Important statistical distributions
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Transcript Important statistical distributions
Lecture 10
Important statistical distributions
What is the probability that of 10 newborn babies at least 7 are boys?
p(girl) = p(boy) = 0.5
Bernoulli distribution
0.3
n k nk
p (k ) p q
k
0.25
p(X)
0.2
0.15
n
p
0.1
0.05
i 0
0
0
2
4
6
8
i
1
10
X
10 7 3 10 8 2 10 9 1 10 10 0
p(k 6) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.172
7
8
9
10
Bernoulli or binomial distribution
n k nk
p (k ) p q
k
0.35
0.3
n x n x
F (k ) p( x k ) p q
x 0 x
k
np
f(p)
0.25
0.2
0.15
0.1
0.05
0
0
2 npq
1
2
3
4
5
p
6
7
8
10
p(k ) 0.2k 0.810 k
k
The Bernoulli or binomial distribution comes from the Taylor expansion of the
binomial
n i n1 n n i
( p q) p q p (1 q) n1
i 0 i
i 0 i
n
n
9
10
Assume the probability to find a certain disease in a tree population is 0.01. A biomonitoring program surveys 10 stands of trees and takes in each case a random sample of
100 trees. How large is the probability that in these stands 1, 2, 3, and more than 3 cases of
this disease will occur?
1000
0.01* 0.99999 0.0004
p(1)
1
1000
0.012 * 0.99998 0.0022
p(2)
2
Mean, variance, standard deviation
1000* 0.01 10
2 1000* 0.01* 0.99 9.9
1000
0.013 * 0.99997 0.0074
p(3)
3
9.9 3.146
1000
0.010 0.991000
p(k 3) 1 p (k 3) 1 0.01 0.99 1
i 0
0
1000
1000
1000
1
999
2
998
0.01 0.99
0.01 0.99
0.0130.99997 0.99
1
2
3
3
i
n i
n k nk
p (k ) p q
k
What happens if the number of trials n becomes larger and larger and p the event
probability becomes smaller and smaller.
np
rp
r
p
q 1 p
1 p
r
r
(r k )!
k
rr
k
1
(r k )!
p( X k )
k !(r 1)! (r ) k ( r ) r k ! r ( r 1)!( r ) k
1
r
lim r
1
e
1
r
(r k )!
lim r
1
k
(r 1)!(r )
r
p( X k )
k
k!
e
Poisson distribution
The distribution or rare events
Assume the probability to find a certain disease in a tree population is 0.01. A biomonitoring program surveys 10 stands of trees and takes in each case a random sample of
100 trees. How large is the probability that in these stands 1, 2, 3, and more than 3 cases of
this disease will occur?
Poisson solution
1000* 0.01 10
Bernoulli solution
p (1) 0.0004
10 10
e 0.00045
1!
102 10
p ( 2)
e 0.0023
2!
103 10
p (3)
e 0.0076
3!
p (1)
p (2) 0.0022
p (3) 0.0074
The probability that no infected tree will be detected
100 10
p(0)
e e 10 0.000045
0!
p(0) e
The probability of more than three infected trees
Bernoulli solution
p(0) p(1) p(2) p(3) 0.00045 0.0023 0.0076 0.019
p(k 3) 1 0.019 0.981
p(k 3) 0.99
p(k)
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
=1
=2
=3
0
1
2
3
4
=4
5
=6
6
7
8
9 10 11 12 13
k
2
Variance, mean
1
Skewness
What is the probability in Duży Lotek to have three times cumulation
if the first time 14 000 000 people bet, the second time 20 000 000,
and the third time 30 000 000?
The probability to win is
p ( 6)
6!43!
1
49!
14000000
1
1 14000000
1
14000000
1
2 20000000
1.428571
14000000
1
3 30000000
2.142857
14000000
The probability of at least one event:
p(k 1) 1 e
10 1
p1 e 0.368
0!
