170B Note
Sangchul Lee
November 17, 2015
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Week 8
1.1
Summary
Bernoulli process
• A Bernoulli process of parameter p is a sequence (X1, X2, X3, · · · ) of i.i.d. Bernoulli RVs of parameter p.
• Inter-arrival times (T1, T2, · · · ) are i.i.d. ∼ Geometric(p).
• The nth arrival time has Pascal distribution of order n and parameter p.
Stopping times and fresh-start
∞ , a random variable T : Ω → {1, 2, 3, · · · } is called a stopping time if for any
• Given a process (X n )n=1
n = 1, 2, · · · , the information {T = n} can be determined only by X1, · · · , X n .
∞ is a Bernoulli process and T is a stopping time, then
• (fresh-start after stopping time) If (X n )n=1
(XT +1, XT +2, · · · ) is another Bernoulli process which is independent of (X1, X2, · · · , XT ).
1.2
Remarks
Bernoulli processes can be characterized in two equivalent ways:
(a) (time-based characterization) X1, X2, · · · are i.i.d. Bernoulli trials (of parameter p).
∞ are i.i.d Geometric random variables, then a process
(b) (occurrence-based characterization) If (Tn )n=1
∞
(X n )n=1 defined by

 1,
Xn = 
 0,

if n = T1 + · · · + Tk for some k.
otherwise
is a Bernoulli process.
Interplay between these two characterizations is an important concept throughout this chapter.
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2
Problems
Exercise 2.1. A certain team wins every match in a tournament with probability 1/2 independently of all
other matches. If the team won only two of the first n matches, find the PMF of the number of wins in the
first 2n matches.
Solution. Let N be the number of wins in the first 2n matches. The number of wins between the (n + 1) th match
and the 2nth match is independent of the first n matches and has Binomial(n, 1/2) distribution. Consequently
N − 2 ∼ Binomial(n, 1/2) and hence
!
n
p N (k) =
pk−2 (1 − p) n−k+2 for k = 2, · · · , n + 2.
k −2
Exercise 2.2. A duck family is trying to cross a single lane.
• Successive car interarrival times are i.i.d. with PDF f T (t).
• They can cross the lane if an interval between successive cars is > d. Let’s call it an opportunity.
• Car lengths and duck sizes are negligible.
• Our experiment begins the instant after the 0th car goes by.
Determine expressions for the probability that
(a) They can cross for the first time just before the nth car goes by.
(b) They shall have had exactly m opportunities by the instant the nth car goes by.
(c) The occurrence of the mth opportunity is immediately followed by the arrival of the nth car.
∞ be defined by
Solution. Let (X n )n=1

 1,
Xi = 
 0,

if they have an opportunity just before the nth car goes by
.
otherwise
Then it is a Bernoulli process of parameter p = P(T > d) =
∞ be inter-arrival times. Then
(Tn )n=1
R
∞
f T (t)
d
∞ be arrival times and
dt. Let (Yn )n=1
(a) P(Y1 = n).
(b) P(X1 + · · · + X n = m).
(c) P(Ym = n + 1).
Exercise 2.3. Let Yk ∼ Pascal(k, p). Compute the PMF of Yk .
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Solution. Understanding Yk as the k th arrival time of a Bernoulli process (X1, X2, · · · ) of parameter p, we find
that
pYk (n) = P(Yk = n) = P(the k th success happens exactly at the nth trial)
∃ exactly (k − 1) successes up to the (n − 1) th trial
=P
and the nth trial is a success
!
= P(X1 + · · · + X n−1 = k − 1 and X n = 1).
Since X1 + · · · + X n−1 ∼ Binomial(n − 1, p) and is independent of X n , we have
!
n−1 k
pYk (n) = P(X1 + · · · + X n−1 = k − 1)P(X n = 1) =
p (1 − p) n−k .
k −1
Exercise 2.4. Let Yn be a Pascals random variable of order n and parameter 1/2, and let pYn be it’s PMF.
Use central limit theorem to estimate
(a) P(Y500 ≤ 990).
P
(b) ∞
k=500 pYk (990)
Solution. (a) We proceed as in the previous exercise.
P(Y500 ≤ 990) = P(the 500th success happens no later than the 990th trial)
= P(∃ at least 500 successes up to the 990th trial)
= P(X1 + · · · + X990 ≥ 500).
Now the random variable N := X1 + · · · + X990 has Binomial(990, 1/2) distribution, thus by CLT
!
!
N − 495 500 − 495
500 − 495
P(Y500 ≤ 990) = P(N ≥ 500) = P √
≤ √
≈Φ √
≈ Φ(0.3178).
247.5
247.5
247.5
(b) We again utilize the idea as in the previous exercise.
∞
X
∞
X
∃ exactly (k − 1) successes up to the 989th trial
pYk (990) =
P
and the 990th trial is a success
k=500
k=500
!
∃ at least 499 successes up to the 989th trial
=P
and the 990th trial is a success
!
= P(X1 + · · · + X989 ≥ 499)P(X990 = 1).
Since the random variable N := X1 + · · · + X989 has Binomial(989, 1/2) distribution, by CLT we have
∞
X
pYk (990) = P(N ≥ 499)P(X990 = 1) ≈
k=500
1
(1 − Φ(0.2862)).
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Exercise 2.5. Let X n be a Bernoulli process of parameter 1/2 and let N be the first time we observe value
different from X1 , that is N = min{n : X n , X1 }. Define the process (Y1, Y2, Y3, · · · ) as
(Y1, Y2, · · · , Yk , · · · ) = (X N −1, X N , · · · , X N +k−2, · · · ).
(a) Show that N is a stopping time.
(b) Compute the probability P(X N +1 = X N +2 = 1).
∞ is not a Bernoulli process.
(c) Show that P(Yk = 1) = 1/2 for all k ≥ 1, but (Yk )k=1
Solution. (a) Notice that
{N = n} = {X1 = X2 = · · · = X n−1 but X n , X1 }
so that the information {N = n} is determined only by X1, · · · , X n . Therefore N is a stopping time.
(b) Since (X N +1, X n+2, · · · ) is a Bernoulli process of parameter 1/2, we have
P(X N +1 = X N +2 = 1) = (1/2) 2 = 1/4.
(c) We know that (Y3, Y4, · · · ) = (X N +1, X N +2, · · · ) is another Bernoulli process. So we have P(Yk = 1) = 1/2 for
any k ≥ 3. For the first two terms,
P(Y1 = 1) = P(X N −1 = 1) = P(X1 = 1) = 1/2
and
P(Y2 = 1) = P(X N = 1) = P(X1 = 0) = 1/2.
On the other hand, we always have Y1 = X N −1 , X N = Y2 and hence Y1 and Y2 are not independent. Therefore
∞ is not a Bernoulli process.
(Yn )n=1
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