following lottery: There are a total of 10 tickets, of which 5
are “win” and 5 are “lose”. You draw tickets (without replacement) until
you draw the first “win”. Drawing one ticket costs $2, 2 tickets $4, 3 tickets
$8, 4 tickets $16, 5 tickets $32 and 6 tickets $64. A winning ticket pays out
$8.
(a) Let X be the number of tickets you draw in the lottery (i.e. the number
of tickets until the first win, including the winning ticket). Calculate
the probability mass function of X.
(b) Calculate the expectation E [X].
(c) What are your expected winnings in this game?
(3) Prove the following claims. Here, X, Y are discrete random variables on
the same sample space, and a, b ∈ R.
(a) E h[aX + b] = aE [X] + b i
2
(b) E (aX + b − E[aX + b]) = a2 E[(X − E[X])2 ]
(c) E[(X − E[X])2 ] = E[X 2 ] − (E [X])2 .
(d) Challenge (not marked) E[X + Y ] = E [X] + E [Y ].
(e) Challenge (not marked) Deduce from (c) that E[X 2 ] ≥ (E [X])2
and that if E[X 2 ] = (E [X])2 then there must exists some number
c ∈ R such that P[X = c] = 1. Hint: Show that if Z is a discrete
random variable satisfying that P[Z ≥ 0] = 1 then E[Z] ≥ 0 and
moreover that if P[Z ≥ 0] = 1 and E[Z] = 0, then P[Z = 0] = 1.
Remark: You are allowed to use part (a) in all other parts, even if you did
not prove part (a). You are allowed to use part (d) in part (c) (but not in
part (a)) even if you did not prove part (d).
(4) Suppose that a discrete random variable X has cumulative distribution
function given by


0
t<1





1/4
1≤t<2

FX (t) := P[X ≤ t] = 1/3 2 ≤ t < 5 .


1/2 5 ≤ t < 100



1
t ≥ 100
2
Find the probability mass function pX (∙) := P[X = ∙] of X.
(5) Suppose that a continuous random variable X has cumulative distribution
function given by

√

t< 3
0
√
FX (t) := P[X ≤ t] = t2 − 3
3≤t<2.


1
t≥2
(a) Find P[X = 1.9].
(b) Find P[1 ≤ X ≤ 1.9].
(c) Find a probability density function of X.
(6) Consider a random graph on n ≥ 3 vertices labeled by [n] := {1, . . . , n}
defined as follows: for each i, j draw an edge between i and j w.p. 1/2 so
that the events Ai,j = {i and j are connected by an edge} are (mutually)
independent for all pairs {i, j} with i 6= j ∈ [n]. Let X be the number of
triangles in the graph. That is, the number of {i, j, k} ⊂ [n] so that i 6= j,
j 6= k and i 6= k and each of the three pairs {i, j}, {i, k} and {j, k} has an
edge between them.
(a) Calculate E [X].
(b) Challenge, part (b) not marked. Calculate E (X − E[X])2 .
(7) Challenge, not marked (yet, important!). Let X be a random variable with a probability density function denoted by fX and a cumulative
distribution function denoted by FX . Assume that there is some interval
(a, b) (possibly a = −∞ or b = +∞ or both) such that fX (x) > 0 for all
x ∈ (a, b) and fX (x) = 0 for all x ∈
/ [a, b] (the last part holds trivially when
(a, b) = (−∞, ∞)).
Let g : (a, b) → R be a strictly increasing function which is differentiable
at every point in (a, b). Let (c, d) := {g(x) : a < x < b} be the image of g.
Let h : (c, d) → (a, b) be the inverse function of g. That is, (g ◦ h)(x) :=
g(h(x)) = x for all x ∈ (c, d) and (h ◦ g)(x) := h(g(x)) = x for all x ∈ (a, b).
Denote Z := g(X).
(a) Show that the cumulative distribution function of Z, denoted by FZ ,
is given by


z≤c
0
FZ (z) := FX (h(z)) c < z < d .


1
z≥d
(b) Show that the probability density function of Z, denoted by fZ , is
given by
(
(h(z))
c<z<d
fX (h(z))h0 (z) = fgX0 (h(z))
.
fZ (z) :=
0
otherwise
Hint: Part (a) is definition chasing. For part (b) recall that fZ can be
recovered from FZ via differentiation. Use your answer to part (a) together
1
with the chain rule and the formula h0 (x) = g0 (h(x))
.
3
(8) Challenge, not marked. In a town, there are on average 2.3 children in a
family and a randomly chosen child has on average 1.6 siblings. Determine
the variance of the number of children in a randomly chosen family.
(9) Challenge, not marked. The purpose of this question is to demonstrate
the sad fact that “on average, your friends have more friends than you”.
Consider a community of n people. Let m1 , . . . , mn−1 ∈ {0, 1, 2, . . . , n}
Pn−1
be such that
i=1 mi = n. Assume that there are precisely mi people
in the community which have i friends in the community (by assumption,
m0 = 0). Assume that being friends is a symmetric relationship (if person
i is a friend of person j then person j is a friend of person i).
Choose uniformly at random a person from the community. Let X be the
number of friends that person has. Now choose a second person uniformly
at random among the friends of the first person which was chosen. Let Y
be the number of friends of that person.
Show that E[X] ≤ E[Y ] and that E[X] = E[Y ] only if everyone has
exactly the same number of friends.
Pn−1
Hint: Since P[X = i] = mni for all i, we have that E[X] = i=1 i mni . Show
that P[Y = i] =
Q 3.
imi
nE[X]
for all i and so E[Y ] =
E[X 2 ]
E[X] .
Finally, use part (e) of
imi
for
The only hard part from the hint is showing that P[Y = i] = nE[X]
all i. To show this, imagine the n people are labeled by 1, 2, . . . , n. Let
Ui be the number of the ith person chosen (i = 1, 2). Argue that there
Pn−1
are i=1 imi (which happens to equal nE[X]) different possibilities that
(U1 , U2 ) can take, and each is obtained with equal probability. Then argue
that exactly imi of them satisfy that U2 has i friends in the community.