STAT 131: Take-Home Test 2

1. [70总分]（“ Exchange悖论”）您在与以下货币游戏对战

2

2
。这使得金额的期望值

1个
2
\$ x
2
</ s> </ s> </ s>
+

1个
2
</ s> </ s> </ s>
（\$ 2 x）= \$ 5x
4
，大于\$ x

M裁判将放置在信封1中，并让X为您在信封中找到的金额

（a）[20分]解释为什么这个问题的出现意味着P（X = m | M = m）= P（X =
2 m | M = m）= 1
2
，并以此来表明
P（M = x | X = x）= p（x）
p（x）+ p

X
2

</ s> </ s> </ s>
M =
X
2

X = x
</ s> </ s> </ s>
=
p

X
2
</ s> </ s> </ s>
p（x）+ p

X
2
。 （1）

E（Y | X = x）= p（x）
p（x）+ p

X
2
（2 x）+ p

X
2
</ s> </ s> </ s>
p（x）+ p

X
2
</ s> </ s> </ s>
X
2
</ s> </ s> </ s>
。 （2）
（b）[40分]假设在这个游戏中，金钱和效用重合（或至少假设

p
X
2
</ s> </ s> </ s>
<2 p（x

(3) for prior distributions that are decreasing (that is, for which p(m2) < p(m1) for m2 > m1). Make
a sketch of what condition (3) implies for a decreasing p. One possible example of a continuous
decreasing family of priors on M is the exponential distribution indexed by the parameter λ, which
represents the reciprocal of the mean of the distribution. Identify the set of conditions in this family
of priors, as a function of x and λ, under which it’s optimal for you to trade. Does the inequality
you obtain in this way make good intuitive sense (in terms of both x and λ)? Explain briefly.
(c) [10 points] Looking carefully at the correct argument in paragraph 2 of this problem, identify the
precise point at which the argument in the first paragraph breaks down, and specify what someone
who believes the argument in paragraph 1 is implicitly assuming about the prior distribution p(m).
2. [210 total points] (practice with joint, marginal and conditional densities) This is a toy problem
designed to give you practice in working with a number of the concepts we’ve examined; in a course
like this, every now and then you have to stop looking at real-world problems and just work on
technique (it’s similar to classical musicians needing to practice scales, in addition to actual pieces
of symphonic or chamber music).
Suppose that the continuous random vector X = (X1, X2) has PDF given by
fX(x) = 
4 x1 x2 for 0 < x1 < 1, 0 < x2 < 1
0 otherwise 
(4)
in which x = (x1, x2), and define the random vector Y = (Y1, Y2) with the transformation (Y1 =
X1, Y2 = X1 X2).
(a) Are X1 and X2 independent? Present any relevant calculations to support your answer. [10
points]
(b) Either work out the correlation ρ(X1, X2) between X1 and X2 or explain why no calculation
is necessary in correctly identifying the value of ρ. [10 points]
(c) Sketch the set S of possible X values and the image T of S under the transformation from
X to Y , and show that the joint distribution of Y = (Y1, Y2) is
fY (y) = 
4
y2
y1
for 0 < y1 < 1, 0 < y2 < y1 < 1
0 otherwise 
, (5)
in which y = (y1, y2). Verify your calculation by demonstrating that RR
T
fY (y) dy = 1. [50
points]
(d) Work out
(i) the marginal distributions for Y1 and Y2, sketching both distributions and checking that
they both integrate to 1;
(ii) the conditional distributions fY1 | Y2
(y1 | y2) and fY2 | Y1
(y2 | y1), checking that they each
integrate to 1; and
(iii) the conditional expectations E(Y1 | Y2) and E(Y2 | Y1); and
2
(iv) the conditional variances V (Y1 | Y2) and V (Y2 | Y1). (Hint: recall that the variance of a
random variable W is just E (W2
) − [E(W)]2
.)
[120 points]
(e) Are Y1 and Y2 independent? Present any relevant calculations to support your answer. [10
points]
(f) Either work out the correlation ρ(Y1, Y2) between Y1 and Y2 or explain why no calculation is
necessary in correctly identifying the value of ρ. [10 points]
3. [100 total points] (moment-generating functions) Distributions may in general be skewed, but
there may be conditions on their parameters that make the skewness get smaller or even disappear.
This problem uses moment-generating functions (MGFs) to explore that idea for two important
discrete distributions, the Binomial and the Poisson.
(a) We saw in class that if X ∼ Binomial(n, p), for 0 < p < 1 and integer n ≥ 1, then the MGF
of X is given by
ψX(t) =
p et + (1 − p)
n
. (6)
for all real t, and we used this to work out the first three moments of X (note that the
expression for E (X3
) is only correct for n ≥ 3):
E(X) = n p , E
X
2

