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Unformatted text preview: SOLUTION FORHOMEWORK 1, STAT 4382 Well, welcome to your ﬁrst homework. In my solutions you may ﬁnd some seeded mistakes
(this is why it is not a good idea just to copy my solution). If you ﬁnd them — please, do
not email or call me. Instead, write down them on the ﬁrst page of your solutions and you
may give yourself an extra credit — but keep in mind that the total for your homeworks
cannot exceed .20 points. I ' Now let us look at your problems. 1. Problem 1.2.1. . This is a nice exercise to polish your understanding about combination of events. DeMor
gan’s Law, which states that (U,E,)C = WEE, and similarly (WE)3 = UiEf, is useful. Also,
using complementary events may be useful. Here we are dealingwith three arbitrary events
E, F and G‘ from the same sample space 8. Remember that we may write EF 2: E n F to
make formulae simpler. (a). Only F occurs is the event F(E U 6")" = PIE“ (1 Ga) = FECGC.
(b) both E and F but not G can be written “as you read it”
_ EFGC.
(c) at least one event occurs is simply
E U F U G
(d) at least two events occur you may write as EFUFGUEG (e) all three events occur is simply
EFG (f) none occurs is the complementary event to at least one occurs,
(E U F U G)“ = ECFCGC. In the last equality I used DeMorgan’s Law.
(g) At most one event occurs is the complementary event to at least two occurs, discussed
in ((31). So we get via DeMorgan’s law (EF)C(FG)C(EG)C. (h) At most two occur is the complementary to all three events occur, discussed in (e), so we get
(Eire)c 1 2. Exercise 1.2.1. The law of total probablity tells us that if E1, . . . ,En is a partition of
the sample space, that is, these events are mutually exclusive and their union is the sample space, then
Pr(A) = Pr'(A n @3130) = Z Pr(A n E1). £11
Note that B and B‘3 are always a partition.
Keeping this in mind, we can verify the formula. Write, Pr_(A) = Pr(An (B U BC)) :: Pr(AB) + Pr(ABC).
Remark: Note that we used one of the DISTRIBUTIVE LAWS (EUF)G=EG'UFG’, EFUG=(EUG)(FUG). 3. Exerc 1.2.2. Let us do it using the total probability and distributive laws. Write
Pr(A) = Pr(AB) + Pr(ABc)
and
Pr(A U B) : Pr((A U B)B) + Pr((A U B)BC)
= Pr(AB U B) + Pr(ABC) = Pr(B) + Pr(ABC). Find Pr(ABC) from the ﬁrst line, substitute in the second, and we get the wished (classical!) equality
Pr(A U B) = Pr(A) + Pr(B) — Pr(AB). You can also use the book’s hint and/ or check using Venn Diagram. 4. Problem 1.2.3. The plot is below: (b) The corresponding density function is:
f(:c) = 01(9: 3 0) + 3$2I(0 < :1: < 1) + 01(23 2 1). (c) The mean is E(X) = [m 33f($)da: = fol m(3x2)d:c : 3(1/4)934 ;=0 = 3/4. —00 5".
V 2
*1? (d) By deﬁnition of the cdf F(:1:), and keeping in mind that X is continuous RV on the
real line, we write ‘ Pr(1/4 g X 53/41) = F(3/4) — F(1/4) : (3/4)3 — (1/4)? 5. Exerc 1.2.4. (a) Below is the plot: m, , g. { ~—————~—
___.__.————'> :9 1’ >
, $3; aa————>
__i
5,
t7
0 'f 2. .3 ,‘r
(b) By deﬁnition
—ZZPZ(Z 0)(+1/4) (1)(1/2)+(2)(1/8)+(3)(1/8) 39/8 (c) By deﬁnition
Var(Z):E{(Z—E(Z D2}: Z( 2—1? )2 pz(z). _ There is also a useful formula Var(Z) 2 3(22) — [Em]? that is often simpler to use. Indeed, the ﬁrst moment is already calculated, so we need to
calculate the second moment: E(ZQ) =Zzgp2(z) 0(1/4)+ (1 )2(1/2) + (2)2(1/8) + (3)2(1/8) = 17/8
Plug in and get 1
meZ) = ”Q7“ ... [38? = 55/64. 6. Exerc. 1.2.5. Well, here we will use the result of Exercise 1.2.2. Write
Pr((A u B) U C) = Pr(A U B) + Pr(0)  Pr((A U B) n C)
[using distributive law for the analysis of the third term]
= [Pr(A) + Pr(B) — Pr(A n B)] + Pr(C) — Pr((A o C) U (B n 0))
[using the result of Exerc. 1.2.2 for the last term] = [Pr(A) + Pr(B) + Pr(C) — Pr(A n B)] — [Pr(A n C) + Pr(B n C) — Pr((A n C) n (B n Clll 3 = [Pr(A) + Pr(B) + Pr(C)] ~— [Pr(A ﬂ 3)] + PI(A ﬂ 0) + Pr(B ﬂ 0)] + Pr(A H B D C).
