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STAT 510 HW #2 – Written: Answers
Due Friday, September 15
You can turn it in in class, or to my office (116B IH) or mailbox in 101 IH, by 4PM.
There are no questions on Mallard this time.
- Suppose X ∼ Beta(α, β). Find
(a) E[X(1 − X)];
Answer:
E[X(1 − X)] =
∫ 1
0
x(1 − x)f (x)dx
Γ(α + β)
Γ(α)Γ(β)
∫ 1
0
x(1 − x) x
α− 1 (1 − x)
β− 1 dx
Γ(α + β)
Γ(α)Γ(β)
∫ 1
0
x
(α+1)− 1 (1 − x)
(β+1)− 1 dx
Γ(α + β)
Γ(α)Γ(β)
Γ(α + 1)Γ(β + 1)
Γ(α + 1 + β + 1)
αβ
(α + β)(α + β + 1)
(Use the fact that Γ(α + 1) = αΓ(α), and such.)
(b) E[X
a (1 − X)
b ] for nonnegative integers a and b.
Answer: Similarly,
E[X
a (1 − X)
b ] =
∫ 1
0
x
a (1 − x)
b f (x)dx
Γ(α + β)
Γ(α)Γ(β)
∫ 1
0
x
(α+a)− 1 (1 − x)
(β+b)− 1 dx
Γ(α + β)
Γ(α)Γ(β)
Γ(α + a)Γ(β + b)
Γ(α + β + a + b)
[(α + a − 1)(α + a − 2) · · · α] × [(β + b − 1)(β + b − 2) · · · β]
(α + β + a + b − 1)(α + β + a + b − 2) · · · (α + β)
- X ∼ Geometric(p) has space X = { 0 , 1 , 2 ,.. .} and pmf
f (x | p) = (1 − p)p
x
for 0 < p < 1.
(a) Find the moment generating function of X.
Answer:
MX (t) = E[e
tX ]
= (1 − p)
∞ ∑
x=
e
tx p
x
= (1 − p)
∞ ∑
x=
(e
t p)
x
1 − p
1 − p e
t
if p e
t < 1
1 − p
1 − p e
t
if t < − log(p),
which is fine, since − log(p) > 0, so the mgf is finite in a neighborhood of 0.
(b) Find the mean and variance of X.
Answer: It is a bit easier to use the cumulant generating function,
cX (t) = log(MX (t)) = log(1 − p) − log(1 − p e
t ).
Then
c
′
X
(t) =
p e
t
1 − p e
t
1 − p e
t
− 1 ⇒ μ 1 = c
′
X
p
1 − p
c
′′
X
(t) =
p e
t
(1 − p e
t )
2
⇒ σ
2 = μ 2 = c
′′
X
p
(1 − p)
2
- Suppose X ∼ M ultinomial(n, p), where X is K × 1. Show that
Cov[X] = n(diag(p) − p p
′ ),
where diag(p) is the K × K diagonal matrix
diag(p) =
p 1
0 p 2
0 0 p 3
0 0 0 · · · p K
(You can use the results on the variances and covariances of the X i ’s that were derived in
class.)
The covariance matrix of the deviations is
Cov[AX] = σ
2 B.
What is the matrix B?
Answer: The covariance is A Cov[X] A
′ = σ
2 AA
′
. If you write out the multiplication, you
find that AA
′ = A, hence B = A.
- Suppose Z 1 and Z 2 are independent N (0, 1)’s.
(a) Find the moment generating function of (Z 1 , Z 2 ).
Answer: The mgf of each Z i is e
t
2 / 2 , and they are independent, so that
M
(Z 1 ,Z 2 ) (t 1 , t 2 ) = e
(t
2 1
+t
2 2
)/ 2 .
(b) Let X = Z 1
+ Z
2 and Y = Z 1
− Z
2
. Find the moment generating function of (X, Y ).
Answer: Try to write it in terms of the mgf from part (a):
M
(X,Y ) (s 1 , s 2 ) = E[e
s 1 X+s 2 Y ]
= E[e
s 1 (Z 1 +Z 2 )+s 2 (Z 1 −Z 2 ) ]
= E[e
(s 1 +s 2 )Z 1 +(s 1 −s 2 )Z 2 ]
= M
(Z 1 ,Z 2 ) (s 1
= e
((s 1 +s 2 ) 2 +(s 1 −s 2 ) 2 )/ 2
= e
s
2 1
+s
2 2 .
(c) Are X and Y in part (b) independent? Why or why not?
Answer: Yes, they are independent because their joint mgf factors:
M
(X,Y ) (s 1 , s 2 ) = e
s
2 1 e
s
2 2 = M X (s 1
)M
Y (s 2
- Set X = cos(θ) and Y = sin(θ), where θ ∼ U nif orm(0, 2 π).
(a) What is the marginal space X of X? The marginal space of Y?
Answer: X = Y = (− 1 , 1) or [− 1 , 1] or [− 1 , 1) (all are ok since the variables are continuous).
(b) For x ∈ X , find F (x) = P [X ≤ x]. (Figure out which θ’s correspond to X ≤ x. The
answer should have a cos
− 1 in it.)
Answer:
F (x) = P [X ≤ x] = P [cos(θ) ≤ x].
Now you have to figure out which θ’s those are, remembering that the space has 0 ≤ θ < 2 π.
So
{θ ∈ (0, 2 π) | cos(θ) ≤ x} = {θ | cos
− 1 (x) ≤ θ ≤ 2 π − cos
− 1 (x)},
where the cos
− 1 (x) is the angle between 0 and π with cosine x. Since θ is uniform,
F (x) = P [cos
− 1 (x) ≤ θ ≤ 2 π − cos
− 1 (x)] =
2 π − cos
− 1 (x) − cos
− 1 (x)
2 π
cos
− 1 (x)
π
(You can check that F (−1) = 0 and F (1) = 1.)
(c) Find the pdf of X.
Answer: The pdf for x ∈ (− 1 , 1) is
f (x) = F
′ (x) = −
π
d
dx
cos
− 1 (x).
To find the derivative, you can either remember it from calculus, or look it up in a book, or
use the following general result. Any strictly monotone (increasing or decreasing) function g
which is differentiable has an inverse g
− 1 so that g
− 1 (g(x)) = g(g
− 1 (x)) = x. The derivative
of x is 1, so by the chain rule,
d
dx
x =
d
dx
g(g
− 1 (x)) = g
′ (g
− 1 (x))
d
dx
g
− 1 (x).
Thus
d
dx
g
− 1 (x) =
g
′ (g
− 1 (x))
Now apply that to g(y) = cos(y), so that g
′ (y) = − sin(y), to obtain
d
dx
cos
− 1 (x) = −
sin(cos
− 1 (x))
1 − x
2
because if the cosine is x, then the sine is
1 − x
2
. So
f (x) =
π
1 − x
2
(d) Let R =
X
2
2
. What is the space of R? What is the space of (θ, R)?
Answer: R =
√
cos
2 (θ) + sin
2 (θ) = 1, so the space of R is R = { 1 }. The space of (θ, R) is
then (0, 2 π) × { 1 }.
