prob review

ECE/CSC 570

Probability Theory Review

Do Young Eun ECE/CSC 570, Fall 2014

Why Probability ?

Extensive use in modeling network traffic Area of mathematics most fundamentally related to

computer networking Leads on to more powerful analysis tools and modeling

paradigms Markov chains Queueing theory, etc.

Deterministic approach is infeasible! You can’t rely on simulations/test-bed for every possible

scenarios We need insight, to be able to predict, optimize system in a

systematic way


Probability and Random Variables Sample space : A set of all possible outcomes (or

sample points) of an experiment

Event: a subset of , denoted as A, B, C, etc.

Random variable : a function of the outcome in a sample space, i.e.,

Probability: A measure of possibility an event of the sample space happened

Probability Axiom


More on Probability

Example: roll a dice = {1,2,3,4,5,6} Event: {2,4,6}, {1,5},

Random variable: e.g., x = 0 if it is even and =1 if odd.

E.g., X(k) = k2 square of the values

Probability: set function (mapping) from an event to [0,1]

Some properties of probability

If A B, then P(A) ≤ P(B) P(A B) ≤ P(A) + P(B)


Discrete Random Variables Discrete sample space, i.e., countable set: finite or

(countable) infinite outcomes Finite sample space – Roulette wheel

Total N slots, each slot is marked with a label picked up from (There are Ni slots marked with label xi

12

43

1

4Now spin the wheel.

X is a r.v. that records the outcome of the spin test. Probability pi that xi appears:


Infinite sample space Count the number of telephone calls Y arrived at a server in time

interval [0,1] Repeating the experiment many times.

We have (check!)

Y is called a Poisson r.v. with parameter


Continuous Random Variables

Example: Exponentially distributed random variable Xwith rate > 0 :

Probability distribution function of X : Properties:F(t) is non-decreasing in t, F() = 1

Probability density function (pdf) of X: Properties:


An event A occurs given that event B has occurred

Examples:1. Roulette Wheel Example

B={1, 2, 3}, A= {1,4}

0)P(for )(P

)(P)|(P

BB

BABA

A B

A and B

41

}{P}{P}|{P

321

1

NNN

NBX

BAXBXAX

ee

YYYYYY

1

}0{P1}1{P

0}P{Y}0 and 1{P}0|1{P

2. Let Y be a Poisson r.v.

Conditional Probability


3. Let X be the lifetime of a communication link that is exponentially distributed with rate , we want to estimate its probability that it will survive for at least ‘t’ more seconds, given that has been alive for ‘a’ seconds.

}{P___________________ ___________________

}|{P

tX

aXtaX

Note: This property is called memoryless, namely,

0

}P{}|{P tXaXtaX

Conditional Probability (cont’d)


Theorem of Total Probability

Let Bi be a partition of , i.e., Bi’s are disjoint and

1B2B

3B 4B

1BA


Bayes’ Theorem Let {Ei, i=1,2, …, N} be a partition of Then:

Updates prob. assigned to Ei in light of occurrence of F F is regarded as new evidence Widely used in inference systems


Example 3 coins are given: two of them are “fair”, one of them is

biased (P{H}=2/3, P{T}=1/3) We don’t know which one is biased; they all look the same Flip three coins in a row and we have observed

The first and second coins both show “heads”, while the third one shows “tail”

Class Exercise: Given the observation, P {the first coin is the biased one} = ?

Further question: flip one more (any of three) costs $1; pick the biased one you earn $2. Do you want to continue?


Another example: Monty Hall problem 3 doors; only one of them leads to prize; the other two have “goats”

(or no prize). You choose one of three, and the other (Monty Hall) open one of the

other two (unselected) and show that there is a goat. Monty asks you: “Do you want to stay on your choice or switch to the

other unopened door?”


Moments of an RV

The CDF (or PDF) provides complete info. about a r.v. X determines probability of any event involving r.v. X

Often, we need/want a lot less info about an experiment: We are not interested in all possible events.

We cannot provide/digest so much information

Instead, some partial information is OK

Moments of the PDF provide useful partial info.


Expected Value

E{X} denotes the expected value (average) of RV X: Definition:

discrete RV:

continuous RV:

Expected value of X is the sum of the values of X weighted by their probabilities

If Y = g(X):

discrete RV:

continuous RV:


Higher Moments

E{Xn} denotes the n-th moment of RV X:

Definition: discrete RV:

continuous RV:

First moment: expected value E{X}

Second moment: E{X2}

So on…


Variance

X2 or Var{X} denote the variance of RV X

Definition: X2 = E{(X-E{X})2}

the second moment around the mean

measure of the spread of X around its mean

extremely important, widely used in statistics

X2 = E{X2} – E2{X}

Standard deviation, X: defined as same dimension (units) as the mean E{X}

more realistic measure of the spread of X


Example

A r.v. X is geometrically distributed with parameter p (0,1)

Check if

pppnppn

pnpnpnpn

ppnpn

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

nn

11

)1(

)1( E{X}

0000

1011

1

1

1

1


Example (2)

Exponentially distributed RV X with rate

E{X} = ? Var{X} = ?

