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Обнинский институт атомной энергетики НИЯУ МИФИ

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Ohrimenko+ / Ross S. A First Course in Probability_ Solutions Manual

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			1000
	149.5 −		1000
	149.5 −
23.	(a) P{149.5 < X < 200.5} = P		6	<
		1000	1 5

			6 6
			6 6
	200.5 −166.7
	= Φ
		5000 / 36
		5000 / 36

	≈ Φ(2.87) + Φ(1.46)


(b) P{X < 149.5} =		149.5 −800(1/ 5)
(b) P{X < 149.5} =	P Z <		1 4
		800	1 4

			5 5
			5 5

200.5 −		1000

		6
Z <		6
Z <			1 5
	1000		1 5

			6 6
			6 6
149.5 −166.7
−Φ
	5000 / 36
	5000 / 36

− 1 = .9258.

=P{Z < −.93}

=1 − Φ(.93) = .1762.

24.With C denoting the life of a chip, and φ the standard normal distribution function we have

	6		1.8×106 −1.4 ×106
P{C < 1.8 × 10		} = φ
P{C < 1.8 × 10		} = φ		5
			3×10	5
			3×10
		= φ(1.33)
		= .9082

Thus, if N is the number of the chips whose life is less than 1.8 × 106 then N is a binomial random variable with parameters (100, .9082). Hence,

	19.5 −90.82
	19.5 −90.82
P{N > 19.5} ≈ 1 − φ		= 1 −φ(−24.7) ≈ 1
P{N > 19.5} ≈ 1 − φ	90.82(.0918)	= 1 −φ(−24.7) ≈ 1
	90.82(.0918)

25. Let X denote the number of unacceptable items among the next 150 produced. Since X is a binomial random variable with mean 150(.05) = 7.5 and variance 150(.05)(.95) = 7.125, we obtain that, for a standard normal random variable Z.

P{X ≤ 10} = P{X ≤ 10.5}
X −7.5		≤	10.5 −7.5
= P	7.125		7.125

≈ P{Z ≤ 1.1239} = .8695

The exact result can be obtained by using the text diskette, and (to four decimal places) is equal to .8678.


27.	P{X > 5,799.5} =		799.5
27.	P{X > 5,799.5} =	P Z >	2,500
			2,500

= P{Z > 15.99} = negligible.

Chapter 5

28.Let X equal the number of lefthanders. Assuming that X is approximately distributed as a binomial random variable with parameters n = 200, p = .12, then, with Z being a standard normal random variable,

		X − 200(.12)
P{X > 19.5} =		X − 200(.12)	>	19.5 − 200(.12)
P{X > 19.5} =	P	200(.12)(.88)	>
		200(.12)(.88)		200(.12)(.88)

≈P{Z > −.9792}

≈.8363

29.Let s be the initial price of the stock. Then, if X is the number of the 1000 time periods in which the stock increases, then its price at the end is

suXd1000-X = sd1000 u X

Hence, in order for the price to be at least 1.3s, we would need that

d1000 du >1.3

X > log(1.3) −1000log(d) = 469.2 log(u / d)

That is, the stock would have to rise in at least 470 time periods. Because X is binomial with parameters 1000, .52, we have

		X −1000(.52)
P{X > 469.5} =		X −1000(.52)	>	469.5 −1000(.52)
P{X > 469.5} =	P	1000(.52)(.48)	>
		1000(.52)(.48)		1000(.52)(.48)

≈P{Z > −3.196}

≈.9993

30.

P{in black} =

P{5

black}α

P{5

black}α + P{5

white}(1−α)

−(5−4)2

/ 8

2π

−(5−4)2 / 8

−(5−6)2

/18

α + (1 −α)

2 2π

3 2π

α e−1 / 8

(1−α)

e−1 / 8

α is the value that makes preceding equal 1/2

Chapter 5

31.

(a)

X − a

] = ∫(x − a)

+ ∫(a − x)

−

a −

( ) =

−1 = 0 a = A/ 2

(b)

X − a

] = ∫a (a − x)λe−λxdx + ∞∫(x − a)λe−λxdx

= a(1−e−λa ) + ae−λa +

e−λa

−

+ ae−λa +

e−λa

− ae−λa

Differentiation yields that the minimum is attained at a where

e−λ

=1/ 2 or a = log 2/λ

(c)

Minimizing a = median of F

32.

(a)

e−1

(b)

e−1/2

33.

e−1

34.

(a)

P{X > 20} = e−1

(b) P{X > 30 X > 10 =

P{X > 30}

1/ 4

= 1/3

3/ 4

P{X >10}

35.

(a)

exp − ∫

λ(t)dt = e−.35

(b)

e−1.21

= e−4

36.

(a)

1 − F(2) = exp −

∫t3dt

(b)

exp[−(.4)4/4] − exp[−(1.4)4/4]

(c)

exp − ∫t3dt = e−15/4

37.(a) P{ X > 1/2} = P{X > 1/2} + P{X < −1/2} = 1/2

(b)P{ X ≤ a} = P{−a ≤ X ≤ a} = a, 0 < a < 1. Therefore,

f X (a) =1, 0 < a < 1

That is, X is uniform on (0, 1).

