HW3 442Solutions
Name:
ID:
Homework 3 Solutions
1. [§6-4] Let X1 , X2 , . . . , X8 be i.i.d. normal random variables with mean µ and standard deviation σ. Define
X −µ
,
T =√
S 2 /n where X is the sample mean and S 2 is the sample variance.
(a) Find τ1 such that P(|T | < τ1 ) = .9; and
(b) find τ2 such that P(T > τ2 ) = .05.
(a) First, we notice that T follows a t7 distribution. Since the t distribution is symmetric about 0, we have
0.9 = P(|T | < τ1 ) = P(T < τ1 ) − P(T < −τ1 ) = 2P(T < τ1 ) − 1.
Thus P(T < τ1 ) = 0.95 and τ1 = 1.895.
(b) Since P(T > τ2 ) = .05, we have P(T < τ2 ) = 0.95. Thus τ2 = 1.895.
2. [§8-4] Suppose that X is a discrete random variable with
2
θ,
3
2
P(X = 2) = (1 − θ),
3
P(X = 0) =
1 θ, 3
1
P(X = 3) = (1 − θ).
3
P(X = 1) =
where 0 ≤ θ ≤ 1 is a parameter. The following 10 independent observations were taken from such a distribution: (3, 0, 2, 1, 3, 2, 1, 0, 2, 1).
(a) Find the method of moments estimate of θ.
(b) What is the maximum likelihood estimate of θ?
(a) In general, let X1 , X2 , . . . , Xn be a random sample drawn from this distribution. Since
2
1
2
1
7
µ1 = E[X] = 0 · θ + 1 · θ + 2 · (1 − θ) + 3 · (1 − θ) = − 2θ.
3
3
3
3
3
we have θ= and the MME for θ is
7 1
− µ1 ,
6 2
7 1 θˆ = − X,
6 2
1∑
Xi . In this case, we have X = 3/2 and n i=1 n where X =
7 1 3
5
θˆ = − · =
.
6 2 2
12
(b) The likelihood function of the sample (3, 0, 2, 1, 3, 2, 1, 0, 2, 1) is lik(θ) = f (3, 0, 2, 1, 3, 2, 1, 0, 2, 1|θ)
( )2 ( )3 (
)3 (
)2
2
1
2
1
= θ · θ ·
(1 − θ) ·
(1 − θ)
3
3
3
3
( )5 ( )5
2
1
5
= θ5 (1 − θ) .
3
3
Since
l(θ) = log lik(θ) = 5 log
1
2
+ 5 log + 5 log θ + 5 log(1 − θ),
3
3
5
5
−
,
θ 1−θ
5
5 l′′ (θ) = − 2 −
< 0, θ (1 − θ)2 l′ (θ) =
we have
( )
1
l
= max l(θ).
0≤θ≤1
2
1
Thus in this case, the MLE is θ˜ = .
2
2
3. [§8-7] Suppose that X follows a geometric distribution,
P(X = k) = p(1 − p)k−1 and assume an i.i.d. sample of size n.
(a) Find the method of moments estimate of p.
(b) Find the mle of p.
(The moments of geometric distribution can be
ID:
Homework 3 Solutions
1. [§6-4] Let X1 , X2 , . . . , X8 be i.i.d. normal random variables with mean µ and standard deviation σ. Define
X −µ
,
T =√
S 2 /n where X is the sample mean and S 2 is the sample variance.
(a) Find τ1 such that P(|T | < τ1 ) = .9; and
(b) find τ2 such that P(T > τ2 ) = .05.
(a) First, we notice that T follows a t7 distribution. Since the t distribution is symmetric about 0, we have
0.9 = P(|T | < τ1 ) = P(T < τ1 ) − P(T < −τ1 ) = 2P(T < τ1 ) − 1.
Thus P(T < τ1 ) = 0.95 and τ1 = 1.895.
(b) Since P(T > τ2 ) = .05, we have P(T < τ2 ) = 0.95. Thus τ2 = 1.895.
2. [§8-4] Suppose that X is a discrete random variable with
2
θ,
3
2
P(X = 2) = (1 − θ),
3
P(X = 0) =
1 θ, 3
1
P(X = 3) = (1 − θ).
3
P(X = 1) =
where 0 ≤ θ ≤ 1 is a parameter. The following 10 independent observations were taken from such a distribution: (3, 0, 2, 1, 3, 2, 1, 0, 2, 1).
(a) Find the method of moments estimate of θ.
(b) What is the maximum likelihood estimate of θ?
(a) In general, let X1 , X2 , . . . , Xn be a random sample drawn from this distribution. Since
2
1
2
1
7
µ1 = E[X] = 0 · θ + 1 · θ + 2 · (1 − θ) + 3 · (1 − θ) = − 2θ.
3
3
3
3
3
we have θ= and the MME for θ is
7 1
− µ1 ,
6 2
7 1 θˆ = − X,
6 2
1∑
Xi . In this case, we have X = 3/2 and n i=1 n where X =
7 1 3
5
θˆ = − · =
.
6 2 2
12
(b) The likelihood function of the sample (3, 0, 2, 1, 3, 2, 1, 0, 2, 1) is lik(θ) = f (3, 0, 2, 1, 3, 2, 1, 0, 2, 1|θ)
( )2 ( )3 (
)3 (
)2
2
1
2
1
= θ · θ ·
(1 − θ) ·
(1 − θ)
3
3
3
3
( )5 ( )5
2
1
5
= θ5 (1 − θ) .
3
3
Since
l(θ) = log lik(θ) = 5 log
1
2
+ 5 log + 5 log θ + 5 log(1 − θ),
3
3
5
5
−
,
θ 1−θ
5
5 l′′ (θ) = − 2 −
< 0, θ (1 − θ)2 l′ (θ) =
we have
( )
1
l
= max l(θ).
0≤θ≤1
2
1
Thus in this case, the MLE is θ˜ = .
2
2
3. [§8-7] Suppose that X follows a geometric distribution,
P(X = k) = p(1 − p)k−1 and assume an i.i.d. sample of size n.
(a) Find the method of moments estimate of p.
(b) Find the mle of p.
(The moments of geometric distribution can be