Poisson r.v. X with rate E{X} ? E{X2}=?


Independence of events Two events A and B are said to be independent if and

only if P(A|B) = P(A). Occurrence of B tells us nothing about occurrence of A

We have

If A and B are independent,


Two discrete r.v.s X and Y are independent

Consider two continuous r.v.s X and Y, Joint distribution function of X and Y:

Joint density function:

The two r.v.s X and Y are independent

},{P yYxX

dxdyyYxXdyxYX},{P),(f

2

,

)(P)(P),(P iiii yYxXyYxX

)(f)(f),(f , yxyx YXYX

Independence of random variables


dxdyyxxyXY YX ),(f}{E ,},{E}{E}{E YXXY

Independence of random variables

Independence implies

But, the converse is not true!!

uncorrelated

Example?


Stochastic Dynamic Systems

Dynamic systems: those that evolve in time Must describe the sequence of states a system enters Example: the length of the queue at a router interface

the queue length changes as packets arrive or leave need a model to describe these changes over time

Example: # of packet/call arrivals over time [0, t] Stochastic process: models dynamic systems

combines the concept of RV with the notion of time informally: a vector RV with infinite dimensions


Arrival (or Counting) process

{N(t), t ≥ 0} be a discrete-space, continuous time stochastic process

Definition: {N(t); t ≥ 0} is a counting (or arrival) process if:1. N(0)=02. N(t) is integer valued.3. N(t) is non-decreasing; i.e., if s ≤ t, then N(s) ≤ N(t).4. For s < t, N(t) - N(s) = # of events (or arrivals) in (s,t] (right continuous

property).

N(t)

Timet1 t2 t3

1. N(t) = # of telephone calls arrived in time interval [0,t]

2. N(t) = # of packets generated in time interval [0,t], etc.


“Independent Increments” Property

(s1,t1] and (s2,t2]: two non-overlapping time intervals Y1 = N(t1) – N(s1), Y2 = N(t2) – N(s2): RVs representing # of events in

each interval If arrivals in two intervals are independent, then RVs Y1 and Y2 are

also independent the process has independent increments (very strong property)

t0 s1 s2t1t2

N(t1)-N(s1) N(t2)-N(s2)


“Stationary Increments” Property

Not so strong property: quite popular assumption! For any t, s ≥ 0, the distribution of N(t+s) – N(t) is

independent of t.

Note: N(t+s) – N(t) ≠ N(s) in general: They have the same distribution, but are not equal.

P{N(t+s) – N(t) = n}

= P{N(t1+s) – N(t1) = n}

= P{N(s) = n}


Poisson Process

Two Definitions: Definition 1: An arrival process {N(t); t ≥ 0} is

said to be a Poisson Process with rate > 0, ifN(0)=0 It has stationary and independent incrementsP{N(h) = 1} = h + o(h)P{N(h) ≥ 2} = o(h): “rules out” simultaneous arrivals

Note: What is P{N(h)=0} then?


Poisson Process (cont’d)

Equivalent Definition: Definition 2: An arrival process {N(t), t ≥ 0} is said to

be a Poisson Process with rate , if N(0)=0 The process has independent increments.

The number of arrivals in any interval of length s is Poisson distributed with mean s. In other words, for all s and t ≥ 0,


Joint Distribution: Complete Characterization

For any increasing time instants (t1 , t2 , …, tk ) and non-decreasing integers (n1 , n2 , …, nk ), we want to compute

For instance, P{N(3)=4, N(5)=6, N(8)=10} = ?


Interarrival times of Poisson Process

Consider a Poisson process with rate

Let Xn be the time between the (n-1)st and the nth event (arrival). Then, {Xn, x ≥1} forms a sequence of interarrivaltimes.

Distribution ?

X1X2 X3

t=0 t=1 t=2 t=3


Interarrival times of Poisson Process

First, what is the distribution of X1?

X1 has an exponential distribution with mean 1/.

Repeating the argument …

In fact, Xn, n ≥ 1, are independent and identically distributed (i.i.d.) exponential random variables with mean 1/.

Another way of constructing a Poisson process?


Sn: Time until the nth Arrival

Sn can be expressed as:

The event {Sn ≤ t}, “ the nth arrival before time t ” is equivalent to the event {N(t) ≥ n}, “the number of arrivals occurring by time t is at least n”.

tSn

…


Memoryless Property

Let { N(t), t ≥ 0} be a Poisson process with rate Let t0 be the time since the last arrival Q: what is the distribution of time till the next arrival?