Chapter 5

38.For both roots to be real the discriminant (4Y)2 − 44(Y + 2) must be ≥ 0. That is, we need that Y2 ≥ Y + 2. Now in the interval 0 < Y < 5.

Y2 ≥ Y + 2 Y ≥ 2 and so

P{Y2 ≥ Y + 2} = P{Y ≥ 2} = 3/5.

39.FY(y) = P{log X ≤ y}

=P{X ≤ ey} = FX(ey)

fY(y) = fX(ey)ey = eye−e y

40.FY(y) = P{eX ≤ y}

=FX(log y)

fY(y) = fX (log y) 1y = 1y , 1 < y < e

Chapter 5

Theoretical Exercises

1.The integration by parts formula ∫udv = uv − ∫vdu with dv = − 2bxe−bx2 , u = −x/2b yields that

	∞∫x2e−bx2 dx = − xe2b−bx2 ∞∫+						1		∞∫e−bx2 dx
	∞∫x2e−bx2 dx = − xe2b−bx2 ∞∫+						2b		∞∫e−bx2 dx
	0				0			0
		=	1		∞∫e−y 2 / 2dy by y = x 2b
		=	(2b)3 / 2		∞∫e−y 2 / 2dy by y = x 2b
					0
		=	2π			1		=		π
			2 (2b)3 / 2						4b3 / 2
	where the above uses that			1		∞∫e−y 2 / 2dy = 1/2. Hence, a =					4b3 / 2
				2π		0					π
						0
	∞	∞ −y
2.	∫P{Y < −y}dy =	∫ ∫ fY (x)dx dy
	0	0 −∞
	= ∫0 −∫x fY (x)dy dx = − ∫0 xfY (x) dx
		−∞ 0							−∞
	Similarly,
	∞∫P{Y > y}dy = ∞∫xfY (x) dx
	0	0
	Subtracting these equalities gives the result.
4.	E[aX + b] = ∫(ax + b) f (x) dx = a∫xf (x)dx + b∫ f (x)dx
						= aE[X] + b
5.	E[Xn] = ∞∫P{X n > t}dt
	0
	= ∞∫P{X n > xn}nxn−1dx by t = xn, dt = nxn−1dx
	0
	= ∞∫P{X > x}nxn−1dx
	0
6.	Let X be uniform on (0, 1) and define Ea									to be the event that X is unequal to a. Since ∩Ea is
											a

the empty set, it must have probability 0.

Chapter 5

7.SD(aX + b) = Var(aX +b) = a2σ2 = a σ

8.Since 0 ≤ X ≤ c, it follows that X2 ≤ cX. Hence,

Var(X) = E[X2] −(E[X])2

≤ E[cX − (E[X])2

=cE[X] − (E[X])2

=E[X](c − E[X])

=c2[α(1 − α)] where α = E[X]/c ≤ c2/4

where the last inequality first uses the hypothesis that P{0 ≤ X ≤ c} = 1 to calculate that 0 ≤ α

≤ 1 and then uses calculus to show that maximum α(1 − α) = 1/4.

0≤α≤1

9.The final step of parts (a) and (b) use that −Z is also a standard normal random variable.

(a)P{Z > x} = P{−Z < −x} = P{Z < −x}

(b)P{ Z > x} = P{Z > x} + P{Z < −x} = P{Z > x} + P{−Z > x}

=2P{Z > x}

(c)P{ Z < x} = 1 − P{ Z > x} = 1 − 2P{Z > x} by (b)

=1 − 2(1 − P{Z < x})

10.With c = 1/( 2πσ )we have

f(x) = ce−( x−µ)2 / 2σ 2

f ′(x) = −ce−( x−µ)2 / 2σ 2 (x − µ) /σ2

f ′′(x) = cσ−4e−( x−µ)2 / 2σ 2 (x − µ)2 −cσ−2e−( x−µ)2 / 2σ 2

Therefore,

f ′′(µ + σ) = f′′(µ − σ) = cσ−2e−1/2 − cσ−2e−1/2 = 0

	2	∞		∞	−λx		2
	2		2	−1	−λx		2	2
11.	E[X	] = ∫P{X > x}2x		dx = 2∫xe		dx =		E[X ] = 2 / λ
11.	E[X	] = ∫P{X > x}2x		dx = 2∫xe		dx =	λ	E[X ] = 2 / λ
		0		0
12.	(a)	b + a
12.	(a)	2
		2

(b)µ

13.(a) all values in (a, b)

(b)µ

Chapter 5

14.P{cX < x} = P{X < x/c} = 1 − e−λx/c

15.	λ(t) =		f (t)	=	1/ a	=	1	, 0 < t < a

		F (t)			(a −t) / a		a −t

16.If X has distribution function F and density f, then for a > 0

FaX(t) = P{aX ≤ t} = F(t/a)

and

fax =

f (t / a)

Thus,

f (t / a)

(t) =

(t / a) .

1− F(t / a)

18.

