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Stat 400 HW 9 Solutions TA: Emily King
- (Book # 11) See the solutions manual.
- (Book # 13) See the solutions manual.
- (Book # 22) Consider a random sample taken from the pdf
f (x; θ) =
(θ + 1) xθ^0 ≤ x ≤ 1 0 o.w. (a) We use the method of moments to find a point estimator for θ.
μX = E(X) =
0
(θ + 1) dx
= θ^ + 1 θ + 2 x(θ+2)|^10
= θ + 1 θ + 2 Now solve for θ:
(θ + 2) μX = θ + 1
⇒ θ(μX − 1) = 1 − 2 μX
⇒ θ = 2 μX − 1 1 − μX So set θˆ = (^12) −XX^1. Plugging in the data, we get
x =
10 (.92 +^ .79 +^...^ +^ .77) =^.^8 ,
obtaining the point estimate θˆ = 2( 1 .−8).− 8 1 = 3.
(b) We now find the maximum likelihood estimator. The joint pdf of a random sample from our given pdf is
f (x 1 , x 2 ,... , xn; θ) =
∏^ n i=
f (xi; θ)
=
[
(θ + 1) xθ 1
] [
(θ + 1) xθ 2
]
[
(θ + 1) xθn
]
= (θ + 1)n^ (x 1 x 2 · · · xn)θ
This is our likelihood function. Our loglikelihood function is then
ln [f (x 1 , x 2 , hdots, xnθ)] = n ln(θ + 1) + θ ln(
∏^ n x=
xi).
Differentiating with respect to θ, we obtain d dθ ln [f^ (x^1 , x^2 , hdots, xnθ)] =^
n θ + 1 + ln(
∏^ n x=
xi)
Setting the derivative equal to zero, we get
0 = n θ + 1
∏^ n x=
xi)
ln( ∏n x=1 xi) n
θ + 1 ⇒ θ + 1 = −n ln( ∏n x=1 xi) ⇒ θ = −n ln(
∏n x=1 xi)^
Since d^2 dθ^2 ln [f (x 1 , x 2 , hdots, xnθ)] = d dθ
n θ + 1
∏^ n x=
xi)
−n (θ + 1)^2 < 0.
So, by the second derivative test, ˆθ = (^) ln(Q−nx=1n xi) − 1 is the MLE. Plugging in the values, we obtain θˆ ≈ 3 .116.
- (WS # 4)
(a) ∫ f (x; θ)dx =
− 1
3 θx^2 dx +
0
3(1 − θ)x^2 dx
= θx^3 |^0 − 1 + (1 − θ)x^3 |^10
= θ + (1 − θ)
= 1
Also since θ > 1 > 0, f (x; θ) ≥ 0. Thus f is a pdf.
(a) ∫ f (x; θ)dx =
0
θ − 1 xθ^ dx
= lim b→∞
∫ (^) b
1
θ − 1 xθ^ dx
= lim b→∞
θ − 1 θ − 1 x−θ+1|b 1
= lim b→∞ −x^1 −θ^ |b 1
= lim b→∞ −b^1 −θ^ − (− 11 −θ^ )
= 0 − (−1)
= 1
Since θ > 1, θ x−θ^1 > 0 for x ≥ 1. So f is a pdf.
(b) Likelihood function:
f (x 1 ,... xn; θ) =
θ − 1 xθ 1
θ − 1 xθ 2
θ − 1 xθn
xi ≥ 1
= (θ − 1)n (x 1 x 2 · · · x 3 )θ^ xi ≥ 1
So the loglikelihood function is
ln f (x 1 ,... xn; θ) = n ln(θ − 1) − θ ln(x 1 x 2 · · · x 3 ) xi ≥ 1
Differentiating with respect to θ, we obtain d dθ ln f (x 1 ,... xn; θ) = n θ − 1 − ln(x 1 x 2 · · · x 3 ) xi ≥ 1
Setting the deriv equal to zero, we get n θ − 1 = ln(x^1 x^2 · · ·^ x^3 )^ xi^ ≥^1 ⇒ 1 + n ln(x 1 x 2... xn) =^ θ By the second derivative test d^2 dθ^2 ln f (x 1 ,... xn; θ) = −n (θ − 1)^2
, so this value is a maximum. The MLE is θˆ = 1 + (^) ln(X 1 Xn 2 ...Xn). (c) Plugging in the values from the worksheet, we get θˆ ≈ 2 .954.