= n p[(1 + (n − 1)p] , (7)
E

X
3

= n p[1 + (n − 2)(n − 1)p
2 + 3 (n − 1)p] , (8)
from which we also found that V (X) = n p(1 − p). Show that the above facts imply that
skewness(X) = 1 − 2 p
p
n p(1 − p)
. (9)
Under what condition on p, if any, does the skewness vanish? Under what condition on n, if
any, does the skewness tend to 0? Explain briefly. [30 points]
(b) In our brief discussion of stochastic processes we encountered the Poisson distribution: if
Y ∼ Poisson(λ), for λ > 0, then the PMF of Y is
fY (y) =  λ
y e−λ
y!
for y = 0, 1, . . .
0 otherwise 
. (10)
(i) Use this to show that for all real t the MGF of Y is
ψY (t) = e
λ(e
t−1)
. (11)
[10 points]
(ii) Use ψY (t) to compute the first three moments of Y , the variance of Y and the skewness
of Y . Under what condition on λ, if any, does the skewness either disappear or tend to
0? Explain briefly. [60 points]
4. [140 total points] (archaeology) Paleobotanists estimate the moment in the remote past when a
given species became extinct by taking cylindrical, vertical core samples well below the earth’s surface and looking for the last occurrence of the species in the fossil record, measured in meters above
the point P at which the species was known to have first emerged. Letting {yi
, i = 1, . . . , n} denote a
sample of such distances above P at a random set of locations, the model (Yi
|θ)
IID∼ Uniform(0, θ) (∗)
emerges from simple and plausible assumptions. In this model the unknown θ > 0 can be used,
through carbon dating, to estimate the species extinction time.
The marginal distribution of a single observation yi
in this model may be written
pYi
(yi
| θ) =  1
θ
if 0 ≤ yi ≤ θ
0 otherwise 
=
1
θ
I (0 ≤ yi ≤ θ) , (12)
where I(A) = 1 if A is true and 0 otherwise.
(a) Briefly explain why the statement {0 ≤ yi ≤ θ for all i = 1, . . . , n} is equivalent to the
statement {m = max (y1, . . . yn) ≤ θ}, and use this to show that the joint distribution of
Y = (Y1, . . . , Yn) (given θ) in this model is
fY1,…,Yn
(y1, . . . , yn | θ) = I(m ≤ θ)
θ
n
. (13)
[20 points]
(b) Letting the observed values of (Y1, . . . , Yn) be y = (y1, . . . , yn), an important object in both
frequentist and Bayesian inferential statistics is the likelihood function `(θ | y), which is obtained from the joint distribution of (Y1, . . . , Yn) (given θ) simply by
(1) thinking of fY1,…,Yn
(y1, . . . , yn | θ) as a function of θ for fixed y, and
(2) multiplying by an arbitrary positive constant c:
`(θ | y) = c fY (y | θ). (14)
Using this terminology, in part (a) you showed that the likelihood function in this problem
is `(θ | y) = θ
−n
I(θ ≥ m), where m is the largest of the yi values. Both frequentists and
Bayesians are interested in something called the maximum likelihood estimator (MLE) ˆθMLE,
which is the value of θ that makes `(θ | y) as large as possible.
(i) Make a rough sketch of the likelihood function, and use your sketch to show that the
MLE in this problem is ˆθMLE = m = max (y1, . . . yn). [20 points]
(ii) Maximization of a function is usually accomplished by setting its first derivative to 0 and
solving the resulting equation. Briefly explain why that method won’t work in finding
the MLE in this case. [10 points]
(c) A positive quantity W follows the Pareto distribution (written W ∼ Pareto(α, β)) if, for
parameters α, β > 0, it has density
fW (w) = 
α βα w
−(α+1) if w ≥ β
0 otherwise 
. (15)
This distribution has mean αβ
α−1
(if α > 1) and variance αβ2
(α−1)2(α−2) (if α > 2). EasyDue™ 支持PayPal, AliPay, WechatPay, Taobao等各种付款方式!

E-mail: easydue@outlook.com  微信:easydue

EasyDue™是一个服务全球中国留学生的专业代写公司