E[Xk] =

∞∫xk λe−λxdx = λ−k ∞∫λe−λx (λx)k dx

= λ−k Γ(k+1) = k!/λk

19.

E[Xk] =

∞∫xk λe−λx (λx)t −1dx

Γ(t)

−k

∞

∫λe−λx (λx)t +k −1dx

Γ(t)

λ−k

Γ(t + k)

Γ(t)

Therefore,

E[X] = t/λ,

E[X2] = = λ−2Γ(t + 2)/Γ(t) = (t + 1)t/λ2

and thus

Var(X) = (t + 1)t/λ2 − t2/λ2 = t/λ2

Chapter 5

		∞
20.	Γ(1/2) = ∫e−x x−1 / 2dx
		0
		∞
	=	2 ∫e− y 2 / 2 dy by x = y2/2, dx = ydy = 2x dy
		0
	= 2 π ∞∫(2π)−1 / 2 e−y 2 / 2dy
		0
	= 2 π P{Z > 0} where Z is a standard normal
	=	π
21.	1/λ(s) =	∫λe−λx (λx)t −1dx / λe−λs (λs)t −1
		x≥s
	= ∫e−λ( x−s) (x / s)t −1dx
		x≥s
	=	∫e−λy (1+ y / s)t −1dy by letting y = x − s
		y≥0

As the above, equal to the inverse of the hazard rate function, is clearly decreasing in s when t ≥ 1 and increasing when t ≤ 1 the result follows.

22.λ(s) = c(s − v)β−1, s > v which is clearly increasing when β ≥ 1 and decreasing otherwise.

23.F(α) = 1 − e−1

24.Suppose X is Weibull with parameters v, α, β. Then

X −v

1 / β

≤ x =

≤ x

= P{X ≤ v + αx1/β}

= 1 − exp{−x}.

25.

We use Equation (6.3).

Γ(a +1) Γ(a + b)

E[X] = B(a + 1, b)/B(A, b) =

Γ(a +b +1) Γ(a)

a + b

Γ(a + 2) Γ(a +b)

(a +1)a

E[X

] = B(a + 2, b)/B(a, b) =

Γ(a +b + 2) Γ(a)

(a +b +1)(a +b)

Thus,

Var(X) =

(a +1)a

−

+b +1)(a +b)

(a +b)2

(a +b +1)(a

+b)2

26.(X − a)/(b − a)

Chapter 5

28.P{F(X ≤ x} = P{X ≤ F−1(x)}

=F(F−1(x))

29.FY(x) = P{aX + b ≤ x}

			x −b
=	P X ≤			when a > 0
=	P X ≤		a	when a > 0
			a
= FX((x − b)/a)				when a > 0.
fY(x) =	1	fX((x − b)/a) if a > 0.
fY(x) =		fX((x − b)/a) if a > 0.
	a

When a< 0, F (x) = P X

fY(x) = − 1 fX x −b . a a

30.FY(x) = P{eX ≤ x}

=P{X ≤ log x}

FX(log x)

fY(x) = fX(log x)/x

=	1	e−(log x−µ)2
x	2πσ

	x −b		x −b
≥		=1	− FX		and so
	a			a

/ 2σ 2

Chapter 5

Chapter 6

Problems

2.(a) p(0, 0) = 138 127 = 14/39,

p(0,

1) = p(1, 0) =

8 5

= 10/39

13 12

5 4

p(1,

1) =

= 5/39

13 12

(b) p(0, 0, 0)

8 7 6

= 28/143

12 11

8 7 5

p(0, 0, 1)

= p(0, 1, 0) = p(1, 0, 0) =

= 70/429

13 12 11

p(0, 1, 1)

= p(1, 0, 1) = p(1, 1, 0) =

8 5 4

= 40/429

13 12 11

5 4 3

p(1, 1, 1)

= 5/143

12 11

3.(a) p(0, 0) = (10/13)(9/12) = 15/26

p(0, 1) = p(1, 0) = (10/13)(3/12) = 5/26 p(1, 1) = (3/13)(2/12) = 1/26

(b)p(0, 0, 0) = (10/13)(9/12)(8/11) = 60/143

p(0, 0, 1) = p(0, 1, 0) = p(1, 0, 0) = (10/13)(9/12)(3/11) = 45/286

p(i, j, k) = (3/13)(2/12)(10/11) = 5/143	if i + j + k = 2
p(1, 1, 1) = (3/13)(2/12)(1/11) = 1/286

4.(a) p(0, 0) = (8/13)2, p(0, 1) = p(1, 0) = (5/13)(8/13), p(1, 1) = (5/13)2

(b)p(0, 0, 0) = (8/13)3

p(i, j, k) = (8/13)2(5/13) if i + j + k = 1 p(i, j, k) = (8/13)(5/13)2 if i + j + k = 2

5.p(0, 0) = (12/13)3(11/12)3

p(0, 1) = p(1, 0) = (12/13)3[1 − (11/12)3]

p(1, 1) = (2/13)[(1/13) + (12.13)(1/13)] + (11/13)(2/13)(1/13)

Chapter 6

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