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Question 12 - Week 3 Practice | Prasnya
Q12
00:00
4 Aug 2024
Q12.
Let
X
1
X_1
X
1
and
X
2
X_2
X
2
be i.i.d., where
X
X
X
has PMF
P
(
X
=
0
)
=
0.2
P(X=0)=0.2
P
(
X
=
0
)
=
0.2
,
P
(
X
=
1
)
=
0.4
P(X=1)=0.4
P
(
X
=
1
)
=
0.4
,
P
(
X
=
2
)
=
0.4
P(X=2)=0.4
P
(
X
=
2
)
=
0.4
. Define
Y
=
X
1
+
X
2
Y=X_1+X_2
Y
=
X
1
+
X
2
. Find the MGF of
Y
Y
Y
.
A
M
Y
(
λ
)
=
(
0.2
+
0.4
e
λ
+
0.4
e
2
λ
)
2
M_Y(\lambda)=(0.2+0.4e^\lambda+0.4e^{2\lambda})^2
M
Y
(
λ
)
=
(
0.2
+
0.4
e
λ
+
0.4
e
2
λ
)
2
B
M
Y
(
λ
)
=
0.2
+
0.4
e
λ
+
0.4
e
2
λ
M_Y(\lambda)=0.2+0.4e^\lambda+0.4e^{2\lambda}
M
Y
(
λ
)
=
0.2
+
0.4
e
λ
+
0.4
e
2
λ
C
M
Y
(
λ
)
=
(
0.4
e
λ
+
0.4
e
2
λ
)
2
M_Y(\lambda)=(0.4e^\lambda+0.4e^{2\lambda})^2
M
Y
(
λ
)
=
(
0.4
e
λ
+
0.4
e
2
λ
)
2
D
M
Y
(
λ
)
=
0.2
+
0.4
e
2
λ
M_Y(\lambda)=0.2+0.4e^{2\lambda}
M
Y
(
λ
)
=
0.2
+
0.4
e
2
λ
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Check
Details
Q12
00:00
4 Aug 2024
Q12.
Let
X
1
X_1
X
1
and
X
2
X_2
X
2
be i.i.d., where
X
X
X
has PMF
P
(
X
=
0
)
=
0.2
P(X=0)=0.2
P
(
X
=
0
)
=
0.2
,
P
(
X
=
1
)
=
0.4
P(X=1)=0.4
P
(
X
=
1
)
=
0.4
,
P
(
X
=
2
)
=
0.4
P(X=2)=0.4
P
(
X
=
2
)
=
0.4
. Define
Y
=
X
1
+
X
2
Y=X_1+X_2
Y
=
X
1
+
X
2
. Find the MGF of
Y
Y
Y
.
A
M
Y
(
λ
)
=
(
0.2
+
0.4
e
λ
+
0.4
e
2
λ
)
2
M_Y(\lambda)=(0.2+0.4e^\lambda+0.4e^{2\lambda})^2
M
Y
(
λ
)
=
(
0.2
+
0.4
e
λ
+
0.4
e
2
λ
)
2
B
M
Y
(
λ
)
=
0.2
+
0.4
e
λ
+
0.4
e
2
λ
M_Y(\lambda)=0.2+0.4e^\lambda+0.4e^{2\lambda}
M
Y
(
λ
)
=
0.2
+
0.4
e
λ
+
0.4
e
2
λ
C
M
Y
(
λ
)
=
(
0.4
e
λ
+
0.4
e
2
λ
)
2
M_Y(\lambda)=(0.4e^\lambda+0.4e^{2\lambda})^2
M
Y
(
λ
)
=
(
0.4
e
λ
+
0.4
e
2
λ
)
2
D
M
Y
(
λ
)
=
0.2
+
0.4
e
2
λ
M_Y(\lambda)=0.2+0.4e^{2\lambda}
M
Y
(
λ
)
=
0.2
+
0.4
e
2
λ
Save
Check
Details
