Math 426: Probability
MWF 1pm, Gasson 310
Homework 8 Selected Solutions
Exercises from Ross, §5 and §6:
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§5,
§5,
§6,
§6,
p.
p.
p.
p.
212
214
271
275
‘Problems’ (40)
‘Theoretical exercises’ (5, 11, 12)
‘Problems’ (2, 9, 10, 15, 16, 23)
‘Theoretical exercises’ (6, 11)
§6, Problem 16. Let P1 , . . . , Pn denote n points that are independently chosen at random on the
circumference of a circle. Let A denote the event that all the points are contained in some semicircle,
and let Ai be the event that all the points lie in the semicircle beginning at the point Pi and going
clockwise for 180◦ , i = 1, . . . , n.
(a) Express A in terms of the Ai .
(b) Are the Ai mutually exclusive?
(c) Find P (A).
Solution (a). Evidently, if all the points lie in a semicircle clockwise from the starting point Pi , then
all the points lie in some semicircle. Thus we have Ai ⊂ A for each i = 1, . . . , n. On the other hand,
if all the points lie in a semicircle, then we may choose the point Ai closest to the counterclockwise
boundary point of the semicircle. The semicircle that proceeds clockwise starting from Ai must contain
each of the other points, so we have A ⊂ ∪ni=1 Ai . Thus
A = ∪ni=1 Ai .
Solution (b). An outcome ω is in both Ai and Aj if and only if all the points are both in the
clockwise semicircle starting from Pi anfd in the clockwise semicircle starting from Pj . In particular,
this occurs if either Pi = Pj , or Pi and Pj are the endpoints of a common diameter and all the points
coincide with either Pi or Pj . Both of these scenarios are possible, so the Ai are not mutually exclusive.
Note, however, that in either case (at least for n > 2) we must have an outcome in which some pair of
points coincide.
Solution (c). Though these events are not mutually exclusive, we nonethless have the following:
P (Ai ∩Aj ) = 0 for any i 6= j. Indeed, we’ve seen in part (b) that Ai ∩Aj is a subset of the event that a pair
of the points coincide. This event has zero probability: P (Pi = Pj ) = P (Pi = q|Pj = q) = P (Pi = q),
since the random variables Pi and Pj are independent, and the probability of a continuous random
variable having a given exact value is 0.
By the inclusion-exclusion formula, we have
P (A) =
n
X
i=1
P (Ai ) +
n
X
X
r+1
(−1)
r=2 i1 <...<ir
P (Ai1 ∩ . . . ∩ Air ) =
n
X
i=1
1
P (Ai ).
The probability P (Ai ) = (1/2)n−1 by independence of the random variables Pj from Pi (for j 6= i), so
we have P (A) = n(1/2)n−1 .
§6, Theoretical Exercise 11. Let X1 , X2 , X3 , X4 , X5 be independent continuous random variables
having a common distribution function F and density function f , and set
I = P (X1 < X2 < X3 < X4 < X5 ).
(a) Show that I does not depend on F .
Hint: Write I as a five-dimensional integral and make a change of variables.
(b) Evaluate I.
(c) Give an intuitive explanation for your answer to (b).
Solution (a) and (b). Because X1 , . . . , X5 are independent and continuous, they have joint density
function g given by
g(x1 , x2 , x3 , x4 , x5 ) = fX1 (x1 )fX2 (x2 )fX3 (x3 )fX4 (x4 )fX5 (x5 ),
where fXi is the density function of Xi . For ease in notation, we will write x for the five-dimensional
vector (x1 , x2 , x3 , x4 , x5 ). Since each of these density functions is given by the same function f , we have
g(x) = f (x1 )f (x2 )f (x3 )f (x4 )f (x5 ).
