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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
)
=
z
0
λ
1
λ
2
e
-
λ
1
τ
e
-
λ
2
(
z
-
τ
)
=
λ
1
λ
2
e
-
λ
2
z
z
0
e
τ
(
λ
2
-
λ
1
)
=
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
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
8

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