Download Mathematical Statistics I - Final Exam | STAT 510 and more Exams Mathematical Statistics in PDF only on Docsity!
NAME
STAT 510 Final Exam
Monday, December 11, 2006
Closed book & Notes. There are facts to know on the last page.
12 points 1. Suppose X and Y are independent, with X ∼ P oisson(1 + θ) and Y ∼ P oisson(1 − θ), where θ ∈ (− 1 , 1). (a) Find constants a and b so that δ 1 (x) = a + bx is an unbiased estimator of θ. What is V ar[δ 1 (X)]? Answer: E[a + bX] = a + b(1 + θ) = θ if a = −1 and b = 1, so δ 1 (x) = x − 1. Its variance is 1 + θ, the variance of X. (b) Find constants c and d so that δ 2 (x, y) = cx + dy is an unbiased estimator of θ. What is V ar[δ 2 (X)]? Answer: E[cX +dY ] = c(1+θ)+d(1−θ) = θ if c = 12 and d = −^12 , so that δ 2 (x, y) = 12 (x−y). Its variance is then
V ar[δ 2 (X, Y )] = V ar
[ 1
(X − Y )
]
(V ar[X] + V ar[Y ]) =
(1 + θ + 1 − θ) =
(c) Sketch the M SE(θ, δi)’s for the estimators found in parts (a) and (b). Is either one always better than the other? Answer: Both estimators are unbiased, so their MSE’s are their variances: M SE(θ; δ 1 ) = 1 + θ and M SE(θ; δ 2 ) = 12. Since θ goes from -1 to 1, the sketch is:
theta
MSE
delta delta
So, no, neither is always better then the other, though generally δ 2 looks nicer.
- Suppose X 1 ,... , Xn are iid N (μ, σ^2 ), where (μ, σ^2 ) ∈ R×(0, ∞). Let U = ∑n i=1(Xi^ −X) 14 points^2 , so that U ∼ σ^2 χ^2 n− 1. (a) Let δc(u) = cu be an estimator of σ^2 for some constant c. Find the mean and variance of δc, and show that the mean square error of δc is
M SE(σ^2 ; δc) = σ^4 ((c(n − 1) − 1)^2 + 2c^2 (n − 1)).
Answer: E[δc(U )] = E[cσ^2 χ^2 n− 1 ] = cσ^2 (n − 1), and
V ar[δc(U )] = V ar[cσ^2 χ^2 n− 1 ] = c^2 σ^4 2(n − 1).
Thus
M SE(σ^2 ; δc) = Bias^2 +V ariance = (cσ^2 (n−1)−σ^2 )^2 +c^2 σ^4 2(n−1) = σ^4 ((c(n−1)−1)^2 +2c^2 (n−1)).
(b) Find the c that minimizes the M SE in part (a).
(d) Using the prior in part (c), what is the posterior distribution of θ given the Xi = xi’s? Answer: The joint pdf is the Gamma(n + 2, 1 + ∑ xi) density without the constants, so the conditional pdf just puts in the Gamma constants, so that
θ | (X 1 = x 1 ,... , Xn = xn) ∼ Gamma(n + 2, 1 +
∑ xi).
(e) What is the posterior mean of θ from part(d)? Is it the same as the MLE of θ? Answer: It is the mean of a Gamma(n + 2, 1 + ∑ xi):
E[θ | (X 1 = x 1 ,... , Xn = xn)] = n + 2 1 + ∑ xi
No, it is not the same as the MLE. (f) What is the posterior mean of 1/θ from part (d)? Is it an unbiased estimator of 1/θ? Answer: It is the mean of one over a Gamma(n + 2, 1 + ∑ xi):
E[1/θ | (X 1 = x 1 ,... , Xn = xn)] =
∑ xi n + 2 − 1
∑ xi n + 1
The mean of each Xi is 1/θ, hence
E
[ 1 + ∑^ X
i n + 1
]
1 + n/θ n + 1
n + 1
θ
n n + 1
θ
So it is not unbiased.
12 points 4. Suppose X and Y are independent, both Discrete U nif orm{ 0 , 1 , 2 }, so that they each have space { 0 , 1 , 2 } and pmf 13. (a) Find the moment generating function MX (t) (which is the same as MY (t)). For what values of t is the mgf finite? Answer: E[etX^ ] =
e^0 t^ +
e^1 t^ +
e^2 t^ =
(1 + et^ + e^2 t). It is finite for all t ∈ R. (b) Find the moment generating function of W = X − Y.