1.4285710 1.428571
p2
e
0.239
0!
2.1428570 2.142857
p3
e
0.117
0!
The zero term of the Poisson distribution
gives the probability of no event
The events are independent:
p1, 2,3 0.368* 0.239* 0.117 0.01
The construction of evolutionary trees from DNA sequence data
T→C
T
C
A→G
A
G→C→G
T
G→C→A
A
A
C
G
T
T
C
A→G
A
G
T
G
C
C
C
T
Probabilities of DNA substitution
We assume equal substitution
Single substitution
probabilities. If the total
probability for a substitution is p:
Parallel substitution
p
A
T
Back substitution
p
p
p
Multiple substitution
C
G
p
p(A→T)+p(A→C)+p(A→G)+p(A→A) =1
The probability that A mutates to T, C, or G is
P¬A=p+p+p
The probability of no mutation is
pA=1-3p
Independent events
The probability that A mutates to T and C to G
is
PAC=(p)x(p)
Independent events
p( A B) p( A) p( B)
p( A B) p( A) p( B)
The probability matrix
T→C
T
C
A→G
A
G→C→G
T
G→C→A
A
A
C
G
T
T
C
A→G
A
G
T
G
C
C
C
T
Single substitution
Parallel substitution
Back substitution
Multiple substitution
T
A
C
G
p
p
p
1 3 p
1 3 p
p
p
p
P
p
p
1 3 p
p
p
p
p
1 3 p
A
T
C
G
What is the probability that after 5 generations A did not change?
p5 (1 3 p)5
The Jukes - Cantor model (JC69) now assumes that all substitution probabilities are equal.
The Jukes Cantor model assumes equal substitution probabilities within these 4
nucleotides.
Arrhenius model
p
p
p
1 3 p
dP (t )
P(t ) P(t ) P(0)e t
p
1
3
p
p
p
P
dt
p
p
1 3 p
p
Substitution probability after time t
p
p
p
1 3 p
A,T,G,C
A
Transition matrix
t
P(t ) P(0)t
The probability that nothing changes is the
zero term of the Poisson distribution
P( A C, T , G) e e4 pt
Substitution matrix
1 3 4 t
e
4 4
1 1 e 4 t
P4 4
1 1 4 t
4 4e
1 1
e 4 t
4 4
1 1 4 t
e
4 4
1 3 4 t
e
4 4
1 1 4 t
e
4 4
1 1 4 t
e
4 4
1 1 4 t
e
4 4
1 1 4 t
e
4 4
1 3 4 t
e
4 4
1 1 4 t
e
4 4
The probability of at least one substitution is
1 1 4 t
e
4 4
1 1 4 t
e
4 4
1 1 4 t
e
4 4
1 3 4 t
e
4 4
P( A C T G) e 1 e4 pt
The probability to reach a nucleotide from
any other is
1
(1 e 4 pt )
4
The probability that a nucleotide doesn’t
change after time t is
1
1 3
P( A A, T , C , G | A) 1 3( (1 e 4 pt )) e 4 pt
4
4 4
P( A, T , G, C A)
Probability for a single difference
1
3 3
P( A A, T , C , G ) 3( (1 e 4 pt )) e 4 pt
4
4 4
What is the probability of n differences after time t?
0.35
0.3
0.25
f(p)
We use the principle of maximum likelihood and
the Bernoulli distribution
x
n x
n
n x
3
3
3
3
p( x, t ) p (1 p) n x e4 pt 1 ( e4 pt )
4 4
x
x 4 4
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
p
6
7
8
9
n
n
3 3
1 3
ln p( x, t ) ln x ln p (n x) ln(1 p) ln x ln e 4 pt (n x) ln e4 pt )
4 4
4 4
x
x
t
1 4x
ln1
4 p 3n
This is the mean time to get x different sites from a sequence of n nucleotides.
