Probability Theory and Poisson Process Counting
Exponential Distribution
• Definition: Exponential distribution with parameter λ: λe−λx x ≥ 0 f (x) =
0
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 probability that a customer will spend more than 15 minutes in the bank? What is the probability that a customer will spend more than 15 minutes in the bank given that he is still in the bank after 10 minutes?
Solution:
P (X > 15) = e−15λ = e−3/2 = 0.22
P (X > 15|X > 10) = P (X > 5) = e−1/2 = 0.604
2
– Failure rate (hazard rate) function r(t) r(t) =
f (t)
1 − F (t)
– P (X ∈ (t, t + dt)|X > t) = r(t)dt.
– For exponential distribution: r(t) = λ, t > 0.
– Failure rate function uniquely determines F (t):
F (t) = 1 − e
3
t
− 0 r(t)dt
.
2. If Xi, i = 1, 2, ..., n, are iid exponential RVs with mean 1/λ, the pdf of n Xi is: i=1 (λt)n−1
,
fX1+X2+···+Xn (t) = λe−λt
(n − 1)! gamma distribution with parameters n and λ.
3. If X1 and X2 are independent exponential RVs with mean 1/λ1, 1/λ2, λ1 .
P (X1 < X2) = λ1 + λ2
4. If Xi, i = 1, 2, ..., n, are independent exponential
RVs with rate µi. Let Z = min(X1, ..., Xn) and
Y = max(X1, ..., Xn). Find distribution of Z and
Y.
– Z is an exponential RV with rate n µi. i=1 P (Z > x) = P (min(X1, ..., Xn) > x)
= P (X1 > x, X2 > x, ..., Xn > x)
= P (X1 > x)P (X2 > x) · · · P (Xn > x) n =
−µi x
e
−(
=e
n i=1 µi )x
i=1
– FY (x) = P (Y < x) =
4
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 number 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 occurred in
• Definition: Exponential distribution with parameter λ: λe−λx x ≥ 0 f (x) =
0
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 probability that a customer will spend more than 15 minutes in the bank? What is the probability that a customer will spend more than 15 minutes in the bank given that he is still in the bank after 10 minutes?
Solution:
P (X > 15) = e−15λ = e−3/2 = 0.22
P (X > 15|X > 10) = P (X > 5) = e−1/2 = 0.604
2
– Failure rate (hazard rate) function r(t) r(t) =
f (t)
1 − F (t)
– P (X ∈ (t, t + dt)|X > t) = r(t)dt.
– For exponential distribution: r(t) = λ, t > 0.
– Failure rate function uniquely determines F (t):
F (t) = 1 − e
3
t
− 0 r(t)dt
.
2. If Xi, i = 1, 2, ..., n, are iid exponential RVs with mean 1/λ, the pdf of n Xi is: i=1 (λt)n−1
,
fX1+X2+···+Xn (t) = λe−λt
(n − 1)! gamma distribution with parameters n and λ.
3. If X1 and X2 are independent exponential RVs with mean 1/λ1, 1/λ2, λ1 .
P (X1 < X2) = λ1 + λ2
4. If Xi, i = 1, 2, ..., n, are independent exponential
RVs with rate µi. Let Z = min(X1, ..., Xn) and
Y = max(X1, ..., Xn). Find distribution of Z and
Y.
– Z is an exponential RV with rate n µi. i=1 P (Z > x) = P (min(X1, ..., Xn) > x)
= P (X1 > x, X2 > x, ..., Xn > x)
= P (X1 > x)P (X2 > x) · · · P (Xn > x) n =
−µi x
e
−(
=e
n i=1 µi )x
i=1
– FY (x) = P (Y < x) =
4
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 number 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 occurred in