Mathematics 4800
Midterm 1- Solution
October 7th 2010
Partial credit will be awarded for your answers, so it is to your advantage to explain your
reasoning and what theorems you are using when you write your solutions. Please answer
the questions in the space provided and show your computations.
Good luck!
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2
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6
Total
Name:
1
I. (10pts) In a certain college, 25% of the students failed mathematics, 15% of the students
failed chemistry and 10% of the students failed both mathematics and chemistry. A student
is selected at random.
1. If he failed chemistry, what is the probability that he failed mathematics?
2. If he failed mathematics, what is the probability that he failed chemistry?
3. What is the probability that he failed mathematics or chemistry?
Solution: Let us denote by
M = {Students who f ailed M athematics} ,
C = {Students who f ailed Chemistry} ,
T
then P (M ) = 0.25, P (C) = 0.15, and P (M C) = 0.10.
1. The probability that a student failed Mathematics given that he (she) failed Chemistry
is
T
0.10
2
P (M C)
=
= .
P (M |C) =
P (C)
0.15
3
2. The probability that a student failed Chemistry given that he (she) failed Mathematics
is
T
P (C M )
0.10
2
P (C|M ) =
=
= .
P (M )
0.25
5
3. The probability that a student failed Chemistry or Mathematics is
P (C
[
M ) = P (M ) + P (C) − P (M
2
\
C) = 0.25 + 0.15 − 0.10 = 0.3 =
3
.
10
II. (10pts) Let X and Y be independent random variables taking values in the positive
1
integers and having the same mass function f (k) = k for k = 1, 2, . . . . Find
2
1. P (min {X, Y } ≤ k).
2. P (Y > X).
3. P (X > kY ) for a given positive integer k.
Solution:
1.
P (min (X, Y ) ≤ k) = 1 − P (min (X, Y ) > k) = 1 − P (X > k, Y > k)
" ∞
#2
X
= 1 − P (X > k) P (Y > k) = 1 − [P (X > k)]2 = 1 −
P (X = m)
m=k+1
"
1−
∞
X
1
2m
m=k+1
#2
"
=1−
∞
X
1
2k+1
1
2m
m=0
#2
On the other side the infinite geometric series
P∞
1
m=0 2m
1
P (min (X, Y ) ≤ k) = 1 − k
2
= 2. Hence
2
.
2.
P (Y > X) =
∞
X
P (Y > m, X = m)
m=1
∞
X
m=1
∞
X
P (Y > m) P (X = m)
∞
X
P (Y = l) P (X = m)
m=1 l=m+1
∞
∞
X
X
1 1
2m 2l
m=1 l=m+1
∞
X
1
2m
m=1
On the other side
P∞
1
l=0 2l
∞
m+1
X
1 X 1
−
2l
2l
l=0
l=0
= 2 and
Pm+1
l=0
1
2l
P (Y > X) =
3
=
!
1−(1/2)m+1
.
1−1/2
∞
X
1
1
= .
2m
2
3
m=1
Hence
Pm+1
l=0
1
2l
=
1
2m
and
3.
P (X > kY ) =
∞
X
P (X > ky, Y = y)
y=1
∞
X
y=1
∞
X
P (X > ky) P (Y = y)
P (Y = y)
y=1
∞
X
P (X = l)
l=ky+1
On the other side using the index shift and the infinite geometric series we get that
∞
X
P (X = l) =
l=ky+1
1
.
2ky
Hence,
∞
X
1 1
P (X > kY ) =
2y 2ky
y=1
=
∞
X
y=1
1
2(k+1)y
4
=
1
2(k+1)
−1
III. (10pts)
Let X be a discrete random variable such that
1
ω∈A
X(ω) =
2 otherwise
1. Find the distribution function of X if P (A) = 13 .
2. Find the expected and the variance of X.
Solution:
1. The distribution function of X is defined as
FX (x) = P (X ≤ x) .
If x < 1 then P (X ≤ x) = 0.
If 1 ≤ x < 2 then P (X ≤ x) = P (X = 1) = P (A) = 13 .
If x ≥ 2 then P (X ≤ x) = P (X = 1 or X = 2) = P (A) + P (Ac ) = 1.
Hence

x<1
 0
1
1≤x<2
FX (x) =
 3
1
x≥2
2.
E(X) =
X
kP (X = k) = 1P (X = 1) + 2P (X = 2)
k
= P (A) + 2P (Ac ) =
2
5
1
+2 = .
3
3
3
On the other side
E(X 2 ) =
X
k 2 P (X = k) = 12 P (X = 1) + 22 P (X = 2)
k
= P (A) + 4P (Ac ) =
1
2
+ 4 = 3.
3
3
Hence,
2
2
5
= .
V ar(X) = E(X ) − (E(X)) = 3 −
3
9
2
2
5
VI. (10pts) Suppose that X and Y are jointly continuous random variables with joint density
x+y
ce
(x, y) ∈ (−∞, 0] × (−∞, 0]
f(X,Y ) (x, y) =
0
otherwise
1. what is the value of c?
2. What is the probability that X < Y ?
3. What are the marginal densities fX and fY ?
Solution:
1. f(X,Y ) is a joint density if f(X,Y ) ≥ 0 and
R∞ R∞
−∞
−∞
f(X,Y ) (x, y)dxdy = 1.
Hence
• c≥0
•
Z
∞
Z
∞
Z
0
Z
0
ex+y dxdy
f(X,Y ) (x, y)dxdy = c
−∞
−∞
−∞ −∞
0
x
Z
e dx
=c
2
= c.
−∞
Hence c = 1.
2. Let us denote by D = {(x, y) ∈ R2 , x < y}
Z Z
P (X < Y ) = P ((X, Y ) ∈ D) =
Z y
Z 0
ex+y dxdy
dy
=
−∞
−∞
Z 0
1
e2y dy =
=
2
−∞
f(X,Y ) (x, y)dxdy
D
3. X and Y are symmetric hence If x > 0 then fX = fY = 0
If x ≤ 0 then
Z ∞
Z 0
ex+y dy = ex .
fY = fX =
f(X,Y ) (x, y)dy =
−∞
−∞
Hence
ex x ≤ 0
0 x > 0,
ey y ≤ 0
0 y > 0.
fX (x) =
fY (y) =
6
V. (10pts) If X is a Poisson random variable with parameter λ, show that
1.
E [X n ] = λE (X + 1)n−1 .
2. Now use this result to compute E [X 3 ].
Solution:
1.
n
E (X ) =
∞
X
n
k P (X = k) =
k=0
=
=
∞
X
k=1
∞
X
k=0
kn
λk −λ
e
k!
λ −λ
e
k!
k n−1
k=1
∞
X
λk
e−λ
(k − 1)!
k n−1
=λ
=λ
kn
k
k=1
∞
X
=λ
∞
X
λk−1 −λ
e
(k − 1)!
(m + 1)n−1
m=0
∞
X
λm −λ
e
m!
(m + 1)n−1 P (X = m)
m=0
= λE (X + 1)n−1 .
2. Let us take n = 3 in the above equality, we get that
E X 3 = λE (X + 1)2 = λE X 2 + 2X + 1
= λ E(X 2 ) + 2E(X) + 1 .
On the other side E(X) = V ar(X) = λ and E(X 2 ) = V ar(X) + (E(X))2 = λ + λ2 .
Hence,
E X 3 = λ3 + 3λ2 + λ.
7
VI. (5pts (Bonus)) X is a Poisson random variable with parameter λ,
Show for r = 2, 3, . . . ,
E [X(X − 1) . . . (X − r + 1)] = λr .
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