It is also a measure of distance that dependents only on the number of
substitutions
10
Homo sapiens
Pan troglodytes
Pan paniscus
Gorilla
Homo
neandertalensis
Phylogenetic trees are the
basis of any systematic
classificaton
t
1 4x
ln1
4 p 3n
Time
Divergence - number of substitutions
A pile model to generate the binomial.
If the number of steps is very, very large the binomial becomes smooth.
Abraham de Moivre
(1667-1754)
f ( x) Ce
The normal distribution is the
continous equivalent to the discrete
Bernoulli distribution
1
f ( x)
e
2
( x 2 )
1 x
2
2
-2 -1.2 -0.4 0.4 1.2
X
0.05
0.04
0.03
0.02
0.01
0
2
Frequency
Frequency
0.05
0.04
0.03
0.02
0.01
0
0.06
0.04
0.02
0
-2 -1.2 -0.4 0.4 1.2
X
Frequency
0.15
0.1
0.05
0
-2 -1.2 -0.4 0.4 1.2
X
Frequency
Frequency
Frequency
The central limit theorem
If we have a series of random variates Xn, a new random variate Yn that is the sum of all Xn
will for n→∞ be a variate that is asymptotically normally distributed.
2
2
-2 -1.2 -0.4 0.4 1.2
X
0.25
0.2
0.15
0.1
0.05
0
-2 -1.2 -0.4 0.4 1.2
X
0.15
0.1
0.05
0
-2 -1.2 -0.4 0.4 1.2
X
2
2
2
X
X
X
0.06
0.05
f(x)
0.04
0.03
0.02
1
f ( x)
e
2
0.01
( x )2
2 2
0
0
0.5
1
1.5
2
2.5
X
3
3.5
4
4.5
5
4
4.5
5
The normal or Gaussian distribution
1.2
1
1
F ( x)
2
f(x)
0.8
0.6
0.4
x
e
( v )2
2
2
dv
0.2
0
0
0.5
1
1.5
2
2.5
X
Mean:
Variance: 2
3
3.5
•
•
Important features of the normal distribution
The function is defined for every real x.
The frequency at x = m is given by
1
0.4
p( x )
2
•
•
The distribution is symmetrical around m.
The points of inflection are given by the second
derivative. Setting this to zero gives
( x ) x
X
X
X
0.06
0.05
f(x)
0.04
-
0.03
0.02
-2
0.01
0.68
0.95
+
+2
0
0
1
2 e
0.5
1 x
2
2
1 x
2
1
1
2
2 e
2
1
e
2
2
1
2 e
2
1
2 e
1
1.5
2
2.5
X
3
3.5
F ( x)
4
1
2
4.5
x
e
( v )2
2
2
5
dv
0.68
1 x
2
1 x
2
2
1 x
2
2
2
0.95
0.5
0.975
Many statistical tests compare observed values with
those of the standard normal distribution and assign
the respective probabilities to H1.
The Z-transform
1
f ( x)
e
2
1 x
2
The standard normal
1 2 Z 2
f ( x)
e
2
1
2
x
Z
The variate Z has a mean of 0 and and
variance of 1.
A Z-transform normalizes every statistical distribution.
Tables of statistical distributions are always given as Ztransforms.
The 95%
confidence limit
0.1
0.04
0.06
0.04
0.02
0
0.05
0
0
2
0.02
0
0 6 12 18 24 30 36 42 48
The
Z-transformed (standardized)
normal distribution
X
X
X
4
6
8
10
0
3
6
9
12 15 18
0.06
0.05
f(x)
0.04
-
0.03
0.02
-2
0.01
0.68
0.95
+
+2
0
0
0.5
1
1.5
2
2.5
X
3
P( - < X < + ) = 68%
P( - 1.65 < X < + 1.65) = 90%
P( - 1.96 < X < + 1.96) = 95%
P( - 2.58 < X < + 2.58) = 99%
P( - 3.29 < X < + 3.29) = 99.9%
3.5
4
4.5
5
The Fisherian
significance levels
Why is the normal distribution so important?
