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Exponential Distribution
- Definition: Exponential distribution with parameter λ: f (x) =
λe−λx^ x ≥ 0 0 x < 0
F (x) =
∫ (^) x
−∞
f (x)dx =
1 − e−λx^ x ≥ 0 0 x < 0
- Mean E(X) = 1/λ.
- Moment generating function:
φ(t) = E[etX^ ] =
λ λ − t
, t < λ
2 dt^2 φ(t)|t=0^ = 2/λ
- V ar(X) = E(X^2 ) − (E(X))^2 = 1/λ^2.
- Properties
- Memoryless: P (X > s + t|X > t) = P (X > s). P (X > s + t|X > t)
=
P (X > s + t, X > t) P (X > t)
=
P (X > s + t) P (X > t)
=
e−λ(s+t) e−λt = e−λs = P (X > s)
- Example: Suppose that the amount of time one spends in a bank is exponentially distributed with mean 10 minutes, λ = 1/ 10. What is the prob- ability that a customer will spend more than 15 minutes in the bank? What is the probability that a customer will spend more than 15 min- utes in the bank given that he is still in the bank after 10 minutes? Solution: P (X > 15) = e−^15 λ^ = e−^3 /^2 = 0. 22 P (X > 15 |X > 10) = P (X > 5) = e−^1 /^2 = 0. 604
- If Xi, i = 1, 2 , ..., n, are iid exponential RVs with mean 1 /λ, the pdf of
∑n i=1 Xi^ is:
fX 1 +X 2 +···+Xn(t) = λe−λt^
(λt)n−^1 (n − 1)!
gamma distribution with parameters n and λ.
- If X 1 and X 2 are independent exponential RVs with mean 1 /λ 1 , 1 /λ 2 ,
P (X 1 < X 2 ) =
λ 1 λ 1 + λ 2
- If Xi, i = 1, 2 , ..., n, are independent exponential RVs with rate μi. Let Z = min(X 1 , ..., Xn) and Y = max(X 1 , ..., Xn). Find distribution of Z and Y. - Z is an exponential RV with rate
∑n i=1 μi. P (Z > x) = P (min(X 1 , ..., Xn) > x) = P (X 1 > x, X 2 > x, ..., Xn > x) = P (X 1 > x)P (X 2 > x) · · · P (Xn > x)
=
∏^ n
i=
e−μix^ = e−(
∑n i=1 μi)x
- FY (x) = P (Y < x) =
∏n i=1(1^ −^ e
−μix).
Poisson Process
- Counting process : Stochastic process {N (t), t ≥ 0 } is a counting process if N (t) represents the total num- ber of “events” that have occurred up to time t. - N (t) ≥ 0 and are of integer values. - N (t) is nondecreasing in t.
- Independent increments : the numbers of events oc- curred in disjoint time intervals are independent.
- Stationary increments : the distribution of the number of events occurred in a time interval only depends on the length of the interval and does not depend on the position.
Interarrival and Waiting Time
- Define Tn as the elapsed time between (n − 1)st and the nth event. {Tn, n = 1, 2 , ...} is a sequence of interarrival times.
- Proposition 5.1 : Tn, n = 1, 2 , ... are independent identically distributed exponential random variables with mean 1 /λ.
- Define Sn as the waiting time for the nth event, i.e., the arrival time of the nth event.
Sn =
∑^ n
i=
Ti.
fSn(t) = λe−λt^
(λt)n−^1 (n − 1)!
gamma distribution with parameters n and λ.
∑n i=1 E(Ti) =^ n/λ.
- Example: Suppose that people immigrate into a terri- tory at a Poisson rate λ = 1 per day. (a) What is the expected time until the tenth immigrant arrives? (b) What is the probability that the elapsed time between the tenth and the eleventh arrival exceeds 2 days? Solution: Time until the 10th immigrant arrives is S 10. E(S 10 ) = 10/λ = 10. P (T 11 > 2) = e−^2 λ^ = 0. 133.
- Proposition 5.2 : {N 1 (t), t ≥ 0 } and {N 2 (t), t ≥ 0 } are both Poisson processes having respective rates λp and λ(1 − p). Furthermore, the two processes are in- dependent.
- Example: If immigrants to area A arrive at a Poisson rate of 10 per week, and if each immigrant is of En- glish descent with probability 1 / 12 , then what is the probability that no people of English descent will im- migrate to area A during the month of February? Solution: The number of English descent immigrants arrived up to time t is N 1 (t), which is a Poisson process with mean λ/12 = 10/ 12. P (N 1 (4) = 0) = e−(λ/12)·^4 = e−^10 /^3.
