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

31.By induction: true when t = 0, so assume for t − 1. Let N(t) denote the number after stage t.

E[N(t) N(t − 1)] = N(t − 1) − E[number selected]

32. E[X1X2 Y = y] = E[X1 Y = y]E[X2 Y = y] = y2

Therefore, E[X1X2 Y] = Y2. As E[Xi Y] = Y, this gives that

Consequently,

Cov(X1, X2) = E[X1X2] − E[X1]E[X2] = Var(Y)

34.(a) E[Tr Tr−1] = Tr−1 + 1 + (1 − p)E[Tr]

(b)Taking expectations of both sides of (a) gives

E[Tr] = E[Tr−1] + 1 + (1 − p)E[Tr]

or

E[Tr] = 1p + 1p E[Tr −1]

(c)Using the result of part (b) gives E[Tr] = 1p + 1p E[Tr −1]

=1/p + (1/p)2 + (1/p)2E[Tr−2]

=1/p + (1/p)2 + (1/p)3 + (1/p)3E[Tr−3]

r

= ∑(1/ p)i + (1/ p)r E[T0 ]

i=1

r

= ∑(1/ p)i since E[T0] = 0.

i=1

37.Let Xn denote the number of white balls after the nth drawing

Taking expectations now yields (a).

To prove (b), use (a) and the boundary condition M0 = a

(c) P{(n + 1)st is white} = E[P{(n + 1)st is white Xn}]

41.(a) No

(b)Yes, since fY(x I = 1) = fX(x) = fX(−x) = fY(x I = 0)

(c)fY(x) = 12 fX (x) + 12 fX (−x) = fX (x)

(d)E[XY] = E[E[XY X]] = E[XE[Y X]] = 0

(e)No, since X and Y are not jointly normal.

42.If E[Y X] is linear in X, then it is the best linear predictor of Y with respect to X.

43.Must show that E[Y 2] = E[XY]. Now

E[XY] = E[XE[X Z]]

=E[E[XE[X Z] Z]]

=E[E 2[X Z]] = E[Y 2]

44. Write Xn = Zi where Zi is the number of offspring of the ith individual of the (n − 1)st

generation. Hence,

E[Xn] = E[E[Xn Xn−1]] = E[µXn−1] = µE[Xn−1]

so,

E[Xn] = µE[Xn−1] = µ2E[Xn−2] … = µnE[X0] = µn

(c) Use the above representation to obtain

E[Xn Xn−1] = µXn−1, Var(Xn Xn−1) = σ2Xn−1

Hence, using the conditional Variance Formula,

Var(Xn) = µ2 Var(Xn−1) + σ2µn−1

(d) π = P{dies out}

= ∑P{dies out Xi = j}p j

j

= ∑π j p j , since each of the j members of the first generation can be thought of as

j

starting their own (independent) branching process.

∞

46. It is easy to see that the nth derivative of ∑(t2 / 2) j / j! will, when evaluated at t = 0, equal 0

j =0

whenever n is odd (because all of its terms will be constants multiplied by some power of t).

d n {tn}/( j!2 j ) plus constants multiplied by powers dtn

E[Z2j] - (2j)!/(j!2j)

47.Write X = σZ + µ where Z is a standard normal random variable. Then, using the binomial theorem,

Now make use of theoretical exercise 46.

48.φY(t) = E[etY] = E[et(aX+b)] = etbE[etaX] = etbφX(ta)

49.Let Y = log(X). Since Y is normal with mean µ and variance σ2 it follows that its moment generating function is

M(t) = E[etY] = eµt +σ 2t 2 / 2

Hence, since X = eY, we have that

E[X] = M(1) = eµ+σ 2 / 2

and

E[X 2] = M(2) = e2µ+2σ 2

Therefore,

Var(X) = e2µ+2σ 2 −e2µ+σ 2 = e2µ+σ 2 (eσ 2 −1)

50.ψ(t) = log φ(t)

ψ′(t) = φ′(t)/φ(t)

ψ′′(t) = φ(t)φ′′(φt)2 (−t)(φ′(t))2

ψ ′′(t) t =0 = E[X 2 ] −(E[X ])2 = Var(X).

51.Gamma (n, λ)

52.Let φ(s, t) = E[esX+tY]

53.Follows from the formula for the joint moment generating function.

54.By symmetry, E[Z 3 ] = E[Z] = 0 and so Cov(Z,Z 3) = 0.

55.(a) This follows because the conditional distribution of Y + Z given that Y = y is normal with mean y and variance 1, which is the same as the conditional distribution of X given that

Y = y.

