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MTH 202 : Probability and Statistics
Homework 3
13th January, 2017
(1) Any point in the interval [0, 1) can be represented by its decimal expansion 0.x1 x2 . . . . Suppose a point is chosen at random
from the interval [0, 1). Let X be the first digit in the decimal
expansion representing the point. Compute the density of X.
[Ref : Exercise-1, Hoel, Port, Stone, Page-77]
Solution : Note that every point x ∈ [0, 1) has a unique decimal expansion except the rational numbers :
x = 0.x1 x2 . . . xN = 0.x1 x2 . . . xN −1 (xN − 1)99 . . .
where x1 , x2 , . . . , xN ∈ {0, 1, . . . , 9} with xN ≥ 1. Hence the
only points which does not have a unique first digit in its decimal expansion (considering the above ambiguous cases) are
1 2
9
, ,...,
10 10
10
corresponding to N = 1, as above. For these numbers we would
disallow the second kind of expansion, which mean the allowed
decimal expansion of i/10 is
i
= 0.i (and not 0.(i − 1)999 . . . )
10
for i = 1, 2, . . . , 9. With this note, the random variable X
(which take values 0, 1, . . . , 9) is well defined. Then X(ω) = i
if ω ∈ [ 10i , i+1
) and hence we can assign the probability P (X =
10
1
i) = 10 which is the length of the interval [ 10i , i+1
). The density
10
function f : R → R is thus defined as
(
1
if x = 0, 1, . . . , 9,
f (x) = 10
0 otherwise
which come from the relation f (x) = P (X = x).
(2) Let N be a positive integer and let
(
c2x if x = 1, 2, . . . , N,
f (x) =
0
otherwise
1
2
Find the value of c such that f is a probability density.
[Ref : Exercise-4, Hoel, Port, Stone, Page-77]
Solution : The values of x where f (x) > 0 are x = 1, 2, . . . , N .
The
PN necessary condition to make f a probability density is
x=1 f (x) = 1. The value c thus satisfy
N
X
c2x = 1 ⇒ 2c(2N − 1) = 1
x=1
and hence c =
1
.
2N +1 −2
(3) Suppose a box has 12 balls labelled 1, 2, . . . , 12.Two independent repetitions are made of the experiment of selecting a ball
at random from the box. Let X denote the larger of the two
numbers on the balls selected. Compute the density of X.
[Ref : Exercise-8, Hoel, Port, Stone, Page-78]
Solution : Denote by X1 and X2 , the random variables which
represent the two independent experiments of selecting a ball
at random from the box. Then X = max(X1 , X2 ). Both of the
random variables X1 , X2 satisfy the uniform distribution on 12
discrete points given by :
(
1
if x = 1, 2, . . . , 12,
P (Xi = x) = 12
0 otherwise
for i = 1, 2. For x = 1, 2, . . . , 12, we have
P (X ≤ x) = P (X1 ≤ x, X2 ≤ x) = P (X1 ≤ x)P (X2 ≤ x)
since X1 and X2 are independent. Now P (Xi ≤ x) =
x 2
hence P (X ≤ x) = 12
. Then,
x
,
12
and
2
x 2
P (X = x) = P (X ≤ x) − P (X ≤ x − 1) = 12
− x−1
=
12
2x−1
. The density function f corresponding to X is thus given
144
by :
(
2x−1
if x = 1, 2, . . . , 12,
f (x) = 144
0
otherwise
3
(4) Let X and Y be independent random variables each having the
uniform density on {0, 1, . . . , N }. Find the densities of :
(a) min(X, Y ), (b) max(X, Y ), (c) |Y − X|. [Ref : Exercise-15,
Hoel, Port, Stone, Page-78]
Solution : Both of the independent random variables satisfy
the density given by
(
1
if x = 0, 1, . . . , N,
f (x) = N +1
0
otherwise
For (a) let Z = min(X, Y ). Then, for x ∈ {0, 1, . . . , N } we have
P (Z ≥ x) = P (X ≥ x, Y ≥ x) = P (X ≥ x)P (Y ≥ x). Hence
2
P (Z ≥ x) = N −x+1
. Then we have :
N
P (Z = x) = P (Z ≥ x) − P (Z ≥ x + 1) =
2(N − x) + 1
(N + 1)2
for x = 0, 1, . . . , N .
For (b) let Z = max(X, Y ). Then, for x ∈ {0, 1, . . . , N } we
have P (Z ≤ x) = P (X ≤ x, Y ≤ x) = P (X ≤ x)P (Y ≤ x).
