Bayesian statistics
3. Bayesian statistics
Probability theory and statistics are important in genomics:
- Evolution itself is stochastic in nature
- Large amounts of data make statistical approaches powerful
- Significance: unlikely things do happen in large genomes
Two schools of thought: Frequentists and Bayesians.
In both cases, the concept of “probability” is used.
It is mathematically formalized in the same way, but has a (slightly) different interpretation.
Jerzy Neyman (1894 – 1981)
Confidence interval
Ronald A Fisher (1890-1962)
Max Likelihood; ANOVA
Rev. Thomas Bayes (1702-1761)
Bayes’ Formula
Axioms of probability theory
Basic concepts are:
* Sample space Ω is the set of all possible outcomes
(examples: {head,tails}; all possible trajectories DJ(t) of the Dow-Jones index over time)
* Events E are subsets of Ω
(examples: {head}; Dow-Jones = 10000 at January 1st = { DJ(t) | DJ(Jan 1st) = 10000 }.)
* Probability measure P, assigns a real number to subsets E, with properties:
- P(Ω) = 1
- P() = 0
- P(A  B) = P(A) + P(B) if A,B are disjunct (do not share elements)
(Technical note: not all subsets E may be allowed; in the case that Ω is very large, e.g. all real numbers, or functions, it turns
out to be necessary to restrict yourself to “well behaved” subsets, known as “measurable” sets.)
(Another technical note: In the case that Ω is not a discrete set, P is a probability density, P(E) dE, defined with respect to
another measure dE)
Bayesian vs. Frequentist
Frequentist:
•
E models actual outcomes of
repeatable experiments
•
P(E) models their frequency of occurrence
•
Adequacy of P assessed by hypothesis testing
•
When the model P depends on a parameter ,
the most likely value is preferred
Bayesian:
•
E models both actual outcomes and underlying hypotheses or parameters
•
For a model P,  is considered a random variable rather than a fixed parameter
•
The interpretation of P(E) is different for the two “components” of E:
– Observables: the frequency of the actual occurrences as before
– Hypotheses / parameters: belief in (plausibility of) truth or value.
•
Adequacy of a hypothesis H is tested by computing its posterior probability
– Bayesian approaches include prior probabilities: beliefs before seeing any data
Difference between the two approaches lies mostly in the interpretation. Many researchers use both.
Bayesian approaches are useful with limited amounts of data, as prior information is included.
With lots of data, the prior does not influence the result, and the two approaches give the same answers.
Bayesian vs. Frequentist: inference
Frequentist:
•
Parameters of the model are considered to have a single true (but unknown) value. P(E) models the
parameter-dependent frequency-of-occurrence of E.
•
As a function of  (rather than of E), P(E) is called the likelihood of E.
•
Parameters are estimated by maximizing the likelihood for a given event E.
Example:
Drawing red and black balls from an urn (with replacement).
Urn contains a proportion  of red balls. Draw N balls.
0.3
0.25
0.2
E = {n red, m=N-n black balls drawn}.
This has a maximum for  = n / (n+m).
n  m n
m
 (1   )
P ( E )  
 n 
0.15
0.1
0.05
0.2
This seems reasonable if n and m are large.
0.4
Plot Binomial 7, 2 x^2 1
0.6
0.8
x ^ 5, x, 0, 1
However, suppose you’ve been asked to estimate the probability that the next ball is red.
If n=1, m=0, the maximum likelihood estimate for  clearly is not a reasonable estimate for this
probability, if it is interpreted as your belief in the prediction. This interpretation is quite natural, for
instance it is the way you interpret the statement “there is a 70% probability that it will rain tomorrow”.
1
Bayesian vs. Frequentist: inference
Bayesian:
•
Parameters of the model are considered to be unknown, but not all parameter values are
(necessarily) equally likely a priori
•
 is part of the sample space; the model specifies the joint probability P(E, ).
•
The posterior P(|E) is computed by Bayes’ rule: P(|E) = P(E|) P() / P(E).
Here, P() is the prior probability of the various values that  can take.
•
Parameter are estimated from the posterior P(|E) for .
Example:
0.3
Drawing red and black balls from an urn (with replacement).
Urn contains a proportion  of red balls. Draw N balls.
0.25
0.2
0.15
0.1
E = {n red, m=N-n black balls drawn}.
n  m n
m
 (1   )
P ( E |  )  
 n 
Multiply with a uniform prior P() = 1d:
Calculate P(E):
n m n
m
P ( E , )  
  (1   ) d 
 n 
1
P(E ) 

0
Dividing (*) by P(E) gives posterior.
Posterior mean value of :
Maximum of posterior:
(n+1) / (n+m+2)
n / (n+m)
0.2
0.4
0.6
0.8
(*)
n  m n
1
m
  n  (1   ) d   n  m  1

0 
1
P ( E ,  )d  
0.05
(n+m+1 possibilities)
(Posterior averaging: typical Bayesian approach).
(Maximum a-posteriori [MAP] estimate)
MAP = MLE when uniform priors are used
1
Bayes’ Rule and conditional probabilities
Straightforwardly, P(E) is the probability of the event E, compared to the “universe” Ω of possibilities.
