Transcript CA660_DA_L5_2013_2014x

DATA ANALYSIS
Module Code: CA660
Lecture Block 5
ESTIMATION /H.T.
Rationale, Other Features & Alternatives
• Estimator validity – how good?
• Basis statistical properties (variance, bias, distributional etc.)
• Bias E (ˆ)   where ˆ is the point estimate,  the true
parameter. Bias can be positive, negative or zero.
Permits calculation of other properties, e.g. MSE  E (ˆ   ) 2
where this quantity and variance of estimator are the same if
estimator is unbiased.
Obtained by both analytical and “bootstrap methods”
Bias 

ˆ j f ( x)  
j
Similarly, for continuous variables
or for b bootstrap replications,
Bias B 
1
b

b
ˆi  
i 1
2
Estimation/H.T. Rationale etc. - contd.
• For any, estimator ˆ , even unbiased, there is a difference
between estimator and true parameter = sampling error
Hence the need for probability statements around ˆ
P{T 1  ˆ  T 2}  
C.L. for estimator = (T1 , T2), similarly to before and  the
confidence coefficient. If the estimator is unbiased, in other
words,  = P{that true parameter falls into the interval}.
• In general, confidence intervals can be determined using
parametric and non-parametric approaches, where parametric
construction needs a pivotal quantity = variable which is a
function of parameter and data, but whose distribution does
not depend on the parameter.
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Related issues in Hypothesis Testing -POWER
of the TEST
• Probability of False Positive and False Negative errors
e.g. false positive if linkage between two genes declared, when
really independent
Fact
H0 True
H0 False
Hypothesis Test Result
Accept H0
Reject H0
1-
False negative
=Type II error=
False positive
= Type I error =
Power of the Test
= 1- 
• Power of the Test or Statistical Power = probability of rejecting
H0 when correct to do so. (Related strictly to alternative
hypothesis and )
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Example on Type II Error and Power
• Suppose have a variable, with known population S.D. = 3.6. From
the population, a r.s. size n=100, used to test at =0.05, that
H 0 :   0  17.5
H1 :   17.5
• critical values of
x C.I for a 2-sided test are:
xi   0  U 
  0  1.96 
n
n
for =0.05 where for xi , i = upper or lower and 0  under H0
• So substituting our values gives:
xU  17.50  1.96 (0.36)  18.21;
xL  17.50  1.96 (0.36)  16.79
But, if H0 false,  is not 17.5, but some other value …e.g. 16.5 say ??
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Example contd.
• Want new distribution with mean  = 16.5, i.e. new distribution
is shifted w.r.t. the old.
• Thus the probability of the Type II error - failing to reject false H0
is the area under the curve in the new distribution which
overlaps the non-rejection region specified under H0
• So, this is
18.21  16.5 
1.71 
16.79  16.5
 0.29
P
U 
U 
  P

0
.
36
0
.
36
0
.
36
0
.
36




 P{0.81  U  4.75}
 1  0.7910  0.209
• Thus, probability of taking the appropriate action (rejecting H0
when this is false) is 0.791 = Power
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Shifting the distributions
Non-Rejection
region
Rejection region
f ( x0 )
Rejection region
/2
/2
16.79
17.5
18.21
f ( x1 )
16.5

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Example contd.
Power under alternative  for given 
Possible values of 
under H1 for H0 false

16.0
16.5
17.0
18.0
18.5
19.0
0.0143
0.2090
0.7190
0.7190
0.2090
0.0143
1-
0.9857
0.7910
0.2810
0.2810
0.7910
0.9857
• Balancing  and  :  tends to be large c.f.  unless original
hypothesis way off. So decision based on a rejected H0 more
conclusive than one based on H0 not rejected, as probability of
being wrong is larger in the latter case.
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SAMPLE SIZE DETERMINATION
• Example: Suppose wanted to design a genetic mapping
experiment, or comparative product survey. Conventional
experimental design - ANOVA), genetic marker type (or product
type) and sample size considered.
Questions might include:
What is the statistical power to detect linkage for certain
progeny size? (or common ‘shared’ consumer preferences, say)
What is the precision of estimated R.F. (or grouped
preferences) when sample size is N?
• Sample size needed for specific Statistical Power
• Sample size needed for specific Confidence Interval
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Sample size - calculation based on C.I.
For some parameter , Normal approximation approach valid, C.I. are
ˆ  U (1 ) / 2ˆ