2
Hence P (Z ≤ x) = x+1
. Then we have :
N
P (Z = x) = P (Z ≤ x) − P (Z ≤ x − 1) =
2x + 1
(N + 1)2
for x = 0, 1, . . . , N .
For (c) let Z = |Y − X|. Notice that if x 6= 0,
P (|Y − X| = x) = P (Y − X = x) + P (Y − X = −x)
Hence for x 6= 0
P (Y − X = x) =
N
X
P (Y − X = x, X = t)
t=0
=
N
X
P (Y = x + t, X = t) =
t=0
N
−x
X
P (Y = x + t)P (X = t)
t=0
Hence
P (Y − X = x) =
(N − x + 1)
(N + 1)2
Similarly,
P (Y − X = −x) =
(N − x + 1)
(N + 1)2
4
Combining these we have
2(N − x + 1)
(N + 1)2
while x 6= 0. While x = 0 we have
P (|Y − X| = x) =
N
X
P (|Y − X| = 0) = P (X = Y ) =
P (X = Y, X = t)
t=0
=
N
X
P (Y = t, X = t) =
t=0
N
X
P (Y = t)P (X = t)
t=0
Hence,
P (|Y − X| = 0) =
1
N +1
(5) Let X and Y be independent random variables having geometric densities with parameters p1 and p2 respectively. Find the
density of (a) min(X, Y ), (b) X + Y . [Ref : Exercise-17, Hoel,
Port, Stone, Page-79]
Solution : (a) The densities of the random variables X and Y
satisfy
(
p1 (1 − p1 )x if x = 0, 1, . . .
P (X = x) =
0
otherwise
and
(
p2 (1 − p2 )y
P (Y = y) =
0
if y = 0, 1, . . .
otherwise
For an integer x ≥ 0 we have,
∞
∞
X
X
P (X ≥ x) =
P (X = t) =
p1 (1 − p1 )t = (1 − p1 )x
t=x
t=x
and similarly P (Y ≥ x) = (1 − p2 )x . Letting Z = min(X, Y )
we have
P (Z ≥ x) = P (X ≥ x, Y ≥ x) = P (X ≥ x)P (Y ≥ x)
and hence P (Z ≥ x) = (1 − p1 )x (1 − p2 )x . This implies,
P (Z ≤ x − 1) = 1 − P (Z ≥ x) = 1 − (1 − p1 )x (1 − p2 )x
Now,
P (Z = x) = P (Z ≤ x) − P (Z ≤ x − 1)
5
= [1 − (1 − p1 )(1 − p2 )](1 − p1 )x (1 − p2 )x = q(1 − q)x
for x ≥ 0, where q = p1 + p2 − p1 p2 . Note that while x = 0,
P (Z = 0) = P (Z ≤ 0) = 1 − (1 − p1 )(1 − p2 ) = q
Hence Z satisfies geometric density with parameter q.
For (b) set Z = X + Y . Then P (Z = x)
x
x
X
X
=
P (X + Y = x, Y = t) =
P (X = x − t)P (Y = t)
t=0
t=0
= p1 p2 (1 − p1 )x
=
Px
t=0
x+1
p1 p2
[(1−p2 )
p1 −p2
1−p2
1−p1
t
−(1−p1 )x+1 ] using the finite geometric sum.
(6) Suppose 2r balls are distributed at random into r boxes. Let
Xi denote the number of balls in box i.
(a) Find the joint density of X1 , . . . , Xr .
(b) Find the probability that each box contains exactly 2 balls.
[Ref : Exercise-21, Hoel, Port, Stone, Page-79]
Solution : (a) Let I := {0, 1, . . . , 2r}, and A ⊆ I r defined by
A = {(x1 , x2 , . . . , xr ) ∈ I r : x1 + x2 + . . . + xr = 2r}
For a vector (x1 , x2 , . . . , xr ) ∈ A, the joint density is given by
P (X1 = x1 , . . . , Xr = xr )
2r 2r−x1
. . . 2r−(x1 +xx2 r+...+xr−1 )
x1
x2
=
r2r
The numerator is equal to C(2r; x1 , . . . , xr ) and calculated using
induction in section 3.4.1, Page-68, Hoel, Port, Stone. This is
given by
(2r)!
C(2r; x1 , . . . , xr ) =
(x1 !) . . . (xr !)
Hence we have
(2r)!
P (X1 = x1 , . . . , Xr = xr ) =
(x1 !) . . . (xr !)r2r
Also P (X1 = x1 , . . . , Xr = xr ) = 0 if (x1 , x2 , . . . , xr ) 6∈ A.
