Week 3 Practice Problems Solution
(Review continuous random variables and their properties + likelihoods for discrete
variates)
1.
Suppose
Y
is an exponential random variable with pdf
f
(
y
)
=
exp{
−
y
/2}/2,
y
>
0.
Find
a)
(0
1)
PY
<<
exp{
−
y
/2}/2
dy
=−
exp{
−
y
/2}
0
1
=
0
1
∫
1
−
exp{
−
1/2}
≈
0.3935
b)
for any
()
(
)
Sy PY y
=>
0
y
>
P(Y > y) = 1F(y) = 1(exp{y/2}) = exp{y/2}
c)
Suppose
are 5 independent copies of
Y
and let
1
,...,
YY
5
Y
15
min{ ,.
.., }
VY
=
.
Find the pdf of
V
by first finding
PV v
>
P(V > v)=
P
(min{
Y
1,.
...,
Y
5
}
>
v
)
=
P
(
Y
i
>
v
)
=
exp{
−
v
/2}
=
exp{
(
−
v
/2}
i
=
1
5
∑
i
=
1
5
∏
i
=
1
5
∏
= exp{5v/2} = S(v)
F(v) = 1 S(v) = 1 – exp{5v/2}
f
(
v
)
=
F
'(
v
)
=
5exp{
−
5
v
/2}/2
2.
Suppose
.
~(
0
,
1
)
ZG
a)
Find
and
(1
.
8
7
PZ
>
)
(
 2)
>
P( Z > 1.87) = 0.0307
P( Z > 2) = P( Z < 2) + P ( Z > 2) = 0.0228 + 0.0228
≈
0.0456
b)
If
, find
32
YZ
=+
(2
4)
<
<
and
(
2  1)
−
>
P( 2 < Y < 4) =P ( 2 < 3 + 2Z < 4) = P( 1/2 < Z < 1/2 ) = 0.6915 – 0.3085
≈
0.383
P( Y2 > 1) = P( Y2 > 1) + P( Y2 < 1) = P( Y > 3) + P( Y < 1)
P( Y > 3) = P ( 3 + 2Z > 3) = P( Z > 0) = 0.5
P( Y < 1) = P ( 3 + 2Z < 1) = P( Z < 1 )
≈
0.1587
P( Y2 > 1) = 0.5 + 0.1587
≈
0.6587
c)
What is the distribution of
24
*
WY
=
−
?
Because W is multiplier and scale of Y and Y is multiplier and scale of Z, W is a
Gaussian distribution.
E(W) = E ( 2  4Y) = E( 2 – 4(3 + 2Z)) = E( 10 – 8Z) = 10 – 8E(Z) = 10
Var(W) = Var( 2 – 4(3 + 2Z)) = Var( 10 – 8Z) = 10 +64Var(Z) = 10+64 = 54
W ~ G(10,
54
)
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 Spring '10
 McKay
 Normal Distribution, probability density function, Maximum likelihood, Likelihood function, Yi

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