ˆ  U / 2ˆ
U =standardized normal deviate (S.N.D.) and range is from lower to
upper limits, i.e. for 95% limits
d LU  3.92  ˆ  2 1.96  ˆ
is just a precision measurement for the estimator
Given a true parameter  ,
2
 ˆ 
or in " Information" terms 
n
1
nI ( )
So manipulation gives:
2
2
2
2
(
2
U
)

 3.92 
(
2
U
)
ˆ
2

  ˆ 
n  

2
2
d
d
d
 LU 
LU
LU I ( )
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Sample size - calculation based on Power
(firstly, what affects power)?
• Suppose  = 0.05,  =3.5, n=100, testing H0: 0=25 when true  =24;
assume H1 : 1 < 25. Sample mean found = 24.45.
One-tailed test (U = 1.645) : shift small, lower limit of original distribution
virtually coincides with actual sample value
xL   0  U  
n
 25  1.645(0.35)  24.43
;
U alt 
x

n

24.43  24
 1.23
0.35
Under H1  Power = 0.50+0.39 = 0.89; correct decision 89% of time
Note: Two-sided test at  = 0.05 gives critical values, under H0 given by
xL  24.31 , xU  25.69 : equivalently  UL= + 0.89, Uu = 4.82 for H1
In general: substitute for x limits & then recalculate for new  = 1
So, P{do not reject H0: =25 when true mean =24} = 0.1867 =  (Type II)
Thus, Power = 1 - 0.1867 = 0.8133
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Sample Size and Power contd.
• Suppose, n=25, other values same. 1-tailed now
Power = 0.4129
xL  23.85, U L   0.22
• Suppose  = 0.01, critical values 2-tailed xL  24.10 , xu  25.90
with, equivalently, UL = + 0.29, UU = +5.43
So, P{do not reject H0: =25 when true mean =24} = 0.1141
Power = 0.8859
FACTORS : , n and type of test (1- or 2-sided), true parameter value
 2 (U   U  ) 2
n
(  0  1 ) 2
where subscripts 0 and 1 refer to null and alternative, and  value
taken as ‘generic’ (either all in one tail, 1-sided test/limit or split
between two, 2-sided test/limit)
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‘Other’ Estimation/Test Methods
NON-PARAMETRICS/DISTN FREE
• Standard Pdfs can not be assumed for data, sampling distributions or
test statistics – uncertain due to small or unreliable data sets, nonindependence etc. Parameter estimation - not key issue.
• Example / Empirical-basis. Weaker assumptions. Less ‘information’
e.g. median used. Simple hypothesis testing as opposed to estimation.
Power and efficiency are issues.
Counts - nominal, ordinal (natural non-parametric data type/ measure).
• Nonparametric Hypothesis Tests - (has parallels to parametric case).
e.g. H.T. of locus orders requires complex ‘test statistic’ distribution, so
need to construct empirical pdf. Usually, assume the null hypothesis
and use re-sampling techniques, e.g. permutation tests, bootstrap,
jack-knife.
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LIKELIHOOD METHOD - DEFINITIONS
• Suppose X can take a set of values x1,x2,…with


L( )  P{ X  x  }

where
  is a2vector of parameters affecting observed x’s
• e.g.   (  ,  ) . So can say something about P{X} if we
know, say, X ~ N ( ,  2 )

