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Queen’s University
Department of Mathematics and Statistics
STAT 353
Midterm Examination February 12, 2009
• Total points = 30. Duration = 58 minutes.
• This is a closed book exam.
• One 8.5 by 11 inch sheet of notes, written on both sides, is permitted.
• A simple calculator is permitted.
• Write the answers in the space provided, continue on the backs of pages if needed.
• SHOW YOUR WORK CLEARLY. Correct answers without clear work showing
how you got there will not receive full marks.
• Marks per part question are shown in brackets at the right margin.
• The last page contains formulas you may find useful. Please check this page first.
Marks: Please do not write in the space below.
Problem 1 [10]
Problem 2 [10]
Problem 3 [10]
Total: [30]
STAT 353 -- Midterm Exam, 2009
Page 2 of 5
1. Let S(x, y) denote the square on the plane which has side length 2 and is centered at the
point (x, y). Let (X1 , Y1 ) be a point chosen uniformly at random from the square S(1, 1)
and let (X2 , Y2 ) be a point chosen uniformly at random from the square S(−1, −1). Find
P ((X2 , Y2 ) ∈ S(X1 , Y1 )) (i.e., the probability that the point (X2 , Y2 ) is inside the square
of side length 2 centered at (X1 , Y1 ).
[10]
Solution: The joint pdf of (X1 , Y1 , X2 , Y2 ) is
(
1
for (x1 , y1 ) ∈ S(1, 1) and (x2 , y2 ) ∈ S(−1, −1)
16
f (x1 , y1 , x2 , y2 ) =
0 otherwise.
The point (X2 , Y2 ) is inside the square S(X1 , Y1 ) if and only if X2 ≥ X1 − 1 (X1 − 1 is
the left edge of S(X1 , Y1 )) and Y2 ≥ Y1 − 1 (Y1 − 1 is the bottom edge of S(X1 , Y1 )). The
relevant integral is
P ((X2 , Y2 ) ∈ S(X1 , Y1 )) = P (X2 ≥ X1 − 1, Y2 ≥ Y1 − 1)
Z 1Z 1Z 0 Z 0
1
dy2 dx2 dy1 dx1
=
0
0
x1 −1 y1 −1 16
Z 1Z 1Z 0
1
=
(1 − y1 )dx2 dy1 dx1
16 0 0 x1 −1
Z 1Z 1
1
=
(1 − y1 )(1 − x1 )dy1 dx1
16 0 0
Z 1
1
1
=
(1 − x1 )dx1
16 0 2
1
1
1
=
× = .
32 2
64
STAT 353 -- Midterm Exam, 2009
Page 3 of 5
2. Let X1 , . . . , Xn be a sequence of independent random variables uniformly distributed on
the interval (0, 1), and let X(1) , . . . , X(n) denote their order statistics. For fixed k let gn (x)
denote the probability density function of nX(k) . Find gn (x) and show that
( k−1
x
e−x for x ≥ 0
(k−1)!
lim gn (x) =
n→∞
0
for x < 0
which is the Gamma(k,1) density.
Solution: The pdf of X(k) is
(
fk (xk ) =
[10]
n!
xk−1 (1
(k−1)!(n−k)! k
− xk )n−k for 0 < xk < 1
0
otherwise.
Let X = nX(k) . Then the set of possible values of X (the sample space of X) is {x : 0 <
x < n}. So for x ∈ (0, n), by the (1-dimensional) change of variable formula the pdf of X
is given by
x k−1 1
x n−k
(n − 1)!
gn (x) = fk (x/n) =
1−
n
(k − 1)!(n − k)! n
n
and gn (x) = 0 for x ∈
/ (0, n). Now fix x > 0 and let n > x. Then
x k−1 (n − 1)!
x n−k
1−
(k − 1)!(n − k)! n
n
(n − 1) × . . . × (n − k + 1)
x −k
1
x n
k−1
=
1
−
x
1
−
nk−1
n
(k − 1)!
n
1
→
xk−1 e−x ,
(k − 1)!
gn (x) =
as desired, since
(n − 1) × . . . × (n − k + 1)
nk−1
n−1
n−k+1
=
× ... ×
n
n
→ 1 × . . . × 1 = 1,
(1 − x/n)−k clearly converges to 1 as n → ∞ (for fixed k), and (1 − x/n)n → e−x as
n → ∞ (from calculus). Since gn (x) = 0 for x < 0 it is obvious that gn (x) → 0 as n → ∞
if x < 0.
STAT 353 -- Midterm Exam, 2009
Page 4 of 5
3. Suppose we take an ordinary deck of 52 cards, randomly shuffled, and flip all 52 cards
over, one at a time. Let X be the number of times we flip a card that has the same face
value as the previously flipped card. Find E[X].
[10]
Solution: Let Xi , i = 2, . . . , 52 be defined by
(
1 if the ith card flipped has the same face value as the previously flipped card
Xi =
0 otherwise.
Then X = X2 + . . . + X52 and so E[X] = E[X2 ] + . . . + E[X52 ], where
E[Xi ] = P (ith and (i − 1)st cards flipped have the same face value).
The above probability can be computed by counting all sequences of 52 cards in which
the cards in positions i − 1 and i have the same face value, then dividing this number by
52!, the total number of sequences. Since the number of sequences in which the cards in
positions i − 1 and i both have face value j is the same as the number of sequences in
which the cards in positions i − 1 and i both have face value k, for any j 6= k, and there
are 13 possible face values, we can write
# of sequences in which the cards in positions i − 1 and i have the same face value
= 13 × # of sequences in which the cards in positions i − 1 and i have face value 2
The number of sequences in which the cards in positions i − 1 and i have face value 2 can
be determined as follows. There are 4 cards with face value 2. Choose 2 of them; there
are 42 = 6 ways to do that. Now order them; there are 2 ways to do that. Now put the
ordered pair into positions i − 1 and i, then place the other 50 cards into the remaining
50 positions; there are 50! ways to do that. So the total number of ways to put the 52
cards into the 52 positions so that positions i − 1 and i both have a 2 is 6 × 2 × 50!. Then
the total number of ways to put the 52 cards into the 52 positions so that positions i − 1
and i have the same face value is 13 × 6 × 2 × 50!. Therefore,
13 × 6 × 2 × 50!
52!
3
1
=
= .
51
17
P (ith and (i − 1)st cards flipped have the same face value) =
Note that this is independent of i, which should be intuitively obvious. Finally,
E[X] =
51
= 3.
17
STAT 353 -- Midterm Exam, 2009
Formulas:
Uniform distribution on the interval (0, 1) has pdf:

1 if 0 < x < 1
f (x) =
0 otherwise,
Page 5 of 5