PROBABILITY SOLUTION 2
1. (a) lim sup An =
Sk
j=1 Bj ,
lim inf An =
Tk
j=1 Bj .
(b) The relation is lim inf An ⊆ lim inf Ank and lim sup Ank ⊆ lim sup An .
2. Fix some ω ∈ Ω.
Ilim inf An (ω) = 1 if and only if there is some N such that for every n > N ω is
in An . This is equivalent to IAn (ω) = 0 only for a finite number of sets An so
lim inf IAn (ω) = 1.
Similarly, Ilim sup An (ω) = 1 is equivalent to IAn (ω) = 1 for infinite number of sets An
and therefore is equivalent to lim sup IAn (ω) = 1.
3. A proof that Z is a random variable:
Let x ∈ R.
{w ∈ Ω : Z(w) ≤ x} = {w ∈ Ω : min(X(w), Y (w)) ≤ x} =
{w ∈ Ω : X(w) ≤ x} ∪ {w ∈ Ω : Y (w) ≤ x}.
Each of these two sets is an event, since X,Y are random variables. Hence, their union
is also an event and Z is a random variable.
The proof for T is very similar.
4. (a) For a ∈ R, if a < 0 then {X 2 ≤ a} = ∅, and if a ≥ 0 then
√
√
{X 2 ≤ a} = {− a ≤ X ≤ a}. The first is always an event, and the second is
an event since we know that X is a random variable.
(b) You saw in class that if X, Y are random variables, then so are X + Y and cX
for c ∈ R. Therefore, both X + Y and X − Y are random variables. From (a)
we know that (X + Y )2 and (X − Y )2 are random variables, which implies that
XY = 14 [(X + Y )2 − (X − Y )2 ] is a random variable.
5. (a) Since the radius is always between zero and one, for r < 0 the probability is zero
2
and for r ≥ 1 the probability is one. For 0 ≤ r < 1, P (Ar ) = πrπ = r2 .
(b) For α < −π the probability is zero and for α ≥ −π the probability is one. For
−π ≤ a < π, the probability is a+π
2π .
1