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Student Number
Queen’s University
Department of Mathematics and Statistics
STAT/MTHE 353
Midterm Examination February 15, 2013
• 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/MTHE 353 -- Midterm Exam, 2013
Page 2 of 6
1. Let X and Y be jointly continuous with joint probability density function given by
(
1
if 1 < x < ∞ and 1 < y < ∞
x2 y 2
f (x, y) =
0 otherwise.
Find the probability density function of U = XY . Hint: Define V = Y and first find the
joint density of U and V . Be careful about the sample space of (U, V ).
[10]
Solution: Let u = xy and v = y. Since x > 1 and y > 1, for given v, u > v. That is, the
sample space of (U, V ) is
Q = {(u, v) : 1 < v < u < ∞}.
The inverse transformation is x = u/v and y = v, with Jacobian
"
#
1
u
−
1
v2
J = det v
= .
v
0 1
The joint density of U and V is then given by
( 2
(
v 1 1
1
for
1
<
v
<
u
<
∞
for 1 < v < u < ∞
1
u2 v 2 v
u2 v
g(u, v) = f (u/v, v) =
=
v
0
otherwise
0
otherwise.
Then the marginal density of U is
1
g(u) = 2
u
for u > 1 and g(u) = 0 for u ≤ 1.
Z
1
u
u
1
1
ln u
dv = 2 ln v = 2 ,
v
u
u
1
STAT/MTHE 353 -- Midterm Exam, 2013
Page 3 of 6
2. Suppose that 5 cards are drawn at random, with replacement, from an ordinary deck of 52
playing cards. What is the probability that in the 5 drawn cards there are exactly 3 red
cards and exactly 3 of the cards are either hearts or clubs?
[10]
Solution: Let A denote the event that exactly 3 of the cards are red and exactly 3 of the
cards are either hearts or clubs. Let X1 be the number of hearts drawn, X2 the number
of diamonds, X3 the number of clubs, and X4 the number of spades. Then the vector
(X1 , X2 , X3 , X4 ) has a Multinomial distribution with parameters 5 and 1/4, 1/4, 1/4, 1/4.
In terms of X1 , X2 , X3 , and X4 , the event A can be expressed as
{X1 = 1, X2 = 2, X3 = 2, X4 = 0}
∪ {X1 = 2, X2 = 1, X3 = 1, X4 = 1}
∪ {X1 = 3, X2 = 0, X3 = 0, X4 = 2}.
(Another way to look at it is that this is the event {X1 + X2 = 3} ∩ {X1 + X3 = 3}).
From the multinomial distribution we have
P (A) = P (X1 = 1, X2 = 2, X3 = 2, X4 = 0)
+ P (X1 = 2, X2 = 1, X3 = 1, X4 = 1)
+ P (X1 = 3, X2 = 0, X3 = 0, X4 = 2)
5 5!
5!
5!
1
+
+
=
4
1!2!2!0! 2!1!1!1! 3!0!0!2!
30 + 60 + 10
100
25
=
=
=
≈ 0.098.
1024
1024
256
STAT/MTHE 353 -- Midterm Exam, 2013
Page 4 of 6
3. A bird slingshots itself towards a pig, which is sitting without moving but is taunting the
bird with snorts. The slingshot seems to be the bird’s only means of flight. The bird
lands at a point (X, Y ). The pig is sitting at point (0,0). The pig has girth. If the bird
lands within a disc of radius 1 centred at (0,0), the pig will be knocked out.
(a) If X and Y are independent and normally distributed with mean 0 and variance 1,
what is the probability that the pig gets knocked out?
[5]
Solution: We want the probability that X 2 + Y 2 ≤ 1. We know that X 2 and Y 2
each have a χ21 distribution and they are independent, so that X 2 + Y 2 has a χ22
distribution, which is the same as an Exponential distribution with parameter 1/2.
Thus,
P (X 2 + Y 2 ≤ 1) = 1 − e−1/2 ≈ 0.393.
STAT/MTHE 353 -- Midterm Exam, 2013
Page 5 of 6
(b) The bird misses. Two of its friends are indignant, and produce a better slingshot.
But the pig has built a wooden shelter around himself. Not being too bright, or
perhaps taunting the birds, he had built the foundation of his shelter out of glass.
If the two birds, using two slingshots, can hit the ground so they both land within
1 unit of distance from one another, they will create a resonance that will shatter
the glass. If bird number 1 lands at point (X1 , Y1 ) and bird number 2 lands at point
(X2 , Y2 ), what is the probability that they will succeed in shattering the glass?
Suppose that X1 , Y1 , X2 , Y2 are independent and normally distributed with mean 0
and variance 1. Note that the squared distance between the two landing points is
(X1 − X2 )2 + (Y1 − Y2 )2 .
[5]
Solution: We want P ((X1 − X2 )2 + (Y1 − Y2 )2 ≤ 1). Now X1 − X2 and Y1 − Y2 are
both normally distributed with mean 0 and variance 2, and they are independent,
√
√
so (X1 − X2 )/ 2 and (Y1 − Y2 )/ 2 are both normally distributed with mean 0 and
variance 1, and they are independent. Then
(X1 − X2 )2 (Y1 − Y2 )2
+
2
2
has a χ22 distribution, which is the same as the Exponential distribution with parameter 1/2. Then we can compute
(X1 − X2 )2 (Y1 − Y2 )2
1
2
2
+
≤
P ((X1 − X2 ) + (Y1 − Y2 ) ≤ 1) = P
2
2
2
= 1 − e−1/4 ≈ 0.221.
STAT/MTHE 353 -- Midterm Exam, 2013
Page 6 of 6
Formulas:
• The standard normal distribution has pdf
1
2
f (x) = √ e−x /2
2π
for x ∈ R