HW4 442Solutions

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Homework 4 Solutions
First Midterm Exam:20/2/15
1. [§8-13] In Example D of Section 8.4, the pdf of the population distribution is 
 1 + αx −1 ≤ x ≤ 1
2
f (x|α) =
,
−1 ≤ α ≤ 1,

0 otherwise and the method of moments estimate was found to be α
ˆ = 3X (where X is the sample mean of the random sample X1 , . . . , Xn ). In this problem, you will consider the sampling distribution of α
ˆ.
(a) Show that the estimate α
ˆ is unbiased.
(b) Find Var[ˆ α]. [Hint: What is Var[X]?]
(b) Use the central limit theorem to deduce a normal approximation to the sampling distribution of α
ˆ . According to this approximation, if n = 25 and α = 0, what is the P(|ˆ α| > .5)?
(a) We notice that
[ ]
E X = E[X1 ] =

∫

1

1 + αx x· dx =
2
−1
( 2
) 1 x αx3 α =
+
= .
4
6
3
−1

Therefore,

∫

1

−1

(

x αx2
+
2
2

[ ]
[ ] α E[ˆ α] = E 3X = 3E X = 3 · = α.
3

(b) First, we have
)
∫ 1( 2
1 + αx x αx3 dx =
+
2
2
2
−1
−1
( 3
) 1
4
x αx 1
=
+
= ,
6
8
3
−1
∫

E[X12 ] =

1

x2 ·

and
Var[X1 ] = E[X12 ] − E[X1 ]2 =

1 ( α )2
3 − α2
−
.
=
3
3
9

)

Thus,
[ n
]
[ n
]
n
∑
[ ]
1∑
1
1 ∑
Xi = 2
Var X = Var
Xi = 2 Var
Var [Xi ] n i=1 n n i=1 i=1 =

1
3 − α2
Var[X1 ] =
.
n
9n

Therefore, we have
[ ]
[ ] 3 − α2
Var[ˆ
α] = Var 3X = 9Var X =
.
n
(
) α 3 − α2
(c) According to the central limit theorem, we have X ∼ N
,
,
3
9n approximately. Therefore, α
ˆ = 3X implies that
(
)
3 − α2 α ˆ ∼ N α,
,
approximately. n In the case α = 0 and n = 25, we have α
ˆ ∼ N (0, 0.12), approximately.
Thus
P(|ˆ α| > .5) = P(ˆ α > .5) + P(ˆ α < −.5)
(
)
(
) α ˆ−0
0.5 − 0 α ˆ−0
−0.5 − 0
=P √
> √
+P √
< √
0.12
0.12
0.12
0.12
≈ P(Z > 1.44) + P(Z < −1.44)
= 0.0749 + 0.0749 = 0.1498.
2. [§8-53] Let X1 , . . . , Xn be i.i.d. uniform on [0, θ].
(a) Find the method of moments estimate of θ, and the mean, variance, bias, and MSE of the MME.
(b) The mle of θ is θˆ = max Xi . The pdf of max1≤i≤n Xi (How do we
1≤i≤n

find this? ) is

 n−1
 nx f (x|θ) = θn 
0

0<x<θ

.

otherwise

Calculate the mean and variance of the mle.