Answer:
MX−Y (t) = E[e(X−Y^ )t] = E[etX^ ] E[e−tY^ ] (by indep.) = MX (t) MY (−t) =
(1 + et^ + e^2 t) ×
(1 + e−t^ + e−^2 t)
=
(1 + e−^1 + e−^2 t^ + et^ + 1 + e−t^ + e^2 t^ + et^ + 1)
=
(3 + 2et^ + 2e−t^ + e^2 t^ + e−^2 t).
(c) Find the space and pmf of the W in part (b). Answer: The space is {− 2 , − 1 , 0 , 1 , 2 }, and the values of the pmf can be found from the coefficients of the eit’s in the mgf of part (b). Thus
f (−2) = f (2) =
, f (−1) = f (1) =
, f (0) =
14 points 5. Suppose X ≡ (X 1 , X 2 , X 3 )′^ is multivariate normal, where the means of the Xi’s are all zero, the variances are all 2, and the covariances are all 1. Let Y = (Y 1 , Y 2 , Y 3 )′, where
Y 1 = X 1 + X 2 + X 3 , Y 2 = X 1 − X 2 , Y 3 = X 1 − X 3.
(a) Write down the covariance matrix of X. Answer:
Cov[X] =
.
(b) Find the matrix A for which Y = AX. Answer:
Y =
X 1 + X 2 + X 3
X 1 − X 2
X 1 − X 3
=
X 1
X 2
X 3
= AX.
(c) What is the distribution of Y? (Be sure to give its covariance matrix.)
(c) Find the joint pdf of (U, V ). Answer: The pdf of (X, Y ) is 1, hence the pdf of (U, V ) is just the Jacobian: f (u, v) = e−u.
(d) Find the marginal space and pdf of U. What is that distribution? [Hint: It is one of those on the page of facts.] Answer: The marginal space of U is (0, ∞). You have to be careful about the joint space in order to find the marginal pdf. You want to integrate out v, but you have to remember that for fixed u, v goes from 0 to u:
fU (u) =
∫ (^) u 0
f (u, v)dv =
∫ (^) u 0
e−udv = ue−u.
You recognize this pdf as that of a Gamma(2, 1). (e) Find the conditional space and pdf of V | U = u. What is that distribution? [Hint: It is one of those on the page of facts.] Answer: As noted in part (d), the conditional space of V given U = u is (0, u). The conditional pdf is fV |U (v | u) = f (u, v) fU (u)
e−u ue−u^
u
Thus V | U = u ∼ U nif orm(0, u).
8 points 7. Suppose θ has space { 0 , 1 }, with P [θ = 0] = P [θ = 1] = 12. (Thus θ ∼ Bernoulli(^12 ).) Also, let X | θ = θ ∼ U nif orm(θ, θ + 1).
(a) What is the marginal distribution of X? Answer: Heuristically, it is half a U nif orm(0, 1) plus half a U nif orm(1, 2), which means it must be U nif orm(0, 2). More formally, the joint density (it is part pdf, part pmf) is
f (x, θ) = I{θ This is the U nif orm(0, 2) pdf if you add in the point { 1 }. If you don’t want to add that in, that’s fine, too.
(b) What is P [θ = 0 | X = .3]? What is P [θ = 0 | X = 1.6]?
Answer: You can answer this without any formulas, since if X = .3, θ would have to be 0, because X has to be between θ and θ + 1. Similarly. if X = 1.6, θ would have to be 1. Thus P [θ = 0 | X = .3] = 1 and P [θ = 0 | X = 1.6] = 0. Or
P [θ = 0 | X = .3] = f (. 3 , 0) fX (.3)
I{ 0 <. 3 < 1 } 12
1 2 I{^0 <.^3 <^1 or^1 <.^3 <^2 }
because the two indicator functions are both 1, and
P [θ = 0 | X = 1.6] = f (1. 6 , 0) fX (1.6)
I{ 0 < 1. 6 < 1 } 12
1 2 I{^0 <^1.^6 <^1 or^1 <^1.^6 <^2 }
because the indicator function in the numerator is 0.
Some possibly useful facts
Poisson. X ∼ P oisson(λ) has space { 0 , 1 , 2 ,.. .}, pmf
fλ(x) = e−λ^ λx x!
and E[X] = V ar[X] = λ. Uniform. X ∼ U nif orm(a, b) has space (a, b), pdf
fa,b(x) =
b − a I{a 1.
Chi-squared. The χ^2 ν random variable has mean ν and variance 2ν. Also, χ^2 ν = Gamma(ν 2 , 12 ).