The normal distribution is often at least approximately found in nature. Many additive or
multiplicative processes generate distributions of patterns that are normal. Examples are body sizes,
intelligence, abundances, phylogenetic branching patterns, metabolism rates of individuals, plant and
animal organ sizes, or egg numbers. Indeed following the Belgian biologist Adolphe Quetelet (17961874) the normal distribution was long hold even as a natural law. However, new studies showed that
most often the normal distribution is only a approximation and that real distributions frequently follow
more complicated unsymmetrical distributions, for instance skewed normals.
The normal distribution follows from the binomial. Hence if we take samples out of a large
population of discrete events we expect the distribution of events (their frequency) to be normally
distributed.
The central limit theorem holds that means of additive variables should be normally distributed.
This is a generalization of the second argument. In other words the normal is the expectation when
dealing with a large number of influencing variables.
Gauß derived the normal distribution from the distribution of errors within his treatment of
measurement errors. If we measure the same thing many times our measurements will not always give
the same value. Because many factors might influence our measurement errors the central limit
theorem points again to a normal distribution of errors around the mean.
In the next lecture we will see that the normal distribution can be approximated by a number of
other important distribution that form the basis of important statistical tests.
The estimation of the population mean from a series of
samples
x,s
x,s
n xi
n
x
n
i
n
n x
Z i 1 n
i 1
n
2
si
n
x,s
x,s
x,s
i 1
n=10
0.25
x,s
f(x)
f(x)
0.2
0.15
0.1
,
0.05
0
0
2
4
6
8
10
x,s
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0.12
n=20
n=50
0.1
Zx
n
0.08
f(x)
0.3
x,s
The n samples from an additive random
variate.
Z is asymptotically normally distributed.
0.06
0.04
0.02
0
0
3
X
6
9 12 15 18
X
0 6 12 18 24 30 36 42 48
X
0.06
x
n
0.05
f(x)
0.04
-
0.03
0.02
-2
0.01
0.68
0.95
Standard error
+
Confidence limit of the estimate
of a mean from a series of
samples.
+2
0
0
0.5
1
1.5
2
2.5
X
3
3.5
4
4.5
5
is the desired probability level.
How to apply the normal distribution
Intelligence is approximately normally distributed with a mean of 100 (by definition)
and a standard deviation of 16 (in North America). For an intelligence study we need
100 persons with an IO above 130. How many persons do we have to test to find this
number if we take random samples (and do not test university students only)?
1
F ( x 130)
2
( v )2
2
2
e
130
1
dv 1
2
a
( z )
F ( x a)
2
130 ( v )
2 2
e
dv
0.03
0.025
f(IQ)
0.02
0.015
0.01
IQ<130
IQ>130
0.005
0
40
60
80
100
IQ
120
140
160
One and two sided tests
We measure blood sugar concentrations and know that our method estimates the
concentration with an error of about 3%. What is the probability that our
measurement deviates from the real value by more than 5%?
Albinos are rare in human populations.
Assume their frequency is 1 per 100000 persons. What is the probability to find 15
albinos among 1000000 persons?
1000000
15
999985
p( X 15)
(0.00001)
(0.99999)
15
=KOMBINACJE(1000000,15)*0.00001^15*(1-0.00001)^999985 = 0.0347
np
2 npq
Home work and literature
Refresh:
Literature:
•
•
•
•
•
•
•
Łomnicki: Statystyka dla biologów
Mendel:
http://en.wikipedia.org/wiki/Mendelian_inheritan
ce
Pearson Chi2 test
http://en.wikipedia.org/wiki/Pearson's_chisquare_test
Bernoulli distribution
Poisson distribution
Normal distribution
Central limit theorem
Confidence limits
One, two sided tests
Z-transform
Prepare to the next lecture:
•
•
•
•
•
•
c2 test
Mendel rules
t-test
F-test
Contingency table
G-test