Stat 426
EXAM 1
Name:
Exam Instructions: Use a separate piece of paper to answer each question. Label the
paper with the question number and your name. Try to show as much work as possible and
explain your thoughts. There are 5 problems on this test. Calculators/computers are
not permitted; one 8.5 by 11 inch sheet of notes can be brought to exam
1) Let Y1 denote the weight (in tons) of toilet paper stocked by a supplier at the beginning
of a week and let Y2 denote the weight of this item sold by the supplier during the week.
The joint probability density function of Y1 , Y2 is
fY1 Y2 (y1 , y2 ) =
K
0
0 ≤ y2 ≤ y1 , 0 ≤ y1 ≤ 1;
elsewhere.
a) Find K. (If you can’t get K give all future answers in terms of K below.)
b) Find P [Y2 > .5].
c) Find the probability density function of Y1 .(Be careful to specify the probability density
function for all −∞ < y1 < ∞.)
d) Find P [Y1 > .75, Y2 < .25].
e) The weight of toilet paper unsold by the end of the week is Y1 − Y2 . Find the chance
that at least 1/2 ton of toilet paper is unsold at the end of the week.
f) The supplier is known to have stocked 3/4 ton of toilet paper. What is the chance that
the supplier sells less than 1/4 of a ton by the end of the week?
2) Boys and girls are sitting in a row.
a) We distinguish between individual boys and we distinguish between individual girls. In
how many ways can 3 boys and 3 girls sit in a row if only the boys must sit together?
b) We do NOT distinguish between individual boys and we do NOT distinguish between
individual girls. In how many ways can 3 boys and 3 girls sit in a row if only the boys
must sit together?
1
3) Suppose that conditional on N , X ∼ Binomial(N, p). N has a Binomial distribution
with parameters (M, r). Derive the marginal distribution of X and give the name of this
distribution.
4) Suppose that X and Y are discrete random variables with joint probability mass function,
fX,Y (x, y). We conduct the following procedure for a given y0 (y0 is a number):
Step 1) We draw X ∼ pX (x)
Step 2) We draw R from a Bernoulli with success probability q where q = pY |X (y0 , |X).
Step 3) If R = 1 set Z = X. If R = 0 return to Step 1) and start over.
The above procedure ends (Hint: this means we got an R = 1). Prove that the distribution
of Z is the same as the distribution of X given that Y = y0 .
5) A density function sometimes used by engineers to model the lengths of life of electronic
components is the Rayleigh density, given by
f (y) =
2y
2
e−y /θ
θ
0
if 0 ≤ y;
elsewhere.
Let U = Y 2 .
a) Find the cumulative distribution function of Y . (Be careful to specify the cumulative
distribution function for all −∞ < u < ∞.)
b) Using the transformation method, derive the probability density function of U . (Be
careful to specify the probability density function for all −∞ < u < ∞.)
c) Using the method of distribution functions, derive the probability density function of U .
(Be careful to specify the probability density function for all −∞ < u < ∞.)
2
Answers
1a)
1=
Z 1 Z y1
0
0
Kdy2 dy1
implies that K = 2.
1b) The density of Y1 , Y2 is nonzero on the triangle with sides y2 = y1 , y1 = 1, y2 = 0.
The event {Y2 > 1/2} becomes the triangle with sides y2 = y1 , y1 = 1, y2 = 1/2. Thus
P [Y2 > 1/2] is
Z
Z
1
y1
1/2
1/2
2dy2 dy1 .
This is equal to 1/4.
1c)
2y1 0 ≤ y1 ≤ 1
0
elsewhere.
1d) The event {Y1 > 3/4, Y2 < 1/4} is the square bordered by y2 = 1/4, y2 = 0, y1 = 1,
and y1 = 3/4. So
fY1 (y1 ) =
P [Y1 > 3/4, Y2 < 1/4] =
Z 1 Z 1/4
3/4
0
2dy2 dy1 .
This is 1/8.
1e) Find P [Y1 − Y2 > 1/2]. This is the triangle bordered by y1 = 1, y2 = y1 − 1/2 and
y2 = 0.
P [Y1 − Y2 > 1/2] =
Z 1 Z y1 −1/2
1/2
0
2dy2 dy1 .
This is 1/4.
1f)
2
0 ≤ y2 ≤ y1 , 0 ≤ y1 ≤ 1
0
elsewhere.
Z 1/4
1
P [Y2 < 1/4|Y1 = 3/4] =
dy2
3/4
0
fY2 |Y1 (y2 |y1 ) =
2y1
This is 1/3.
2a)
!
4
∗ 3! 3! or equivalently 4! ∗ 3!
1
2b)
!
4
1
3
P
3) We know that pX (x) = n pX|N (x|n)pN (n) by the law of total probability. Note that
pX|N (x|n) = 0 if n < x so this constrains the sum over n. We have
pX (x) =
=
=
=
=
X
n
M
X
pX|N (x|n)pN (n)
!
!
M n
n x
r (1 − r)M −n
p (1 − p)n−x
n
x
n=x
M
px X
M!
(1 − p)n−x rn (1 − r)M −n
x! n=x (n − x)!(M − n)!
( let v = n − x)
−x
px rx M ! MX
(M − x)!
(1 − p)v rv (1 − r)M −(v+x)
x!(M
−
x)!
v!(M
−
(v
+
x))!
v=0
!
M
(pr)x ((1 − p)r + 1 − r)M −x
x
The last equality follows by the Binomial Series. So X ∼ Binomial(M, pr).
4) Since the procedure stopped we know we have that R = 1. We have that
P [Z = z] = P [X = z|R = 1]
P [X = z, R = 1]
=
P [R = 1]
P [R = 1|X = z]P [X = z]
=
P [R = 1]
P [R = 1|X = z]P [X = z]
= P
x P [R = 1|X = x]P [X = x]
But P [R = 1|X = z] = pY |X (y0 |z) and P [X = z] = pX (z) so we obtain
pY |X (y0 |z)pX (z)
P [Z = z] = P
x pY |X (y0 |x)pX (x)
pY,X (y0 , z)
=
pY (y0 )
= pX|Y (z|y0 )
5a)
2
1 − e−y /θ if y ≥ 0;
0
elsewhere.
√
2
2
−1
5b) Since U = Y we have g(y) = y and g (u) = u. Also
FY (y) =
d −1
1
g (u) = u−1/2 .
du
2
4
Thus,
d −1
fU (u) = fY (g√−1 (u)) du
g (u)
2 u u/θ 1 −1/2
= θ e 2u
,
so
fU (u)) =
1
e−u/θ
θ
0
5c) If u < 0 then P [U ≤ u] = 0. If u ≥ 0 then
if u ≥ 0;
elsewhere.
= P [Y 2 ≤
q u]
= P [Y ≤ (u)]
P [U ≤ u]
q
= FY ( (u))
= 1 − e−u/θ .
Thus
fU (u) =
1
e−u/θ
θ
0
5
if u ≥ 0;
elsewhere.