What was wished to show. 7. Exerc. 1.2.6. (a) Here Z := max(X, Y) where X and Y are independent with known
distributions. Now we make the following classical steps. For any z E (—00, 00) we can write,
FZ(z) :: Pr(Z S 2:) : Pr(maX(X, Y) S 2) I Pr([X S 2) (“I (Y _<_ z)) [using independence of X and Y we continue]
= Pr(X 5 z)Pr(Y S 2) = FX(z)FY(z). What was wished to prove. (b). Here W := min(X, Y). To get the cdf of W, we use the same technique only here it
is convinient to work with the survivor function GW (10) :2 Pr(W > w) z: 1 — F W(w). For
any real to we can write GW(w) := Pr(W > w) = Pr(min(X, Y) > w) = Pr((X > w) H (Y S w))
[using independence of X and Y we continue]
= Pr(X > w)Pr(Y > z) = GX(w)GY(w).
Now remember the relationship between the cdf and the survivor function, and get
FW(w)=1— GW(w) = 1 — (1 * FX(w))(1— FY(w)).
8. Exerc. 1.2.7. Here we consider a random variable X with the density
fX(a:) = T$T_1I(0 g a: S 1). Note that I use 7“ in place of the book’s R because I prefer to use upper case letters for
random variables and lower case letters for numbers (parameters).
(a) The distribution function is FX(:IJ)‘I= [:0fX(u)du= (0)I(3: S 0)+331"I(0 < :1: g1)+I(x> l). (b) Write 1 1
E(X) =/0 me(33)d33 = r/O fab: m T/(T + 1).
(c) For the variance I use Var(X) = E(X2) — [E(X)]2. Write, E(X2) = [01x2[r:r:r‘1ld2: z r/(r + 2). This allows us to conclude that Var(X) = r/(r + 2) — [T/(T' + 1)]2. 4 9. Exerc. 1.2.9. Here the following pdf of X is given: fX(x)=:tI(US:I:S1)+(2—a:)I(1§;c<2). (a) The cdf is = (0)I(:r S 0) + (1/2):C2I(0 < :1: S 1) + [(1/2) +23; .. (1/2)“):2 —3/2]I(1 < x g 2) +I(3: > 1).
(b) The expectation is E(X)=[mex(:c)dm=[olrc2dx+/12x(2—a:)dm = (1/3)+1932 — (1/3)$3l§=1 = (1/3) + [3  (7/3)] = 1 (c) We again use the moments to calculate variance: E(X2) =/1 x3d$+f12m2(2—x)dm 0 2 (1/1) + [new —(1/4)w41:1 = (1/1) + [(2/3)? — (1/4105) = 11/12 = 1/6. This yields
Var(X) = E(X2) « [E(X)]2 = 7/6 — 12 n 1/6. 10. Exerc. 1.2.10. This is a nice problem to improve our understanding of the indicator
function and complementary event. Please note that Pr(X E A) :2 E{I (X E A)}, so it is
useful to be familiar with the indicator. In what follows I write I (A) in place of 1A because
the former is more traditional. (a). Let us prove that
I(A) = 1 — I(A°). Indeed, if the event A occures then AC does not. As a result, if A occures then the left side
is 1, and the right is 1 — 0 = 1. If A does not occur, then the left side is zero, and the right
side is 1 — 1 = 0. What was wished to show. There is another way to prove this equality _
(which is also the underlying idea of the total probabiliy law). Write for the sample space S
(remember that it this even always occures) 1 21(8) =I(AUAC)
[using the fact that A and Ac are mutually exclusive]
= I(A) + NA“). This is a direct proof.
(b) Let us show that
I(Aﬂ B) = I(A)I(B). 5 Indeed, the left side is 1 if and only if both A and B occur, otherwise the left side is zero.
The same is true for the right side. Next, let us show that I(A)I(B) = min(I(A), 1(3)). Indeed, the left side is 1 if and only if both A and B occur, otherwise the left side is zero.
The same is true for the right side. What was wished to show. (e) Let us show that
I(A U B) : max(I(A), I(B)). Indeed, the left side is 1 if at least one of the two events occur, otherwise it is zero. And the
same conclusion holds for the right side. What was Wished to Show. ' ...
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 Fall '11
 krwaw

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