Understanding Random Variables, Poisson & Exponential Distributions - Prof. Rudra Dutta, Study notes of Computer Systems Networking and Telecommunications

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Probability

Probability and Random Variables

A Random Variable, X, is a variable whose value depends upon the outcome of a random experiment.

Since the outcomes of our random experiments

are represented as points ω ∈ S then to each such

outcome ω , we associate a real number X( ω ), which is in fact the value the random variables takes on when the experiment outcome is ω.

P { X = x (^) i } = Pi

And (^0) ≤ Pi ≤ 1 for all i

And

∑

n i 1 Pi = 1

Say, we have “A” set of values that x can take

P{x (^) ∈ A} =

xi ∈ A

Pi

The probability that x takes a value in A is defined as the sum of the probabilities Pi of all the values x (^) i in A.

Example:

A ={1,3}

P{x (^) ∈ A} = P 1 + P (^3)

P 1 =2/6=1/3 and P 3 = 1/

P{x (^) ∈ A} = 1/6 +1/3 = ½

A roulette wheel has only ‘finite’ number of outcomes.

We will now consider a case with ‘infinite’ number of outcomes.

1

4 3

1

2 4

Example of infinite number of outcomes

Count the number of photons Y that hit a photo detector in a time interval of 1 minute.
Repeat the experiment many number of times
The fraction of experiments when Y = n is

Pn =

α (^) e − α n

n

! (^) , n ≥ (^0) & α (^) is a + re. Number

Note:

α e (^) = (^)!

0 m

m

m

α ∑

∞

=

Therefore, (^0) ≤ Pn ≤ 1 for all n

And ∑

∞

n = 0

Pn = 1

Say, the lifetime of a communication link

P{x > t } = e −^ λ t , where t ≥ 0 &^ λ^ is a + re. No.

P{ x ≤ t} = 1 - P {x>t}

t e

− λ

The above random variable is said to be

exponentially distributed with rate^ λ^.

P { X ∈ ( t , t + ε )} = P { X > t }− P { X > t + ε}

− λ −λ( + ε) −

t t e e

= {. }

e − e e

t t

= {^1 }

e − e

t

for very small^ ε^ and using^ λε e −^ λε^ ≈ 1 −

ε λε

λ { ( , )}.

t P X t t e

− ∈ + = = f ( t ) ε

for t ∈^ (^ −∞,∞) and^0 ≤ t ≤^1

f(t) is called the Probability Density Function (p.d.f) of the random variable x

Thus the pdf of an exponentially distributed random variable with rate^ λ^ is:

f ( t )=λ e −^ λ^ t for t ≥^0

= 0 for t <^0

Example 2:

Let Y be a poisson random variable with parameter^ λ^ , Then, P[Y=1 | Y >0) = P{Y=1 & Y>0}/P{Y ≥^ 0}

=P{Y=1}/[1-P{Y=0}] =^ λ

λ λ −

− − e

e 1

Example 3: Let x be the exponential lifetime of a light bulb. We want to estimate the probability that a light bulb that has been on for “a” seconds will survive for at least t more seconds. We want to calculate: P[x>x+t | x>a]

= P[x>a+t | x>a] = P{x>a+t}/P{x>a}

{ }

( ) e P x t

t a

a t e

e (^) = − = > −

− + λ λ

λ

Expectation

The expected value (also called the average or mean value) of a random variable is the arithmetic mean of the values observed when the random experiment is replicated many times.

Let X be a random variable. Then, the expected value E{X} of X is defined as the sum of values of X weighted by their probabilities.

E { X }= (^) ∑ i 1 xi P { X = xi }

∞

Example: A random variable X is geometrically distributed with parameter p ∈ [0,1]

P{X=n} = (1-p)p n-1^ , for n = 1,2,3, …

E { X }= (^) ∑ i 1 nP { X = n }

∞

p

n

E X i n p p −

∞ − = (^) ∑ (^) = − = 1

(^11)

Independence

Two random variables are independent if knowing the value of one does not provide information about the value of the other.

Two random variables X&Y are said to be independent when,

P{X=x (^) i and Y= y (^) j } = P{X=x (^) i }P{Y= y (^) j } (1)

From this relation we can conclude:

E{X,Y}=E{X}E{Y} (2)

However, the opposite is not true, i.e. if equation (2) holds this does not mean that any 2 random variables are independent.

Regenerative Method

Assumptions: o A packet is correctly transmitted with a probability (1-p) o A packet in error must be re-transmitted o Transmission of packets is independent o X = number of transmissions needed to get a correct packet.

P[X=n]=p n-1^ (1-p)

X = 1 with probab. 1-p 1+Y with probab. P

where X&Y are equal in distribution (since transmissions are independent) So, E(X)=(1-p)1 + p(1+E(Y)), note, E(X)=E(Y) E(X)=1-p+p+pE(X) E(X) = 1/(1-p)

Free

1 day 5 days

3 days

Prison cell

X = 1 prob. 1/ 2+Y 1 prob. 1/ 4+Y 2 prob. 1/ 6+Y 3 prob. 1/

where: X, Y (^) 1 ,Y (^) 2 ,Y 3 are equal in distribution

Therefore, E(X)=E(Y 1 )=E(Y (^) 2)=E(Y (^) 3)

So, E(X) = ¼[1+(2+E(Y (^) 1))+(4+E(Y (^) 2))+(6+E(Y (^) 3))

E(X) = ¼[13+3E(X)]

E(X) = 13