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

Since X1 + … + Xn is binomial with parameters n, x, we have

k

≥ ∑i P{X = k}

i=1

=P{X = k}k(k + 1)/2

≥k2 P{X = k} 2

7.Take logs and apply the central limit theorem

8.It is the distribution of the sum of t independent exponentials each having rate λ.

9.1/2

10.Use the Chernoff bound: e−tiM(t) = eλ(et −1)−ti will obtain its minimal value when t is chosen to satisfy

λet = i, and this value of t is negative provided i < λ.

Hence, the Chernoff bound gives

P{X ≤ i} ≤ ei−λ(λ/i)i

11.e−tiM(t) = (pet + q)ne−ti and differentiation shows that the value of t that minimizes it is such that

Using this value of t, the Chernoff bound gives that

=iiqi (n −i)n

12.1 = E[eθX] ≥ eθE[X] by Jensen’s inequality. Hence, θE[X] ≤ 0 and thus θ > 0.

Chapter 9

Problems and Theoretical Exercises

1.(a) P(2 arrivals in (0, s) 2 arrivals in (0, 1)}

=P{2 in (0, s), 0 in (s, 1)}/e−λλ2/2)

= [e−λs(λs)2/2][e−(1−s)λ]/(e−λλ2/2) = s2 = 1/9 when s = 1/3

(b)1 − P{both in last 40 minutes) = 1 − (2/3)2 = 5/9

2.e−3s/60

3.e−3s/60 + (s/20)e−3s/60

8.The equations for the limiting probabilities are:

∏c = .7∏c + .4∏s + .2∏g ∏s = .2∏x + .3∏s + .4∏g ∏g = .1∏c + .3∏s + .4∏g ∏c + ∏s + ∏g = 1

and the solution is: ∏c = 30/59, ∏s = 16/59, ∏g = 13/59. Hence, Buffy is cheerful 3000/59 percent of the time.

9.The Markov chain requires 4 states:

0 = RR = Rain today and rain yesterday

1 = RD = Dry today, rain yesterday

2 = DR = Rain today, dry yesterday

3 = DD = Dry today and dry yesterday with transition probability matrix

The equations for the limiting probabilities are:

∏0 = .8∏0 + .4∏2 ∏1 = .2∏0 + .6∏2 ∏2 = .3∏1 + .2∏3 ∏3 = .7∏1 + .8∏3

∏0 + ∏1 + ∏2 + ∏3 = 1

which gives

∏0 = 4/15, ∏1 = ∏2 = 2/15, ∏3 = 7/15.

Since it rains today when the state is either 0 or 2 the probability is 2/5.

10.Let the state be the number of pairs of shoes at the door he leaves from in the morning. Suppose the present state is i, where i > 0. Now after his return it is equally likely that one door will have i and the other 5 − i pairs as it is that one will have i − 1 ant the other 6 − i. Hence, since he is equally likely to choose either door when he leaves tomorrow it follows that

Pi,i = Pi,5−i = Pi,i−1 = Pi,6−i = 1/4

provided all the states i, 5 − i, i − 1, 6 − i are distinct. If they are not then the probabilities are added. From this it is easy to see that the transition matrix Pij, i, j = 0, 1, …, 5 is as follows:

Since this chain is doubly stochastic (the column sums as well as the row sums all equal to one) it follows that ∏i = 1/6, i = 0, …, 5, and thus he runs barefooted one-sixth of the time.

11.(b) 1/2

(c)Intuitively, they should be independent.

(d)From (b) and (c) the (limiting) number of molecules in urn 1 should have a binomial distribution with parameters (M, 1/2).

Chapter 10

1.(a) After stage k the algorithm has generated a random permutation of 1, 2, …, k. It then puts element k + 1 in position k + 1; randomly chooses one of the positions 1, …, k + 1 and interchanges the element in that position with element k + 1.

(b)The first equality in the hint follows since the permutation given will be the permutation

after insertion of element k if the previous permutation is i1, …, ij−1, i, ij, …, ik−2 and the random choice of one of the k positions of this permutation results in the choice of position j.

2.Integrating the density function yields that that distribution function is

e2x / 2 , x > 0 1−e−2s / 2, x > 0

which yields that the inverse function is given by

Hence, we can simulate X from F by simulating a random number U and setting X = F −1(U).

3.The distribution function is given by

Hence, for u ≤ 1/4, F−1(u) is the solution of

x2/4 − x + 1 = u

that falls in the region 2 ≤ x ≤ 3. Similarly, for u ≥ 1/4, F−1(u) is the solution of

x − x2/12 − 2 = u

that falls in the region 3 ≤ x ≤ 6. We can now generate X from F by generating a random number U and setting X = F−1(U).

4.Generate a random number U and then set X = F−1(U). If U ≤ 1/2 then X = 6U − 3, whereas if U ≥ 1/2 then X is obtained by solving the quadratic 1/2 + X 2/32 = U in the region 0 ≤ X ≤ 4.

5.The inverse equation F−1(U) = X is equivalent to

or

1 − e−αX β = U

X = {−log(1 − U)/α}1/β

Since 1 − U has the same distribution as U we can generate from F by generating a random number U and setting X = {−log(U)/α}1/β.

6.If λ(t) = ctn then the distribution function is given by

1 − F(t) = exp{−ktn+1}, t ≥ 0 where k = c/(n + 1)

Hence, using the inverse transform method we can generate a random number U and then set X such that

exp{−kXn+1} = 1 − U

or

X = {−log(1 − U)/k}1/(n+1)

Again U can be used for 1 − U.

7.(a) The inverse transform method shows that U1/n works.

(b)P{MaxUi ≤ v} = P{U1 ≤ x, …, Un ≤ x}

=∏P{Ui ≤ x} by independence

=xn

(c)Simulate n random numbers and use the maximum value obtained.

8.(a) If Xi has distribution Fi , i = 1, …, n, then, assuming independence, F is the distribution of MaxXi. Hence, we can simulate from F by simulating Xi, i = 1, …, n and setting

X= MaxXi.

(b)Use the method of (a) replacing Max by Min throughout.

9.(a) Simulate Xi from Fi, i = 1, 2. Now generate a random number U and set X equal to X1 if

U< p and equal to X2 if U > p.

(b)Note that

F(x) = 13 F1(x) + 23 F2 (x)

where

F1(x) = 1 − e−3x, x > 0, F2(x) = x, 0 < x < 1