By the definition of the joint density function, we have
ZZZZ
I = P (X1 < X2 < X3 < X4 < X5 ) =
f (x1 )f (x2 )f (x3 )f (x4 )f (x5 ) dx1 dx2 dx3 dx4 dx5 .
x1 <x2 <x3 <x4 <x5
Note that the region we are integrating over is the set of x where x1 < x2 < x3 < x4 < x5 . In other
words, this is the set of x where x5 is between −∞ and ∞, x4 is between −∞ and x5 , x3 is between
−∞ and x4 , x2 is between −∞ and x3 , and x1 is between −∞ and x2 . This allows us to rewrite the
integral as an iterated integral:
ZZZZ
I=
f (x1 )f (x2 )f (x3 )f (x4 )f (x5 ) dx1 dx2 dx3 dx4 dx5
x1 <x2 <x3 <x4 <x5
Z∞ Zx5 Zx4 Zx3
Zx2
=
f (x1 )f (x2 )f (x3 )f (x4 )f (x5 ) dx1 dx2 dx3 dx4 dx5
−∞ −∞ −∞ −∞ −∞
Z∞
Zx5
=
f (x5 )
−∞
Zx4
Zx3
f (x4 )
−∞
Zx2
f (x3 )
−∞
f (x2 )
−∞
f (x1 ) dx1 dx2 dx3 dx4 dx5 .
−∞
Z
a
Each of the Xi has the same cumulative distribution function F (a) =
f (x) dx, so the last line gives
−∞
Z∞
I=
Zx5
f (x5 )
−∞
−∞
Zx4
f (x4 )
Zx3
f (x3 )
−∞
f (x2 )F (x2 ) dx2 dx3 dx4 dx5 .
−∞
Making the change of variable u2 = F (x2 ), we have du2 = F 0 (x2 ) dx2 = f (x2 ) dx2 . Since
lim F (x2 ) = 0
x2 →−∞
2
(by the properties of the cumulative distribution function), the limits of the integral after the change of
variables go from 0 to F (x2 ), so that
Z∞
I=
Zx5
f (x5 )
−∞
Z∞
=
f (x4 )
−∞
Zx5
f (x5 )
−∞
FZ(x3 )
Zx4
f (x3 )
−∞
Zx4
−∞
0
f (x3 ) ·
f (x4 )
u2 du2 dx3 dx4 dx5
1
F (x3 )2 dx3 dx4 dx5 .
2
−∞
Repeating the same change of variable several times (letting ui = F (xi ) for each i = 3, 4, 5) we compute:
Z∞
I=
Zx5
f (x5 )
−∞
Z∞
=
−∞
1
f (x4 ) ·
2
−∞
1
f (x5 ) ·
3·2
FZ(x4 )
u23
Z∞
du3 dx4 dx5 =
u34 du4 dx5 =
0
f (x5 )
−∞
0
FZ(x5 )
Z∞
Zx5
f (x4 ) ·
1
F (x4 )3 dx4 dx5
3·2
−∞
1
1
F (x5 )4 dx5 =
f (x5 ) ·
4·3·2
4·3·2
−∞
Z1
u45 du5 =
1
5!
0
Solution (c). Since the Xi are independent variables with the same distribution functions, the
likelihood of any one ordering (e.g. X1 < X2 < X3 < X4 < X5 ) is the same as any other ordering—this
is because switching some of the names of the variables doesn’t change the probability of this event
happening. There are exactly 5! possible orderings of the Xi , and they have equal probabilities that
sum to 1, so each has probability 1/5!.
(There is something not totally trivial happening in this argument: Why is the sum of these probabilities equal to 1? This is so only because the probability of any pair being equal is 0, and that is
because they are all continuous random variables. For instance, we have
ZZ
P (X1 = X2 ) =
f (x1 )f (x2 ) dx1 dx2 ,
(x1 ,x2 ):x1 =x2
but this double integral is being evaluated over the line x1 = x2 , which has zero area. Thus the integral
vanishes, and P (X1 = X2 ) = 0 as claimed.)
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