Suppose you know that the outcome (an element of E) will be from a set C  Ω. Other than that, the situation is
the same. Outcomes that are not in C are simply excluded. The probability of event E in this case is called the
conditional probability of E given (or conditional on) C, symbolically P(E | C)
Note that not all outcomes specified by E are necessarily in C. They may even be disjunct (non-overlapping), in
which case the probability of E given C (or: conditional on C) is 0.
Example: there is a certain probability, on any given day, that the sun shines. However, if we know that it was
sunny yesterday, the probability is higher.
The formula for the conditional probability is
P(E | ) 
P ( E , )
P ( )
Here, P(E,θ) = P(Eθ) is the probability that both E and θ happened (i.e. the outcome is in both E and θ).
Conditional Probabilities – Monty Hall problem
There are 3 doors. Behind one door is a prize; there is nothing behind the other two.
You pick one of the doors – but Money doesn’t open it yet.
To show you how lucky you were, Monty then opens one of the doors you didn’t choose without the prize.
Which of the two unopened doors is more likely to hide the prize?
Bayes’ Rule and conditional probabilities
Bayes’ Rule is the formula that tells you how to convert a conditional probability of (say) data conditional on (say)
an unknown parameter, into the probability of that parameter conditional on the data.
P ( | E ) 
P ( E |  ) P ( )
P(E )
The factor P(E| θ) is called the likelihood (when considered as a function of θ).
The probability P(E) is the marginal probability of the event E; it acts as a normalizing constant. Sometimes this is
hard to compute; luckily, for many applications you can do without.
The factor P(θ) is the prior probability of θ (since it does not take the data E into account)
The left-hand side P(θ|E) is the posterior of θ.
Bayes’ rule can be derived from the definition of conditional probability,
to remember.
P ( E , )
P ( E |  )  , which
P ( )
is all you need
Bayes’ Rule and conditional probabilities
P ( | E ) 
P ( E |  ) P ( )
P(E )
Let the random variable θ denote where the prize is (1,2,3)
Initially there is nothing to distinguish the doors – so without loss of generality (W.L.O.G), say you chose door 1.
Again W.L.O.G, suppose Monty opened door 2 to reveal no prize. Let E denote this event.
Conditional probabilities of E given the 3 possibilities for θ are:
P( E| θ=1 ) = 1/2
P( E| θ=2 ) = 0
P( E| θ=3 ) = 1
We have no reason to favour any of the 3 doors, so the prior for θ is uniform, 1/3 for all three possibilities.
The probability of E is intuitively ½ (Monty could only choose from 2 doors), but we do not need intuition:
P( E ) = θ P(E,θ) = θ P(E | θ) P(θ) = 1/2*1/3 + 1*1/3 = ½
The posterior probability of θ=3 is now
P(θ=3 | E) = P(E | θ=3) P(θ=3) / P( E ) = (1 * 1/3) / ½ = 2/3
Equilibrium, reversibility, and MCMC
Very often, you have some complicated model P(D, θ) describing the joint distribution of data D
and parameters θ, and you’re interested in the posterior P(θ|D) for some observation D.
you often want to compute the average of some function over the posterior. This is often
impossible to calculate directly.
The trick with MCMC is to construct a Markov chain whose equilibrium distribution is precisely
P(θ|D). Now, computing the equilibrium distribution as the left eigenvector of 0 is
impossible for Markov model with many states if there is no special structure. So it is very
hard to construct a Markov chain with a desired equilibrium distribution.
For reversible models however, the equilibrium distribution is easy to determine. Conversely, you
can design Markov models to have any particular equilibrium distribution, as long as the
model is reversible.
Equilibrium, reversibility, and MCMC
Equilibrium distribution:
v R = 0 or x vx Rxy = 0 for all y.
vx Rxy
Reversibility: vx Rxy = vy Ryx for all x and y.
x
y
vy Ryx
Reversibility (of v and R) implies equilibrium:
x vx Rxy = x vy Ryx = vy x Ryx = 0
So reversibility (for v) is a stronger condition than being an equilibrium distribution.
Intuitively, it says that the total “flow” across any arrow of the Markov chain is 0:
there are no “loops”.
It is called “reversible” because if this condition holds, you cannot tell the direction of
time by looking at a system (in equilibrium). The laws of physics have this
property (at thermal equilibrium).
Metropolis-Hastings algorithm
Now suppose you want a Markov chain whose equilibrium is vx = P(x). Here is the recipe:
A. Decide between which pairs of states x,y you will allow transitions. This must be (i) symmetric (if xy then yx),
(ii) all states must be reachable from any other, and (iii) the network must be “aperiodic”.
B. Decide on a “proposal distribution” Q(y|x), which proposes transitions to y for a current state x.
The following rule (Metropolis-Hastings algorithm) generates transitions with the right probabilities:
1.
2.
Draw an y from Q(y|x)
Replace x by y with probability min(1, P(y) Q(x|y) / P(x) Q(y|x) )
Proof: The equilibrium probability of x, multiplied by the probability of the transition xy is
P(x) Q(y|x) min(1,P(y) Q(x|y) / P(x)Q(y|x) ) =
min( P(x) Q(y|x), P(y) Q(x|y) )
x
The expression for the reverse transition is the same (because this expression is symmetric in x and y). So the
system is reversible with P(x) as equilibrium distribution.
Although the equilibrium distribution will be P(x), it may take a while to reach equilibrium (the chain may “mix
badly”). MCMC is about choosing the right Q for MCMC chains to mix well.
y