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STAT 510 Practice Problems for the Final
December, 2006
The final will be comprehensive, but be weighted somewhat towards the material since the last midterm. So you should study all the homework, and the previous exams. The final will have about 6-7 problems. Also, the final will give the necessary pdf’s. It is closed book & notes.
10 points 1. Suppose X 1 , ..., Xn are iid with pdf
f (xi | λ) = λe−λxi^ , for xi > 0 ,
where λ > 0. Then E(Xi) = 1/λ and V ar(Xi) = 1/λ^2. (a) Find a method-of-moments estimator of λ.
Answer: Set x = Ê[X] = 1/λ̂ , so that λ̂ = 1/x.
(b) Find the variance of the estimator in (a). Answer: The Xi’s are independent Gamma(1, λ)’s, so their sum is Gamma(n, λ). Thus
E[λ̂ ] = E
[ 1
X
] = E
[ (^) n ∑ Xi
] = n E
[ 1 Gamma(n, λ)
] .
The expected value of λ̂ is then (with u as Gamma(n, λ))
n
∫ (^) ∞
0
λn Γ(n)
u−^1 un−^1 e−λu^ = n
λn Γ(n)
Γ(n − 1) λn−^1
= λ
n n − 1
Similarly,
E[ λ̂^2 ] = n^2 E
[ 1 Gamma(n, λ)^2
] = n^2
λn Γ(n)
Γ(n − 2) λn−^2
= λ^2
n^2 (n − 1)(n − 2)
Thus V ar[λ̂ ] = λ^2
n^2 (n − 1)(n − 2)
( λ
n n − 1
) 2
n^2 (n − 1)^2 (n − 2)
λ^2.
10 points 2. Suppose X 1 ,... , Xn represent the weights (in ounces) of n randomly chosen (independent) rocks, where the distribution of the weights of the population of rocks is N (θ, 1). There are two possible experiments:
I The rocks are weighed on a scale that gives the exact weight of the rock. So the data are the regular X 1 ,... , Xn.
II The rocks are weighed on a scale that only goes up to 100 ounces. It gives the exact weight if the weight is less than 100; otherwise it gives 100. So the data are Y 1 ,... , Yn, where Yi = min{Xi, 100 }.
(a) For experiment I, is X an unbiased estimator of θ? (This is not a trick question.)
Answer: Yes.
(b) Give the distribution of Yi for experiment II.
Answer: Yi is a mixed-type distribution, where it is N (θ, 1) up until 100, then is has P [Yi = 100] = P [Xi > 100] = 1 − Φ(100 − θ), where Φ is the distribution function of a N (0, 1). Thus the distribution function of Yi can be given as
Fθ(y) =
{ Φ(y − θ) if y < 100 1 if y ≥ 100.
(c) Show that for experiment II,
E[Yi] = θ − (θ − 100)Φ(θ − 100) − φ(θ − 100),
where Φ is the distribution function and φ is the pdf for a N (0, 1). Is Y an unbiased estimator of θ?
Answer: Note: Getting the expected value for this problem takes more work than you’d have to do on an exam. To get the expected value, integrate y up until 100, then put a probability at 100:
E[Yi] =
∫ (^100)
−∞
y φ(y − θ)dy + 100 (1 − Φ(100 − θ)).
Look at the integral part:
∫ (^100)
−∞
y φ(y − θ)dy =
∫ (^) ∞
−∞
y φ(y − θ)dy −
∫ (^) ∞
100
y φ(y − θ)dy = θ −
∫ (^) ∞
100
y φ(y − θ)dy,
since the mean of a N (θ, 1) is θ. Now in that final integral, put z = y − θ:
∫ (^) ∞
100
y φ(y − θ)dy =
∫ (^) ∞
100 −θ
(z + θ) φ(z)dz = θ (1 − Φ(100 − θ)) +
∫ (^) ∞
100 −θ
z φ(z)dz.
Now that final integral is easy to do making the substitution u = z^2 /2 (so that du = zdz):
∫ (^) ∞
100 −θ
z
2 π
e−^
1 2 z^2 dz =
∫ (^) ∞
(100−θ)^2 / 2
z
2 π
e−udu =
2 π
e−^
1 2 (100−θ)^2 = φ(100 − θ).
Answer: e−λ^
λx x!
Γ(α)
λα−^1 e−λ^ =
x!
Γ(α)
λx+α−^1 e−^2 λ.
(b) Without doing any integrations explicitly, find the conditional distribution of λ | X = x. Answer: It is Gamma(x + α, 2).
(c) Find E[λ | X = x]. Answer: It is (x + α)/2.
10 points 5. Suppose X 1 ,... , Xn are iid U nif orm(θ, θ + 1), where θ ∈ R. (Assume n > 2.)
(a) Find the likelihood function L(θ; x 1 ,... , xn). Answer: The pdf for Xi is
f (xi | θ) =
{ 1 if θ < xi < θ + 1 0 otherwise
so L(θ; x 1 ,... , xn) =
∏^ n
i=
f (xi | θ) =
{ 1 if θ < xi < θ + 1 for all i = 1,... , n 0 otherwise.
