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Unformatted text preview: Statistics 24400  Winter 2016 Midterm Solutions
February 3 and 4, 2016 Name (print): On my honor, I will not discuss this exam with ANY PERSON before 18:30 CST
(12:30 GMT) February 4, 2016. Signature 1. Please print your name in the space provided.
2. Do not sit directly next to another student.
3. Do not turn the page until told to do so. 4. This is a closed book, closed notes examination. You are permitted to have a calculator.
Devices capable of connnunication (laptops, tablets, phones) must be powered down. 5. Please provide the answers in the space and blank pages provided. If you do not have
enough space, please use the back of a nearby page, clearly indicating the identity of
the continued problem. 6. Be sure to Show your calculations. In order to receive full credit for a problem, you
must show your work and explain your reasoning. Good work can receive substantial
partial credit even if the final answer is incorrect. 7. Read through the exam before answering any questions. Our scale of credit for questions
may not correlate with the level of difﬁculty you experienceHuse your time wisely! 1. For parts (a)—(h), circle T for true or F for false. Each question is worth two points,
but you don’t start getting points until you have two correct answers. In other words.
if you answered 71 questions correctly, your total number of points 1) = @(2  (n .. 2)).
where 9(1) = 0. II: S 0, @(rc) = .17, .1; > 0. Hence 12 points are possible. (a) ( 2 pts) T Q)
The pdf of a standarc normal random variable X ~ NH), 1) is given by
L
1 s2 / :3:
n6 W 6 1’ (b) (2 pts) (T) F
The 13ml of a binomial random variable X N B(n,p) is given by nl {EMU}? _ $)l7)1(1—p)"—JI @ r [/chKXHU“‘/" (kl—RI") (C) (2pts) L/tH‘(X)+ Q’ﬁ/arf‘r) + (”/(aVCXJ F) —'— L/QJ‘K (X)
Coax y) — _—[Var( (X + 23’) — Va1(X  2W] ”W w c V)
(d) (2 W) T G.) + Lisgmz/ (23 F)
1/7 if! 1:3 1‘1. w 12. NS) F (ﬂat/(X) U
z _,i/ : ., __.._____ = 2
for“  ,ﬂ; 9. mam) ﬁle's '
[IKE [4(1) (e) (2pm) T (Q) Let X and Y be ran 11 valiables withjoint Lclfjm y). Lot f(:r $1.39 f( .r 1;)ci11. ( _ )i
may f_;)j (.1: y)c.1: Then ii 552% S ropes) : /
/‘ 00 co 6 /H,t_. ,
" foo /m f (:r)g(y)dmdy= / / j'(.1:,y)ri:rdy ‘ 7/ 5 ° for all .1‘ and y, then X and Y are indepenclgni
‘ (ll) (21315) T CID /’/Q, 6) H/Ic—Lig I pwr‘2Q5 1/03 L (9)” If 0 maximizes the Likelihood function L(6), then log(6) maximizes log 15(0). (g) (213%) @ F m 5 E : Mu a— + ( gums)?" “gal“L/f
If the MSE of all estimator is zero, then the bias is zero also. 9 C9“ (h) (2])LS) CV F If you know any two of f(0), f(.1:  0), and f(0 I :r), you can ﬁnd the third. yg9/ g (gcciaéQ 1C/39 Ag) :2. ‘C/X/O) W67) __,_._.______________—
‘l l : «5 feta0.99019 2. A Poisson Process (30 pts) Customers appear at a box ofﬁce according to a Poisson
process with rate parameter A. Let us denote the number of customers arriving in the interval [0,t) by NE. (a) (5 pts) Find P(N2 = 1 I M; = 3). (l3) (7 ptS) Find PUVZ = 1 I N4 — N3 = 3). (c) (9 pts) Find the expectation and variance of N20 —— N10.