• But not usually case, i.e. observe x’s, knowing nothing of 
• Assuming x’s a random sample size n from a known
distribution, then 
n



likelihood for 
L( )  L( x1, x 2,....xn) 
L( xi  )

i 1
• Finding most likely  or  s for given data is equivalent to
Maximising the Likelihood function, (where M.L.E. is ˆ )
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LIKELIHOOD –SCORE and INFO. CONTENT
• The Log-likelihood is a support function [S()] evaluated at point,  ´
say
• Support function for any other point, say  ´´ can also be obtained – basis for
computational iterations for MLE e.g. using Newton-Raphson
• SCORE = first derivative of support function w.r.t. the parameter
d [ S ( )] or, numerically/discretely, S (   )  S ( )


d
• INFORMATION CONTENT evaluated at (i) arbitrary point = Observed
Info. (ii)support function maximum = Expected Info.
2
 

   log L ( / x )    
  
 
I ( )  E
 2

  2 log L ( / x ) 
 

E
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Example - Binomial variable
(e.g. use of Score, Expected Info. Content to determine type of mapping
population and sample size for genomics experiments)
Likelihood function

n x
L( )  L(n, p )  P{ X  x / n, p}    (1   ) n  x
 x
Log-likelihood

n
Log{L( )}  Log    xLog  (n  x) Log (1   )
 x
Assume n constant, so first term can be ignored for given x - invariant
Log[ L( p)]  xLogp  (n  x) Log (1  p)
Maximising w.r.t. p i.e. set the derivative of S w.r.t. to 0 so
SCORE
x
nx


0
ˆ
ˆ

1
x
so M.L.E. ˆ  pˆ 
n
How does it work, why bother?
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Numerical Examples
See some data sets and test examples:
Basics:
http://www.unc.edu/~monogan/computing/r/MLE_in_R.pdf
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.74.671&rep=r
ep1&type=pdf
Context:
http://statgen.iop.kcl.ac.uk/bgim/mle/sslike_1.html All sections useful,
but especially examples, sections 1-3 and 6
Also, e.g. for R
http://www.montana.edu/rotella/502/binom_like.pdf
for SPSS – see e.g. tutorial for data sets or
http://www.spss.ch/upload/1126184451_Linear%20Mixed%20Effects
%20Modeling%20in%20SPSS.pdf general for mixed Linear Models
For SAS – of possible interest also for Newton-Raphson
http://blogs.sas.com/content/iml/2011/10/12/maximum-likelihoodestimation-in-sasiml/
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Bayesian Estimation- in context
• Parametric Estimation - in “classical approach” f(x,) for a r.v. X of
density f(x) , with  the unknown parameter  dependency of
distribution on parameter to be estimated.
• Bayesian Estimation-  is a random variable, so can consider the
density as conditional and write f(x| )
Given a random sample X1, X2,… Xn the sample random variables are
jointly distributed with parameter r.v.  . So, joint pdf

f X 1 , X 2 ,... Xn , ( x1, x 2,...xn,  )
• Objective - to form an estimator that gives value of  , dependent
on observations of the sample random variables. Thus conditional
density of  given X1, X2,… Xn also plays a role. This is the posterior
density
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Bayes - contd.
• Posterior Density
f ( x1, x2,...., xn)
• Relationship - prior and posterior:
f ( x1, x 2,....xn ) 
 ( )

n
 f ( xk  )
k 1
n


f ( xk  )  d
 ( )

k 1

 


where () prior density of 
• Value: Close to MLE for large n, or for small n if sample values
compatible with the prior distribution. Also, has strong sample
basis, -(simpler to calculate than M.L.E.)
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Estimator Comparison in brief.
• Classical: uses objective probabilities, intuitive estimators, additional
assumptions for sampling distributions: good properties for some
estimators.
• Moments : { less calculation, less efficient. Despite analytical solutions
& low bias, not well-used for large-scale data because less good
asymptotic properties; even simple solutions may not be unique.}
• Bayesian - subjective prior knowledge, sample info. , close to MLE
under certain conditions - see earlier.
• LSE - if assumptions met,  ’s unbiased + variances obtained, {(XTX)-1} .
Few assumptions for response variable distributions, just
expectations, variance-covariance structure. (Unlike MLE where need
to specify joint prob. distribution of variables). Requires additional
assumptions for sampling distns. Close to MLE if these are met.
Computation easier.
1
β̂ ~ N (β, I(β) 1 ), I  2 XT X

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