For (b) setting (x1 , x2 , . . . , xr ) = (2, 2, . . . , 2) we have
P (X1 = 2, . . . , Xr = 2) =
(2r)!
2r r2r
6
(7) Use the Poisson approximation to calculate the probability that
at most 2 out of 50 given people will have invalid driver’s licences if normally 5% of the people do. [Ref : Exercise-23,
Hoel, Port, Stone, Page-80]
Solution : The probability of the event with Binomial distribution with parameters (50, p), where p = 0.05 (probability of
success) is given by
50 50
50 49
50 48
p +
p (1 − p) +
p (1 − p)2
50
49
48
The Poisson approximation as discussed in section 3.4.2, Page69, Hoel, Port, Stone, is given by
50 k
(50p)k −50p
p (1 − p)50−k ≈
e
k
k!
Hence the Poisson approximation to above probability is given
by
e−5/2 [1 + 5/2 + 1/2 × (5/2)2 ] = (53/8)e−5/2
(8) Let X be a non-negative integer valued random variable whose
2
probability generating function is given by ΦX (t) = eλ(t −1) ,
where λ > 0. Find fX . [Ref : Exercise-33, Hoel, Port, Stone,
Page-81]
Solution : Recall that the probability generating function of
a non-negative integer valued random variable X is given by
∞
X
P (X = x)tx
(−1 ≤ t ≤ 1)
ΦX (t) =
x=0
Here we have the Taylor series expansion
(λt2 )k
+ ...]
k!
Now equating the coefficients of the Taylor expansions we get
2 −1)
ΦX (t) = eλ(t
= e−λ [1 + (λt2 ) + . . . +
P (X = x) = e−λ
λx/2
(x/2)!
when x is a non-negative even integer, and P (X = x) = 0
otherwise. Hence the density function is given by
(
λx/2
e−λ (x/2)!
if x = 0, 2, 4, . . .
P (X = x) =
0
otherwise
7
(9) Let X and Y be independent random variables having Poisson
densities with parameters λ1 and λ2 respectively. Find P (Y =
y|X + Y = z) for y = 0, . . . , z. [Hint : Use Theorem-1(iii),
Page-75, Hoel, Port, Stone]
[Ref : Exercise-35, Hoel, Port, Stone, Page-81]
Solution : Using Theorem-1(iii), Page-75, Hoel, Port, Stone,
we can see that X + Y satisfies a Poisson distribution with
parameter λ1 + λ2 . Now, P (Y = y|X + Y = z) is
P (Y = y, X + Y = z)
P (X = z − y)P (Y = y)
=
=
P (X + Y = z)
P (X + Y = z)
since X and Y are independent. Further plugging the parameter values we have this
−(λ1 +λ2 )
−1
e−λ1 λ1 z−y e−λ2 λ2 y
e
(λ1 + λ2 )z
=
×
×
(z − y)!
y!
z!
z−y y
λ2
z
λ1
for y = 0, . . . , z
=
λ1 + λ2
λ1 + λ2
y
(10) Let X, Y and Z be independent random variables having Poisson densities with parameters λ1 , λ2 and λ3 respectively. Find
P (X = x, Y = y, Z = z|X + Y + Z = x + y + z)
for nonnegative integers x, y and z. [Hint : Use Theorem-1(iii),
Page-75, Hoel, Port, Stone]
[Ref : Exercise-36, Hoel, Port, Stone, Page-81]
Proof : As in the previous exercise, using Theorem-1(iii), Page75, Hoel, Port, Stone, we can see that X + Y + Z satisfies a
Poisson distribution with parameter λ1 + λ2 + λ3 . Now,
P (X = x, Y = y, Z = z|X + Y + Z = x + y + z) is
P (X = x, Y = y, Z = z, X + Y + Z = z)
=
P (X + Y + Z = x + y + z)
P (X = x)P (Y = y)P (Z = z)
=
P (X + Y + Z = x + y + z)
8
since X, Y and Z are independent.
parameter values we have this
Further plugging the
e−λ1 λ1 x e−λ2 λ2 y e−λ3 λ1 z
e−(λ1 +λ2 +λ3 ) (λ1 + λ2 + λ3 )(x+y+z)
=
×
×
×
x!
y!
z!
(x + y + z)!
(x + y + z)!
λ1 x λ2 y λ3 z
=
×
x!y!z!
(λ1 + λ2 + λ3 )x+y+z
!−1