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University
Second Hour Exam: Solutions
ECE 313
of Illinois
Page 1 of
1
Summer 2003
1.
TRUE
F
X
(b) is right continuous.
FALSE
It is possible that F
X
(a) = F
X
(b).
TRUE
This is the certain event.
FALSE
This need not hold.
TRUE
This is true, since
X
is a continuous random variable.
TRUE
This has to hold for all density functions.
2.
P{

X
– 4

> 3} = P{
X
> 7} + P{
X
< 1} = 1 –
Φ
−
4
2
7
+
Φ
−
4
2
1
= 1 –
Φ
(1.25) +
Φ
(–0.25)
= 1 –
Φ
(1.25) + 1 –
Φ
(0.25) = 2 – 0.8944 – 0.5987 = 0.5069.
P{
X
< 3

X
> 2} = P{2 <
X
< 3}/P{
X
> 2} = 2(
Φ
(0.25) –
Φ
(0)) = 2
Φ
(0.25) – 1 = 1.1974– 1 = 0.1974
3.(a)
The PDF of
X
has value 0.2 for –1
≤
u
≤
4. Using LOTUS we have
E[
Y
] =
E[
X
–1] =
.
∫∫
−
−
=
−
+
−
=
−
+
−
1
1
4
1
4
1
2
1
1
2
3
.
1

)
2
/
(
2
.
0

)
2
/
(
2
.
0
)
1
(
2
.
0
)
1
(
2
.
0
u
u
u
u
du
u
du
u
(b)
Y
takes on values in the range [0,3]. Thus, for v < 0 or v > 3, F
Y
(v) = 0. For any v, 0
≤
v
≤
2,
F
Y
(v) = P[
Y
≤
v] = P[1 – v
≤
X
≤
v + 1] = 0.2(v + 1 – (1 – v)) = 0.4v, while for any v, 2
≤
v
≤
3,
F
Y
(v) = P[
Y
≤
v] = P[– 1
≤
X
≤
v + 1] = 0.2(v + 1 – (– 1)) = 0.2(v+2). Differentiating, we obtain
f
Y
(v)
=
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This note was uploaded on 09/08/2009 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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