Exponential distribution is memoryless residual life paradox Exponential dist. is the only one with the memoryless property!


Paradox of Residual Life

Suppose that buses arrive at a stop according to a Poisson process with rate .

Assume that you arrive at that bus-stop at some arbitrary point in time.

Question: How long would you expect to wait? Two logical answers.

t

Xn Xn+2Xn+1


Answer 1: Average wait = 1/(2)

Note that average time between bus arrivals = 1/

On average, you will arrive exactly in the middle of an inter-arrival time.

Your average wait for the next bus = ½ E{Xi} = 1/(2)

t

Xn Xn+2Xn+1

Bus arrives Bus arrives


Answer 2: Average wait = 1/ Since buses arrive according to a Poisson process with rate ,

the time you have to wait is independent of how long it has taken since the last bus, i.e., the average wait = 1/.

In fact, by the second argument, if buses have been operational for a long time, then by the memory-less property, the expected time since the last bus arrival is also 1/

This implies that the expected time between the last bus and the next bus arrival in your interval = 2/

So, which answer is correct?


Paradox of Residual Life

Answer 2 is in fact correct! This is because the interval that you arrive in is not a typical

interval, i.e., it is in fact more likely that you will arrive in a larger time interval.

In the case of a Poisson process, if you are sufficiently far from the origin, this interval that you arrive is in fact two times as long as the average interval.

Inspection Paradox: also applies to other arrival processes

tSN(t) SN(t)+1


Utilities of Poisson Processes It has been shown that the number of call (or connection)

arrivals received in a network system in a time interval [0,t] is well modeled by a Poisson process.

Connection with Binomial distribution

Many nice properties


Poisson as a Limit of Binomial Distribution

Binomial random variable Xi: i.i.d., P{Xi=1} = 1 – P{Xi=0} = p

Bn has binomial distribution with parameters (n, p)

Now, n large, p=p(n) small, such that

Then, as n gets large, Bn converges to a Poisson random variable with mean

# of phone call (packet) arrivals at central switch over small time interval I (of size O(1/n)) Xk = 1 if there is a call arrival over I with probability p from kth source

Total n sources, while np


Utilities of Poisson Processes

Superposition of two independent Poisson processes are again a Poisson process. Let N1(t) be a Poisson process with rate 1, and N2(t) be a Poisson

process with rate 2, and they are independent.

Then, N1(t)+N2(t) is also a Poisson process with rate 1 + 2

Statistically splitting Poisson arrival into two arrival processes (using a Bernoulli r.v. for splitting) Poisson processes !


Other Arrival Processes

Renewal Process: A counting process for which the interarrival times are i.i.d. with an arbitrary distributionNote: if i.i.d. exponential dist. Poisson process !


Example 1: To transmit packets from source to destination.

SENDER RECEIVERSEQ. # CRC

PACKET

ACKTIMEOUT

retransmitif no ACK

acknowledge packetif no errors

time time

SourceDestination

Packet

Regenerative Method


Regenerative Method (2) Assumptions:

When destination receives a correct packet, an ACK would be generated, otherwise a NACK would be generated and sent back to source.

A packet is correctly transmitted with probability (1-p) A packet in error must be retransmitted. Transmission of packets is independent

X = # of transmissions needed for destination to get a correct packet.

Y = # of transmissions needed for destination to get a correct packet given that the first transmission is error.

-p) (pn}{X n- 1P 1

-p)(

pYX 1y probabilit with , 1y probabilit with , 1


Coin Flip Analogy

X = # of flips until tail shows up…

…

Y


Regenerative Method (3)

Since Y and X have the same distribution (Y is statistical “replica” of X):

Note: The purpose of regenerative method is to introduce a simple algebra calculation method instead of complex calculation

-p)/(XX p(-p

Yp(-p)(X

11 }E{})E{11

})E{1 11 }E{


Example 2 Example description: A prison cell has 4 doors (see figure).

Assumptions: It takes a prisoner 1 day to get to each door. A prisoner immediately forgets which door he takes every time. A prisoner chooses each door with equal probability.

Only 1 door lets to freedom, the other 3 doors, if a prisoner enters, he would be trapped and kept inside for 1 day, 3 days, 5 days each. Prison Cell

FREE

1 day

3 days5 days

Door-4

Door-1

Door-2

Door-3


X: # of days he takes to reach freedomE{X}: average days he takes to reach freedom Yi : # of days he takes if his previous choice is door-i other

than the one leads to freedom

In class exercise: E{X} = ?

Example 2 (cont’d)

prob review

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