(b) Suppose the data are 1. 1 , 1. 3 , 1. 2 , 1 .5.
(i) Sketch the likelihood based on these data. Answer: The likelihood is 1 if θ < 1. 1 , 1. 3 , 1. 2 , 1. 5 < θ + 1, and 0 otherwise. Which means it is 1 only if 0. 5 < θ < 1 .1 (because θ < 1 .1 and 1. 5 < θ + 1).
(ii) Is there a unique value of θ that maximizes this likelihood? Answer: No, anything between 0.5 and 1.1 has the same maximum value.
- Suppose gk(u) is a pdf for each k, and U has the pdf g(u) = ∑∞ 10 points k=0 pkgk(u) for some constants pk ≥ 0. (a) Show that
∑∞ k=0 pk^ = 1. [Hint: Integrate both sides.] Answer: Each
∫ gk(u)du = 1, so 1 =
∫ g(u)du =
∑ pk.
(b) Suppose the mean of the random variable with pdf gk(u) is μk. Show that
E(U ) =
∑^ ∞
k=
pkμk.
(You can assume the summation converges.) Answer: Integrate:
E[U ] =
∫ ug(u)du =
∑ pk
∫ ugk(u)du =
∑ pkμk.
(c) Suppose now that U ∼ χ^2 ν (∆), so that gk is the pdf of a χ^2 ν+2k, and
pk = exp(−∆/2)
(∆/2)k k!
which is the P oisson(∆/2) pmf. Show that E(U ) = ν + ∆. [You can use the facts that the mean of a χ^2 a is a, and the mean of P oisson(λ) is λ.] Answer: Here, μk = E[χ^2 ν+2k = ν + 2k, hence
E[U ] =
∑ (ν + 2k)pk = ν + 2
∑ kpk = ν + 2(∆/2) = ν + ∆,
because
∑ kpk is the expected value of P oisson(∆/2).
10 points 7. Let X ∼ N (0, 1), Y ∼ U nif orm(0, 1), where X and Y are independent. [So X has pdf
fX (x) =
2 π
exp(−
x^2 2
), −∞ < x < ∞,
and Y has pdf fY (y) = 1 for 0 < y < 1.] Let U = X/Y and V = Y. (a) What is the space of (U, V )? (b) Give X and Y in terms of U and V. (c) Find the Jacobian. (d) Find the joint pdf of U and V. (e) Show that the marginal pdf of U is
fU (u) =
2 π
u^2
( 1 − exp(−
u^2 )
) .
It is called the Slash distribution. [You may want to make the substitution w = v^2 /2 in the integral. Be careful to get the limits of integration correct.]
10 points 8. If x and y are vectors of length n, then Kendall’s distance between x and y is defined by
d(x, y) =
∑ ∑ 1 ≤i Answer: Their space is a rectangle, and the joint pdf is 16 , which is indeed the product of (^12) and 13.
(e) What is the moment generating function of V? Answer: (1 + et^ + e^2 t)/3.
(f) What is the moment generating function of W? Answer: (1 + et)/2.
(g) What is the moment generating function of V + W? It is the mgf of U , so call it MU (t). Answer:
(1 + et^ + e^2 t)(1 + et)/6 = (1 + et^ + e^2 t^ + et^ + e^2 t^ + e^3 t)/6 =
et^ +
e^2 t^ +
e^3 t.
(h) Write MU (t) as MU (t) = p 0 + p 1 et^ + p 2 e^2 t^ + p 3 e^3 t. What are the pi’s? Answer: See the answer to (g). Thus the pi’s are 16 , 13 , 13 , 16 , respectively.
(i) From MU (t), can you see what the pmf of U must be? (What is the mgf of the random variable with pmf given by the pi’s?) Answer: Yes, fU (i) = pi from part (h).
10 points 9. Suppose X 1 , X 2 , X 3 , and Z are independent, with Xi ∼ N (0, σ^2 X ), i = 1, 2 , 3, and Z ∼ N (0, σ^2 Z ), where σ^2 X > 0 and σ^2 Z > 0. (a) What is the distribution of
X 1
X 2
X 3
Z
?
Answer: It is multivariate normal, with mean 0 4 and covariance matrix diagonal with diag- onals σ^2 X , σ^2 X , σ^2 X , σ^2 Z.
(b) Let Y 1 = X 1 + Z, Y 2 = X 2 + Z, and Y 3 = X 3 + Z. Find the matrix A so that
Y 1
Y 2
Y 3
= A
X 1
X 2
X 3
Z
.
Answer:
A =
.
(c) Show that
Cov
Y 1
Y 2
Y 3
= σ 2
1 ρ ρ ρ 1 ρ ρ ρ 1
for some σ^2 and ρ, and give σ^2 and ρ in terms of σ X^2 and σ^2 Z. Answer:
Cov[Y ] = A Cov[x − z] A′
σ^2 X 0 0 0 0 σ X^2 0 0 0 σ X^2 0 0 0 σ^2 Z
σ^2 X + σ Z^2 σ Z^2 σ^2 Z σ^2 Z σ^2 X + σ^2 Z σ^2 Z σ^2 Z σ Z^2 σ X^2 + σ Z^2
,
so that σ^2 = σ^2 Z + σ^2 Z and ρ = σ^2 Z /(σ^2 Z + σ Z^2 ).
