|
|
|
ece302 e2 fa2008 sol.pdf
ece302_e2_fa2008_sol.pdf
Showing 5-8 out of 9
ece302 e2 fa2008 sol.pdf-Purdue University School ...
ece302_e2_fa2008_sol.pdf-Purdue University School of Electrical and
ece302 e2 fa2008 sol.pdf-Purdue Uni...
ece302_e2_fa2008_sol.pdf-Purdue University School of Electrical and
Page 5
(c) We have that
f
X
2
|
X
1
>X
2
(
x
2
|
x
1
>x
2
)
=
∞
x
2
f
X
1
,X
2
(
x
1
,x
2
)
dx
1
P
(
X
1
>X
2
)
=
λ
2
e
-
x
2
(
λ
1
+
λ
2
)
λ
2
/
(
λ
1
+
λ
2
)
=(
λ
1
+
λ
2
)
e
-
x
2
(
λ
1
+
λ
2
)
,
when
x
2
>
0, and
f
X
2
(
x
2
) = 0, when
x
2
≤
0.
(d) Since
X
1
and
X
2
are independent, we have that
f
Z
(
z
)=
f
X
1
(
x
1
)
*
f
X
2
(
x
2
). Thus,
f
Z
(
z
)=
∞
-∞
f
X
1
(
τ
)
f
X
2
(
z
-
τ
)
dτ.
Since
f
X
1
(
x
1
) = 0 for
x
1
≤
0, and similarly for
f
X
2
(
x
2
), we
have 2 cases:
z
≤
0 and
z>
0. When
z
≤
0, we have that
f
Z
(
z
) = 0, but when
z>
0,
f
Z
(
z
)
=
z
0
f
X
1
(
τ
)
f
X
2
(
z
-
τ
)
dτ
=
z
0
λ
1
λ
2
e
-
λ
1
τ
e
-
λ
2
(
z
-
τ
)
dτ
=
λ
1
λ
2
e
-
λ
2
z
z
0
e
τ
(
λ
2
-
λ
1
)
dτ
=
λ
1
λ
2
e
-
λ
2
z
λ
2
-
λ
1
(
e
(
λ
2
-
λ
1
)
z
-
1) =
λ
1
λ
2
(
e
-
λ
1
z
-
e
-
λ
2
z
)
λ
2
-
λ
1
Therefore,
f
Z
(
z
)=
0
,
z
≤
0
λ
1
λ
2
(
e
-
λ
1
z
-
e
-
λ
2
z
)
λ
2
-
λ
1
,
z>
0
5
Page 6
Problem 3.
(
15 points
) A bag has
N
fair coins, where
N
is equally likely
to be any integer between 1 and 10. All the coins in the bag are tossed
simultaneously.
What is the expected number of heads?
Show your
reasoning.
Solution:
We have that if the number of coins,
N
, is known, that
the number,
n
H
, of heads is a binomial random variable with
parameters
n
=
N
and
p
=1
/
2 (since it is a fair coin). Since the
expected value of a binomial random variable is
np
, then we have
that
E
[
n
H
|
N
]=
N/
2. Now
E
[
n
H
]=
10
N
=1
N
n
H
=0
n
H
P
(
n
H
∩
N
)
=
10
N
=1
N
n
H
=0
n
H
P
(
n
H
|
N
)
P
(
N
)
=
10
N
=1
P
(
N
)
N
n
H
=0
n
H
P
(
n
H
|
N
)
=
10
N
=1
P
(
N
)
E
(
n
H
|
N
)
=
10
N
=1
1
10
N/
2=
1
20
10
N
=1
N
=
55
20
=
11
4
6
Page 7
Problem 4.
(
20 points
) A stick of length 1 is broken at two points,
X
1
and
X
2
, each of which is uniform on [0,1].
X
1
and
X
2
are independent.
(a) (
7 points
) What is the PDF of the length of the left-most piece?