- Conversely : Suppose {N 1 (t), t ≥ 0 } and {N 2 (t), t ≥ 0 } are independent Poisson processes having respec- tive rates λ 1 and λ 2. Then N (t) = N 1 (t) + N 2 (t) is a Poisson process with rate λ = λ 1 + λ 2. For any event occurred with unknown type, independent of every- thing else, the probability of being type I is p = (^) λ 1 λ+^1 λ 2 and type II is 1 − p.
- Example: On a road, cars pass according to a Poisson process with rate 5 per minute. Trucks pass accord- ing to a Poisson process with rate 1 per minute. The two processes are indepdendent. If in 3 minutes, 10 veicles passed by. What is the probability that 2 of them are trucks? Solution: Each veicle is independently a car with probability 5 5+1 =^
5 6 and a truck with probability^
1
- The probabil- ity that 2 out of 10 veicles are trucks is given by the binomial distribution: ( 10 2
- If N (t) = n, what is the joint conditional distribution of the arrival times S 1 , S 2 , ..., Sn?
- S 1 , S 2 , ..., Sn is the ordered statistics of n independent random variables uniformly distributed on [0, t].
- Let Y 1 , Y 2 , ..., Yn be n RVs. Y(1), Y(2),..., Y(n) is the ordered statistics of Y 1 , Y 2 , ..., Yn if Y(k) is the kth smallest value among them.
- If Yi, i = 1, ..., n are iid continuous RVs with pdf f , then the joint density of the ordered statistics Y(1), Y(2),..., Y(n) is fY(1),Y(2),...,Y(n)(y 1 , y 2 , ..., yn)
=
n!
∏n i=1 f^ (yi)^ y^1 < y^2 <^ · · ·^ < yn 0 otherwise
f (s 1 , s 2 , ..., sn | N (t) = n) =
n! tn 0 < s 1 < s 2 · · · < sn < t Proof f (s 1 , s 2 , ..., sn | N (t) = n)
=
f (s 1 , s 2 , ..., sn, n) P (N (t) = n))
=
λe−λs^1 λe−λ(s^2 −s^1 )^ · · · λe−λ(sn−sn−^1 )e−λ(t−sn) e−λt(λt)n/n! =
n! tn^
, 0 < s 1 < · · · < sn < t
- For n independent uniformly distributed RVs on [0, t], Y 1 , ..., Yn: f (y 1 , y 2 , ..., yn) =
tn^
- Proposition 5.4 : Given Sn = t, the arrival times S 1 , S 2 , ..., Sn− 1 has the distribution of the ordered statis- tics of a set n − 1 independent uniform (0, t) random variables.
- Compound Poisson process : A stochastic process {X(t), t ≥ 0 } is said to be a compound Poisson pro- cess if it can be represented as
X(t) =
N ∑ (t)
i=
Yi , t ≥ 0
where {N (t), t ≥ 0 } is a Poisson process and {Yi, i ≥ 0 } is a family of independent and identically distributed random variables which are also indepen- dent of {N (t), t ≥ 0 }.
- The random variable X(t) is said to be a compound Poisson random variable.
- Example: Suppose customers leave a supermarket in accordance with a Poisson process. If Yi, the amount spent by the ith customer, i = 1, 2 , ..., are indepen- dent and identically distributed, then X(t) =
∑N (t) i=1 Yi, the total amount of money spent by customers by time t is a compound Poisson process.
- Find E[X(t)] and V ar[X(t)].
- E[X(t)] = λtE(Y 1 ).
- V ar[X(t)] = λt(V ar(Y 1 ) + E^2 (Y 1 ))
- Proof
E(X(t)|N (t) = n) = E(
N ∑ (t)
i=
Yi|N (t) = n)
= E(
∑^ n
i=
Yi|N (t) = n)
= E(
∑^ n
i=
Yi) = nE(Y 1 )
E(X(t)) = EN (t)E(X(t)|N (t))
=
∑^ ∞
n=
P (N (t) = n)E(X(t)|N (t) = n)
∑^ ∞
n=
P (N (t) = n)nE(Y 1 )
= E(Y 1 )
∑^ ∞
n=
nP (N (t) = n)
= E(Y 1 )E(N (t)) = λtE(Y 1 )
V ar(X(t))
= E(X^2 (t)) − (E(X(t)))^2
=
∑^ ∞
n=
P (N (t) = n)E(X^2 (t)|N (t) = n) − (E(X(t)))^2
∑^ ∞
n=
P (N (t) = n)(nV ar(Y 1 ) + n^2 E^2 (Y 1 )) − (λtE(Y 1 ))^2
= V ar(Y 1 )E(N (t)) + E^2 (Y 1 )E(N 2 (t)) − (λtE(Y 1 ))^2
= λtV ar(Y 1 ) + λtE^2 (Y 1 )
= λt(V ar(Y 1 ) + E^2 (Y 1 ))
= λtE(Y 12 )