(b)Because Y + Z and Y are both linear combinations of the independent normal random variables Y and Z, it follows that Y + Z, Y has a bivariate normal distribution.

(c)µx = E[X] = E[Y + Z] = µ

σx2 = Var(X) = Var(Y + Z) =Var(Y) + Var(Z) = σ2 + 1

(d) and (e) The conditional distribution of Y given X = x is normal with mean

Chapter 8

Problems

1.P{0 ≤ X ≤ 40} = 1 − P{ X − 20 > 20} ≥ 1 − 20/400 = 19/20

2.(a) P{X ≥ 85} ≤ E[X]/85 = 15/17

(b)P{65 ≤ X ≤ 85) = 1 − P{ X − 75 > 10} ≥ 1 − 25/100

3.Let Z be a standard normal random variable. Then,

= P{Z > −1.006} ≈ .8428

where the above uses that an exponential with mean 5 has variance 25.

8.If we let Xi denote the life of bulb i and let Ri be the time to replace bulb i then the desired

524.75 and variance 2500 + 99/48 = 2502 it follows that the desired probability is approximately equal to P{N(0, 1) ≤ [550 − 524.75]/(2502)1/2} = P{N(0, 1) ≤ .505} = .693 It should be noted that the above used that

9.Use the fact that a gamma (n, 1) random variable is the sum of n independent exponentials with rate 1 and thus has mean and variance equal to n, to obtain:

≈ P{N(0,1) > .01 n} = 2P{N(0,1) > .01 n}

Now P{N(0, 1) > 2.58} = .005 and so n = (258)2.

10.If Wn is the total weight of n cars and A is the amount of weight that the bridge can withstand then Wn − A is normal with mean 3n − 400 and variance .09n + 1600. Hence, the probability of structural damage is

P{Wn − A ≥ 0} ≈ P{Z ≥ (400 −3n) / .09n +1600}

Since P{Z ≥ 1.28} = .1 the probability of damage will exceed .1 when n is such that

400 − 3n ≤ 1.28 .09n +1600

The above will be satisfied whenever n ≥ 117.

12.Let Li denote the life of component i.

Now apply the central limit theorem to approximate.

(c) Using that SD(Y − X ) = 196 / 64 +196 / 25 ≈ 3.30 we have

P{Y − X > 2.2} = P{Y − X}/ 3.30 > 2.2 / 3.30} ≈ P{Z > .67} ≈ .2514

(d)same as in (c)

14.Suppose n components are in stock. The probability they will last for at least 2000 hours is

where Z is a standard normal random variable. Since

.95 = P{Z ≥ −1.64} it follows that p ≥ .95 if

2000 −100n ≤ −

30 n

1.64

or, equivalently,

(2000 − 100n)/ n ≤ −49.2

and this will be the case if n ≥ 23.

18.Let Yi denote the additional number of fish that need to be caught to obtain a new type when

20. g(x) = xn(n−1) is convex. Hence, by Jensen’s Inequality

E[Y n/(n−1)] ≥ E[Y])n/(n−1) Now set Y = X n−1 and so E[X n] ≥ (E[X n−1])n/(n−1) or (E[X n])1/n ≥ (E[X n−1])1/(n−1)

21. No

22. (a) 20/26 ≈ .769

(b) 20/(20 + 36) = 5/14 ≈ .357

(d)p ≈ P{Z ≥ (25.5 − 20)/ 20 } ≈ P{Z ≥ 1.23} ≈ .1093

(e)p = .112184

Theoretical Exercises

1.This follows immediately from Chebyshev’s inequality.

(d)answer = µ /σ

4.For ε > 0, let δ > 0 be such that g(x) − g(c) < ε whenever x − c ≤ δ. Also, let B be such that g(x) < B. Then,

E[g(Zn)] = ∫x−c ≤δ g(x)dFn (x) +∫x−c >δ g(x)dFn (x)

≤ (ε + g(c))P{ Zn − c ≤ δ} + BP{ Zn − c > δ}

In addition, the same equality yields that

E[g(Zn)] ≥ (g(c) − ε)P{ Zn − c ≤ δ} − BP{ Zn − c > δ}

Upon letting n → ∞ , we obtain that

lim sup E[g(Zn)] ≤ g(c) + ε lim inf E[g(Zn)] ≥ g(c) − ε

The result now follows since ε is arbitrary.

5.Use the notation of the hint. The weak law of large numbers yields that

lim P{ (X1 +... + Xn ) / n −c > ε} = 0

n→∞