Simple stochastic models 1
Random variation 1
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Genetic, physiological
Environmental
Measurement error
Random sampling
Random variation 2
• Random variability as nuisance
• Random variability of primary interest
Terminology
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Trial
Sample space
Event
Probability
Random variable
Discrete random variable X
Takes values:
x0, x1, x2, ….
with corresponding probabilities P(X=xk)=pk
p0, p1, p2, …
p0 + p 1 + p2 + … = 1
Discrete - integer random
variable X
Takes values:
0, 1, 2, ….
with corresponding probabilities:
P(X=k)=pk
p0, p1, p2, …
p0 + p 1 + p 2 + … = 1
Mean and variance of discrete
random variable
Mean:
E ( X )   pi xi
i
Variance:
V ( X )   pi [ xi  E ( X )]  E ( X )  E ( X )
2
i
2
2
Binomial Distribution
(Bernoulli trials)
n – number of trials
p – probability of success
P(k successes in n trials) = nCk pk (1-p)n-k
n
 n
n!
Ck    
 k  k!(n  k )!
Binomial distribution
X ~ binomial(n,p)
E(X) = n p
V(X) = n p (1 – p )
n = 10, p = 0.5
0.3
0.25
P(X=k)
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
k
6
7
8
9
10
Poisson distribution
X ~ poisson()
k – number of events
P( X  k )  e


k
k!
E(X)= , V(X)= 
=5
0.18
0.16
0.14
P(X=k)
0.12
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
10
k
12
14
16
18
20
When number of Bernoulli trials n is large,
and probability of success p is small, the
distribution of number of successes
becomes Poisson.
Examples
• 30 % of women in Germany are smokers.
We take a random (representative) sample
of 20 women. The distribution of number of
smokers among them is
X ~ binomial(20,0.3)
• Probability that an accident leading to
injury happens in a factory is p=0.0001 per
day. Number of accidents X over a ten year
period (n=3650) is Poisson, X~Poisson(),
with
 = np = 0.365
Generating function of discrete integer random variables
Discrete - integer random variable
X:
0, 1, 2, ….
p0, p1, p2, …
Generating function:
P(s) = p0 + s p1 +
s2
p2 + …=
s
k
k
pk
Properties
P(1) = 1
P’(1) = E(X)
P’’(1) = E[X(X-1)]
Binomial and Poisson distrib.
Binomial: X ~ binomial(n,p), q=1-p
 n  k nk
n
P( s)   s   p q  q  sp 
k 0
k
Poisson: X ~ Poisson( )
n
k

P( s )   s e
k 0
k 
k
k!
e
  s
Continuity theorem for
generating functions
Two – dimensional discrete –
integer random variables
Joint probability distribution
PXY(X=i, Y=k) = pik
Two – dimensional generating function:
PXY ( s1 , s2 )   p s s
i k
ik 1 2
i ,k
Marginal distributions: PX(s)=PXY(s,1)
PY(s)=PXY(1,s)
Independent discrete – integer
random variables
X: PX(X=i) = pi
Y: PY(Y=k) = pk
PXY(X=i, Y=k) = pi pk
Sum
PXY(X=i, Y=k) = pik
Z=X+Y
P ( Z  m) 
p
i , k :i  k  m
ik
Expectation and variance of sums
of independent random variables
E(X+Y) = E(X) + E(Y)
X, Y - independent
V(X+Y) = V(X) + V(Y)
Sum
Z=X+Y
PZ(s) = PXY(s,s)
X,Y – independent:
PZ(s) = PX(s) PY(s)
Example
X1 ~ binomial(1,p) (one Bernoulli trial)
Xn ~ binomial(n,p)
PX1(s)=q+sp
PXn(s)=(q+sp)n
Example
X ~ Poisson(), Y~Poisson()
Z=X+Y
PZ(s) = PX(s) PY(s)
PZ (s)  e
(    )( s 1)
Z ~ Poisson(+)
Can we generalize generating
function method to non integer
discrete r.v. ?
P(X=xk)=pk:
x0, x1, x2, ….
p0, p1, p2, …
P ( s )   pk s x k
k