Download Relation between Binomial and Poisson Distributions

Poisson Processes
• Model for times of occurrences (“arrivals”) of rare phenomena where
λ – average number of arrivals per time period.
X – number of arrivals in a time period.
• In t time periods, average number of arrivals is λt.
• How long do I have to wait until the first arrival?
Let Y = waiting time for the first arrival (a continuous r.v.) then we have
Therefore,
which is the exponential cdf.
• The waiting time for the first occurrence of an event when the number of
events follows a Poisson distribution is exponentially distributed.
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Expectation
• In the long run, rolling a die repeatedly what average result do you expact?
• In 6,000,000 rolls expect about 1,000,000 1’s, 1,000,000 2’s etc.
Average is
• For a random variable X, the Expectation (or expected value or mean) of X
is the expected average value of X in the long run.
• Symbols: μ, μX, E(X) and EX.
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Expectation of discrete random variable
• For a discrete random variable X with pmf
whenever the sum converge absolutely
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Examples
1) Roll a die. Let X = outcome on 1 roll. Then E(X) = 3.5.
2) Bernoulli trials
and
. Then
3) X ~ Binomial(n, p). Then
4) X ~ Geometric(p). Then
5) X ~ Poisson(λ). Then
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Expectation of continuous random variable
• For a continuous random variable X with density
whenever this integral converge absolutely.
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Examples
1) X ~ Uniform(a, b). Then
2) X ~ Exponential(λ). Then
3) X is a random variable with density
(i) Check if this is a valid density.
(ii) Find E(X)
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4) X ~ Gamma(α, λ). Then
5) X ~ Beta(α, β). Then
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Theorem
For g: R  R
• If X is a discrete random variable then
Eg  X    g x  p X x 
x
• If X is a continuous random variable
Eg  X    g x  f X x dx


• Proof:
We proof it for the discrete case. Let Y = g(X) then
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Example to illustrate steps in proof
• Suppose Y  X 2 i.e. g x   x 2 and the possible values of X are
X : 1,  2,  3 so the possible values of Y are Y : 1, 4, 9 then,
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Examples
1. Suppose X ~ Uniform(0, 1). Let Y  X 2 then,
2. Suppose X ~ Poisson(λ). Let Y  e X , then
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Properties of Expectation
For X, Y random variables and a, b  R constants,
• E(aX + b) = aE(X) + b
Proof: Continuous case
• E(aX + bY) = aE(X) + bE(Y)
Proof to come…
• If X is a non-negative random variable, then E(X) = 0
if and only if X = 0 with probability 1.
• If X is a non-negative random variable, then E(X) ≥ 0
• E(a) = a
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Moments
• The kth moment of a distribution is E(Xk). We are usually interested in 1st
and 2nd moments (sometimes in 3rd and 4th)
• Some second moments:
1. Suppose X ~ Uniform(0, 1), then
 
E X2 
1
3
2. Suppose X ~ Geometric(p), then
  x

E X
2
2
pq x 1 
x 1
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Variance
• The expected value of a random variable E(X) is a measure of the “center”
of a distribution.
• The variance is a measure of how closely concentrated to center (µ) the
probability is. It is also called 2nd central moment.
• Definition
The variance of a random variable X is

 
Var  X   E  X  E  X   E  X   
2
2

2
• Claim: Var  X   E X 2   E  X   E X 2    2
Proof:
• We can use the above formula for convenience of calculation.
• The standard deviation of a random variable X is denoted by σX ; it is the
square root of the variance i.e.  X  Var X  .
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Properties of Variance
For X, Y random variables and are constants, then
• Var(aX + b) = a2Var(X)
Proof:
• Var(aX + bY) = a2Var(X) + b2Var(Y) + 2abE[(X – E(X ))(Y – E(Y ))]
Proof:
• Var(X) ≥ 0
• Var(X) = 0 if and only if X = E(X) with probability 1
• Var(a) = 0
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Examples
1. Suppose X ~ Uniform(0, 1), then E  X  
2
1 1
1
Var  X      
3 2
12
1
1
and E X 2   therefore
3
2
1 q
1
2


E
X

2. Suppose X ~ Geometric(p), then E  X   and
therefore
2
p
p
Var  X  
1 q 1
q
1 p



p2
p2
p2
p2
3. Suppose X ~ Bernoulli(p), then E X   p and E X 2   12 p  0 2 q  p
therefore,
Var  X   p  p 2  p1  p 
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Example
• Suppose X ~ Uniform(2, 4). Let Y  X 2. Find PY  9 .
• What if X ~ Uniform(-4, 4)?
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