Probability A - Tutorial 2
Solution to Exercise 7
Elia Bisi
Department of Statistics
University of Warwick
April 23, 2015
Question. Find a space Ω̃ (simpler than Ω) and a distribution P̃ over Ω̃ such that, for
some à ⊆ Ω̃,
P̃(Ã) = P(Am )
∀m = 4, . . . , 13 .
Solution. Consider the space of permutations on 4 elements
n
o
Ω̃ := ω̃ = (ω̃0 , ω̃1 , ω̃2 , ω̃3 ) permutations of {0, 1, 2, 3} ,
equipped with the uniform probability distribution, i.e.
1
P̃ {ω̃} =
4!
∀ω̃ ∈ Ω̃ .
Set
à := {ω̃ ∈ Ω̃ : ω̃3 , 0} .
Then P̃(Ã) = P(Am ) for all m = 4, . . . , 13 (identifying ball m with 0 - this is just to remark
that P(Am ) does not actually depend on m). To calculate P̃(Ã) explicitly, we note that Ã
has 3 · 3! elements, so
#Ã 3 · 3! 3
P̃(Ã) =
=
= .
4!
4
#Ω̃
Alternatively, we may consider the random permutation (X̃0 , X̃1 , X̃2 , X̃3 ), where
X̃k : Ω̃ → R,
X̃k (ω̃) = ω̃k ,
∀k = 0, 1, 2, 3 ,
and note that the distribution of X̃3 is uniform on {0, 1, 2, 3} by symmetry, hence
P̃(Ã) = P̃(X̃3 , 0) = 1 − P̃(X̃3 = 0) = 1 −
1 3
= .
4 4
There is an equivalent way to compute P(Am ) without introducing further probability
−1
spaces. Let (X1−1 , . . . , X13
) be the inverse permutation of (X1 , . . . , X13 ), i.e. Xi−1 = j if and only
−1
−1
−1
−1
if Xj = i (intuitively, Xi represents the instant when ball i is drawn). Let (X(1)
, X(2)
, X(3)
)
−1
be the random vector (X1−1 , X2−1 , X3−1 ) sorted in ascending order (intuitively, X(1)
is the first
−1
−1
instant when a white ball is drawn, X(2)
is the second one and X(3)
is the third one), so
1
−1
−1
−1
−1 < X −1 }.
that X(1)
< X(2)
< X(3)
with probability 1. According to this notation, Am = {Xm
(3)
We note that one and only one of the following events happens:
−1
−1
−1
−1
E0 := {Xm
< X(1)
< X(2)
< X(3)
};
−1
−1
−1
−1
E1 := {X(1)
< Xm
< X(2)
< X(3)
};
−1
−1
−1
−1
E2 := {X(1)
< X(2)
< Xm
< X(3)
};
−1
−1
−1
−1
E3 := {X(1)
< X(2)
< X(3)
< Xm
}.
Since the probability distribution on the (inverse) permutations is uniform, the probabilities of all events Ei are equal. Since P(E0 ) + P(E1 ) + P(E2 ) + P(E3 ) = 1, the common value is
1/4. Consequently,
1 3
P(Am ) = 1 − P(E3 ) = 1 − = .
4 4
2