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Unformatted text preview: 1. Please deﬁne the following terms: (a) A partition of a sample space S]: A partition of a sample space (2 is a collection of pairwise dis—
joint events A1, A2, ..., A” whose union is 0. That is, Al, A2, ..., A”
form a partition of S2 if (i) A; 0 A; = (a for all 2' 7E j, and (ii)
AlUAgU...UAn=Q. (b) A discrete random variable: A random variable is called a discrete
random variable if... ....if the number of values that it can take is’either ﬁnite or
countably inﬁnite. (0) Cumulative distribution function: The cumulative distribution function F (:5) of a random vari—
able X is deﬁned as F(:r) = P(X S as) (d) Expected value: The expected value of a discrete random variable X is EX = Z$P(X = m) where the sum is over all values of :r with P(X = 3:) > 0. 2. (15 points, 5 points each) Give the probability mass functions for each
of the following: (a) The Geometric distribution: __ ’ . ._  :(‘l 
PCX) : P<K 3‘ X) : /" (Lip) P X: £213.” _,
NOLO K {S 444.1 RV Comak5 «f’l/uz nJWLLef 01C ‘lm‘colﬁ “WM “Hm ‘gVS'l SKCQZXS.) 6t» L06 6241’ ' ”(Uml ”K
‘ VC/‘LCL a)? Wadi; ﬁnalV 'W’rba‘f 41g; .wbentfl} a?
$35 M {QC/LL M4 vs F F ‘CLI
(b) The Poisson distribution:
[.er I: be a; RV VJLLILL ﬂ POPSSOA clgsixilﬁumlfcl MDH’i/k
PM” A Ell/LEAK
. ”a 5‘
p(xl“—P(X=X)= A ’32! (c) The binomial distribution:
X €001ka «Hui 4F 0? Sou'coi‘m m A MCLL’P. Nah}
«Back “3% FmLaJarlrg P oi ”(0253: The“. POO :3 PCX: K) : (3:1)PXO'PY‘K .Lr xiii/Jr)“ 3. (20 points, 10 points each) Suppose that X and Y are two independent
random variables with probability mass functions given by the table below.
(8.) Find and graph the corresponding distribution function for X. F(x):\ 0 N<‘”l .K‘) NY:©  Cg" rlEX‘lC)
0:6: 0 ix“)
.032 , Ova) lax:1
70...]   s
. ‘05} ,_ I 3
T7
1:: ;
Hi} (b) Find var(X + Y). Make sure to explain your reasoning. 3. (b) Find var(X + Y ). Make snre to explain your reasoning. Since X and Y are independent, var(X + Y) = V3.1‘(X) + var(Y). To
compute var(X), we ﬁnd EX and EX 2 ﬁrst. _ EX = Z kP(X = k) = (—1)(0.1)+ 0(0.5) + 1(o.3) + 2(0.1) = 0.4.
k 5X2 = 218130: = k) = (—1)2(o.1) + 0203.5) + 12(0.3) + 22(0.1) = 0.8.
h Then vaJ:(X) = EX2 — (EX)2 = 0.8 — (0.4)2 = 0.64. Similarly, we can ﬁnd
var(Y) as follows: EY = Z kP(Y = k) = (—1)(0'.2) + 0(0.1) + 1(0) + 2'(0.7) = 1.2. 51/2 = Z k2P(Y = k) = (—1)2{0.2) + 02(0.1) + 12(0) + 22(07) = 3.
k
var(Y) = EY2 — (EY)2 = 3 — (1.2)2 = 1.56. Therefore, var(X + Y) = var(X) + var(Y) = 0.64 + 1.56 = 2.2. 4. Compute the variance for a geometrically distributed random variable X ,
assuming that we already know that EX = i. 'We use the fact that for real number q such that q < 1, Z Mk i Ugh—2 = .
H (1 e (1)3
First we compute E[X(X — 1)]. E[X(X _ 1)] = flew # 1)P(X = k) k=1
: iMh— 1)P(X = k)
: Zke— 1)(1 —p)k*1p
= (1 ‘19)19: WC — 1)(1 — 120H
,2 2
= (1 ‘1’)?” (1 — <1 —p»3
= 210(1 *2?)
p3
= 2(1 —P)
102 Notice that
E[X(X — 1)] = E[X2 — X] : EX2 — EX.
So EX2 : E[X(X ~ 1)] + EX. Hence,
var(X) = EX2 # (EX)2
= E[X(X — 1)] + EX — (EX)2 :2(1p)+1_i p2 p :32
#2(1cp) 1—19
g p2 _ p2 1—10 5. (24 points, 8 points each part) Let’s look at the following rare event: Suppose that the probability of rain on any given day of the summer    1
1n Los Angeles 1s E' (3.) Using both the binomial distribution and the Poisson approxima
tion to the binomial distribution, ﬁnd the probability of k rainy
days over a 90 day period of the summer. 0 accordini to the binomial distribution:
n :51 O
l 3 S F (34:1): <0]:\ 0 We?) (135) 0 according to the Poisson distribution: \=“r= 10(%ﬁ3
Fo<= \<\ =' e: (3%)" \< ‘. MN (b) What is the expectation and variance for the number of rainy days
in this period: 0 according to the binomial distribution:
7. Ego: “F : QC (xzr ":i: \WWA: “MW” 130 ' ill—5‘ 0 according to the Poisson distribution:
>\ : EX: vat/(XS = hf
, ,r 1:: {1/— X): l
E >\ v L 6 3 (c) Thinking about the binomial distribution, What is n and what is
p here? If we keep n constant, what will happen to the variance
of the binomial distribution as 1) becomes very very small? How
does this agree with what we discussed in class? ' ‘r’\ ‘2 VMmbm’ £4; drug '3
F: flUVJcJQiU‘h/s at a VEGA“ dam) ' r '1
m» ﬁve—w WW”
’2
go as P be COVWZS ‘w~€v'l_r\ Six/v1!» ﬂu; +erm IA"? {3 \oecomes IWcjh‘gcdole Compared to n J 90 wit/[K59 n‘o
“"[i‘vxé m imam and. The; _\,—(Lv1‘(w\C.€ (ire ﬂu; gifting
“CCV' ’n/kL POnggn I chew1.30mi? 31A (“WI/CKIW'HCM ‘f‘E Eiﬂﬁvmal 5% "W17 bu ’ﬂ“m\ , 6. (11 points) Let’s suppose that Octomom is a carrier for hemophilia,
and that starting now she will continue to have'children until one has
hemophilia, at which point she will stop. Suppose further that each
one of her children'has a different father, that hemophilia in the father
population is Poisson distributed with parameter 0.01 (Note: I have no
idea if this is realistic), and that it is equally likely that she has a boy
or a girl. What is the probability of Octomom having k more children?
Please explain all of your reasoning carefully. hamm— allelLvadad
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P V: P(Cl/Luld l/UKS MMOPHIIJEL): P(Hem ’61: s’i) F(Gl>fl)‘1‘ POitm ’533R/%
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: magi,2 ~L Mali”‘—
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b you gilgrl e/CL'R th‘l' 5):” some. a: a.“ a; #05 ’jwLQW». I L‘Civ"4’ Nf(apdd +0 ngw't CPLL‘l' P( aHﬁgwl‘eFllmﬁDai
Les}; “35““ l 3 OOl sme’k' P’CL £00): +(01).,; 'L 5:755 COLA P(OC40W\DWL has K HA0“ (Ll/”Hm A): (jg/08>KL'JLEC’“ ...
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 Spring '07
 SCHONMANN

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