Solution:
We have that the length,
L
L
, of the left-most piece is
described by
L
L
= min(
X
1
,X
2
). We have then that
F
L
L
(
l
)=
P
(
L
L
≤
l
) is described by integrating the joint PDF over the
area shown (which reduces to finding its area):
0
l
1
0
l
1
X
1
X
2
Therefore, we have that
F
L
L
(
l
)
=
1
-
(1
-
l
)
2
⇒
f
L
L
(
l
)
=
dF
L
L
dl
= 2(1
-
l
)
We may also solve the problem as follows:
1
-
F
L
L
(
l
)=
P
(
L
l
>l
)=
P
(min(
X
1
,X
2
)
>l
)
=
P
(
X
1
,X
2
>l
)
=
P
(
X
1
>l
)
P
(
X
2
>l
)
.
Now
P
(
X
1
>l
)=1
-
P
(
X
1
≤
l
)=1
-
l
(and similar for
P
(
X
2
>l
)), so
1
-
F
L
L
(
l
) = (1
-
l
)
2
⇒
F
L
L
=
1
-
(1
-
l
)
2
=2
l
-
l
2
⇒
f
L
L
(
l
)=
dF
L
L
dl
=
2(1
-
l
)
(b) (
7 points
) What is the PDF of the length of the middle piece?
Solution:
We have that the length,
L
M
, of the middle piece is
described by
L
M
=
|
X
2
-
X
1
|
, or the distance between
X
1
and
7
Page 8
0
l
1
0
l
1
X
1
X
2
X
2
. We have, then, that
F
L
M
(
l
)=
P
(
L
M
≤
l
) is described by
the shaded area shown:
Therefore, we have that
F
L
M
(
l
)
=
1
-
(2)
1
2
(1
-
l
)
2
⇒
f
L
M
(
l
)
=
dF
L
M
dl
= 2(1
-
l
)
We may also solve the problem as follows:
If we let
Z
=
X
2
-
X
1
, then
L
M
=
|
Z
|
, and
f
L
M
(
l
)=
f
Z
(
l
)+
f
Z
(
-
l
), for
l
≥
0, and
f
Z
(
z
)=
f
X
1
(
x
1
)
*
f
X
2
(
-
x
2
) is the convolution of
the two functions shown below:
-1.5
-1
-0.5
0
0.5
1
1.5
0
0.2
0.4
0.6
0.8
1
1.2
x
1
f
X
1
(x
1
)
*
-1.5
-1
-0.5
0
0.5
1
1.5
0
0.2
0.4
0.6
0.8
1
1.2
x
2
f
X
2
(-x
2
)
=
-1.5
-1
-0.5
0
0.5
1
1.5
0
0.2
0.4
0.6
0.8
1
1.2
z
f
Z
(z)
f
X
1
(
x
1
)
f
X
2
(
-
x
2
)
f
z
(
z
)
Since
f
L
M
(
l
)=
f
Z
(
l
)+
f
Z
(
-
l
) for
l>
0, we have that
f
L
M
=
2(1
-
l
), for
l>
0, and zero otherwise.
(c) (
6 points
) What is the PDF of the length of the right-most piece?
Solution:
Similar to part (a), we now have that the length,
L
R
,
of the right-most piece is described by
L
R
=1
-
max(
X
1
,X
2
).
We have, then, that
F
L
R
(
l
)=
P
(
L
R
≤
l
) is described by the
shaded area shown:
8
Ace your assessments! Get Better Grades
Browse thousands of Study Materials & Solutions from your Favorite Schools
Purdue University-Main Ca...
Purdue_University-Main_Campus
School:
Probabilistic_Methods_in_Electrical_and_Computer
Course:
Great resource for chem class. Had all the past labs and assignments
Leland P.
Santa Clara University
Introducing Study Plan
Using AI Tools to Help you understand and remember your course concepts better and faster than any other resource.
Find the best videos to learn every concept in that course from Youtube and Tiktok without searching.
Save All Relavent Videos & Materials and access anytime and anywhere
Prepare Smart and Guarantee better grades
Know more
Students also viewed documents
lab 18.docx
lab_18.docx
Course
Course
3
Module5QuizSTA2023.d...
Module5QuizSTA2023.docx.docx
Course
Course
10
Week 7 Test Math302....
Week_7_Test_Math302.docx.docx
Course
Course
30
Chapter 1 Assigment ...
Chapter_1_Assigment_Questions.docx.docx
Course
Course
5
Week 4 tests.docx.do...
Week_4_tests.docx.docx
Course
Course
23
Week 6 tests.docx.do...
Week_6_tests.docx.docx
Course
Course
106