( d) (9 pts) Find the joint distribution of (N3, N5). Given the observation that (N3, N5)
(n3, n5), ﬁnd a maximum likelihood estimate of A based on these observations. ll Solution. (a) P(N2=1N4=3) —P(N2=mN423) P(N4 = 3)
_ P(N2=lﬂN4—N2=2)
_ P(N4 = 3)
= W < from independence >
P(N4 = 3)
6e—2*(2A)e2*(2A)2
Z 26*4"(4A)3 3
8 PUV2 = 1 l N4 — N3 = 3) = P(N2 = 1) < from independence >
2A5” (c)N20 ~ N10 ~ Poisson( 10A). The expectation and and variance are both therefore
equal to 10A. (d) P(N3=n3ﬂN5=n5) : P(N3=TL30N5‘N3 :n5in3)
P(N3 = n3)P(N5 — N3 = n5 W 72.3) < from independence >
€73A(3A)N3e—2A(2A)n5—ng n3!(n5 — 713)! ll Let, f()\) = IOEW = H5A + n5log/\ + K, where K does not
depend on A. Differentiating f (A) w.r.t A and equating it to zero gives ;\ = 715/5. We also check that f”()\) < 0, which shows that the maximum likelihood estimate of A is “F5. 3. Bayesian Inference (30 pts) Recall that ifA ~ Gamma(a,ﬁ) with a, 13 > 0, then its
pdf is given by (a) (b) f()\;oz.ﬁ)=15;)91e,“‘3A .\>o. Use the fact that a pdf integrates to 1 to ﬁnd E [A], and then using calculus, locate
the mode of the pdf. You may need the identity I‘(a + 1) = amen). EA =/ Af(/\  cumin
0
BC! foo {\(a+l)_1€_ﬁAdA
0 : ma)
)6“? (a + 1)
=0)“ 60—“ )fowmld+lﬁ)d)\
_ 3
H
To ﬁnd the mode, set the ﬁrst derivative of the pdf equal to zero, and solve for A.
—f(/\ l a 6): F1: )[(a w DAG—26"“ — ﬁA“_le_ﬁ’\] = 0.
0:
Since A04 and e"3A are both positive, we may divide by them so that
a — 1
A =
[3 Note that a > 1, because /\ > 0, but no points were taken off for not stating this.
(But kudos to those who did, as well as those who justiﬁed that this is indeed the mode!) Suppose that causes of death are reviewed in detail for a city in the United States
for a single year. It is found that 3 persons, out of a population of 200, 000, died
of asthma, giving a crude estimated asthma mortality rate in the city of 1.5 cases
per 100,000 persons per year. Let X be the number of deaths due to asthma in a
year. It is believed that X ~ Poisson(NA), where N is the size of the population
in 100, 000’s. So, for a city of a population of 200,000, X ~ Poisson(2A). Taking Gamma(3, 5) as a prior distribution for A, use the expression derived in the
ﬁrst part to compute the prior mean and mode. (5 pts for the mean, 5 pts for the
mode) EA: ;the mode is — UHFO (c) Derive the posterior density for A, and compute the posterior mean and mode. (2.5
pts for the mean, 2.5 pts for the mode) (10 pts for the derivation of the posterior density, 2.5 pts for the mean, 2.5 pts for
the mode) X m Poisson(2A), X = 3. f()\ I X = 3) oc f(X = 3  A)f()\) oc (2A)3€_2A  A3_1e_5’\ oc A6"le‘7’\, which we recognize as Gamma(6, 7) density. 5 E[A  X = 3] = g; the posterior mode is 7. 4. Rolling the Dice (22 pts) (a) (22pts) You simultaneously roll 0! fair dice. Let the random variable X be the
outcome of the highest scoring die and Y be the outcome of the second highest
scoring die with the convention that the second—highest score equals the highest
score in the case that two or more dice yield the highest score. What is the joint
probability mass function of X and Y? Solutions: The probability P(X = LY = i) is equal the probability that two or more dice
yield the highest score 2' and the other dice score less than 2'. Hence (HM) (“ll—r
( l (1) (,_,)3_3_ (“liddt‘llq
0 6 6 6
ﬁ1)d_ d(' _1)d l
The probability P(X = i, Y = j) for i > 3' is equal the probability that exactly one 6d
die yields the highest score i, at least one die yields the second— highest score 3' and
the other dice score less than 3'. Hence, for i > j, <W:< )(>“‘<e>“““
= (3% l: t) t)" (is) (2%)“? P(X=i,Y=i) = _,
All a.
M IFSII r
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