(d) Are Y 1 and Y 2 independent? Why or why not? Answer: No, their corrlation is not 0, since σ Z^2 > 0.
10 points 10. Let (X, Y ) have pdf fX,Y (x, y) = 2 and space
W = {(x, y) | x > 0 , y > 0 , and x + y < 1 }.
(a) Give Yx, the conditional space of Y given X = x. Answer: Since x + y < 1, it must be that y < 1 − x, hence Yx = (0, 1 − x).
(b) Find the marginal space and pdf of X. Answer: The marginal space is X = (0, 1), and for x ∈ X ,
fX (x) =
∫
y∈Yx
f (x, y)dy =
∫ (^1) −x
0
2 dy = 2(1 − x).
(c) Find the conditional pdf of Y | X = x.
(a) Find the mean and variance of a DE(0, θ). Answer: C(t) = log(MDE (t)) = − log(1 − θ^2 t^2 ). So
C′(t) =
2 θ^2 t 1 − θ^2 t^2
, hence E[X] = C′(0) = 0.
C′′(t) =
(1 − θ^2 t^2 )2θ^2 − 2 θ^2 t(− 2 θ^2 t) 1 − θ^2 t^2
, hence V ar[X] = C′′(0) = 2θ^2.
(b) The mgf of an Exponential(1) random variable is M (t) = 1/(1−t) for t < 1, Suppose X 1 and X 2 are independent Exponential(1)’s. Let Y = X 1 − X 2. Find MY (t), the mgf of Y , and say for which values t it is finite. Answer:
MY (t) = E[etY^ ] = E[et(X^1 −X^2 )] = E[etX^1 ]E[e−tX^2 ] = M (t)M (−t) =
1 − t
1 − (−t)
1 − t^2
for − 1 < t < 1.
(c) Show that the Y in part (b) is DE(0, θ), and give the θ. Answer: The mgf of Y is clearly that of DE(0, 1), so Y ∼ DE(0, 1).
10 points 13. Suppose Y | X = x ∼ N (2x + 1, σ^2 ) and X ∼ N (μ, 1). What are the mean and variance of Y? What is the distribution of Y? Answer: E(x) = E[Y |X = x] = 2x + 1 and v(x) = V ar[Y |X = x] = σ^2. Thus
E[Y ] = E[e(X)] = E[2X + 1] = 2μ + 1,
and V ar[Y ] = E[v(X)] + V ar[e(X)] = E[σ^2 ] + V ar[2X + 1] = σ^2 + 4.
10 points 14. Suppose X ∼ U nif orm(− 1 , 1), so that it has pdf f (x) = 12 for − 1 < x < 1, and 0 elsewhere. Let Y = X^2. (a) Find E(X) and E(Y ). Answer: E[X] = 0, and E[Y ] =
∫ (^1)
− 1
x^2 dx =
(b) Show that Cov(X, Y ) = 0.
Answer: Cov[X, Y ] = E[XY ] − E[X]E[Y ] = E[X^3 ] − 0 = 0.
(c) Are X and Y independent? Why or why not? Answer: No, the space is a parabola, not a rectangle.
10 points 15. Let ( X 1 X 2
) ∼ N 2
(( 0 0
) ,
( 1 ρ ρ 1
)) ,
for some ρ ∈ (− 1 , 1). Set Y 1 = X 1 + X 2 and Y 2 = X 1 − X 2.
(a) Find the distribution of
( Y 1 Y 2
) .
Answer: It is multivariate normal, since it is a linear transformation of a multivariate normal, and the mean is clearly 0 2. The covariance matrix is ( 1 1 1 − 1
) ( 1 ρ ρ 1
) ( 1 1 1 − 1
)
( 2(1 + ρ) 0 0 2(1 − ρ)
) .
(b) Are Y 1 and Y 2 independent? Why or why not? Answer: Yes, their covariance is 0 and they are multivariate normal.
10 points 16. Suppose X and Y are independent, with X ∼ N (0, 1) and Y ∼ Exponential(1), so that they have pdfs fX (x) =
2 π
e−^
(^12) x 2 and fY (y) = e−y^ for ˙y > 0.
Let U = X/
Y and V = Y. (So g(x, y) = (x/
y, y).) (a) Find x and y as functions of u and v. (b) Show that the space of (U, V ) is (−∞, ∞) × (0, ∞). (c) Find the Jacobian Jg− 1 (u, v). (d) Find the joint pdf of (U, V ). (e) Find the marginal pdf of U.
10 points 17. This is the same as Question 14. Suppose X ∼ U nif orm(− 1 , 1), so that it has pdf f (x) = 12 for − 1 < x < 1, and 0 elsewhere. Let Y = X^2. (a) Find E(X) and E(Y ). (b) Show that Cov(X, Y ) = 0.
