Chapter 3: Uncertainty "variation arises in data generated by a model" "how to transform knowledge of this variation into statements about the uncertainty surrounding the model parameters" Confidence intervals via the frequentist/repeated sampling/classical approach

Parameter  T(Y 1,...,Y n ) estimate of  Var(T) =  2 /n nV   2 in probability as n   V 1/2 standard error (estimated s.d.) for T Average. mean  and variance  2 /n S 2 =  (Y i - ) 2 /(n-1) estimates  2 V 1/2 = n -1/2 S s.e. for

pivot - function of data and parameter whose distribution is known distribution of Z(  0 ) does not depend on  0 Exponential. Pr(Y j /  0  u) = 1 - exp(-u), u>0 Z(  0 ) =  Y j /  0 is gamma parameters 1 and n a sum Approximate. Z(  0 ) = (T -  0 )/V 1/2  N(0,1) in distribution Pr(T - V 1/2 z 1-    0  T - V 1/2 z  )   where  (z  ) =  Approximate (1-2  )  100 % CI for  0 interval estimate

Birth data. approx 95% CI for  0 based on normal Z(  0 ) = n 1/2 ( -  0 )/s day 1 data n=16 = 8.77 s 2 = s = 4.30  =.025 z.025 = (6.66,10.87) hrs of labor Pr(T - V 1/2 z 1-    0  T - V 1/2 z  )  

Binomial ditribution. parameters m,  observation R = R/m var( ) =  (1-  )/m s.e. { (1- )/m} 1/2 pivotal quantity ( -  )/{ (1- )/m} 1/2 approx N(0,1) Suppose m = 1000 =.35 approx 95% CI 35  1.96 .015 Margin of error

Delta method. Gauss Method of linearization T n available, but interested in h(T n )  available, but interested in h(  ) (T n -  )/var(T n ) 1/2  Z in distribution n var (T n )   2 in probability T n =  + n -1/2  Z n continues

h( ) smooth h(y)  h(x)+(y-x)h'(x) for y near x h(T n ) = h(  + T n -  )  h(  ) + (T n -  ) h'(  ) = h(  ) + n -1/2  Z n h'(  ) h(T n )  N( h(  ),h'(  ) 2 var(T n ) )

Poisson, Y. mean , variance  Many techniques "expect" constant variance, normality, linear model Seek h( ) such that var(h(Y)) = 1 h'(  ) 2  = 1 h'(  ) = 1/  1/2 h(  ) = 2  1/2 Work with Y 1/2  N(  1/2, 1/4) Approx 95% CI for  1/2 : ( Y 1/2 - z.975 /2, Y 1/2 - z.025 /2 ) Square up y = 16 births day 1, CI for  1/2 : 4  1.96/2 Square up (9.1,24.8)

Tests. Null hypothesis H 0 : supposeaverage labor time  0 = 6 hours Alternative H A :  > 6 hours Oxford = 8.77 hours n = 16 Is this extreme? Is average time longer in Oxford? Pivot t = ( -  0 )/(s/n 1/2 ) 2.58 = t obs Pr 0 (T  t obs )  1 -  (2.58) =.005 P-value, significance level Choice

Normal model. N( ,  2 ) mean , variance  2 Standard normal Z = (Y-  )/  ~ N(0,1) Density  (z) cdf  (z) Y =  +  Z

Chi-squared distribution. Z 1,...,Z ~ IN(0,1) W = Z Z 2 degrees of freedom additive qchisq() qchisq(.975,14) = (1 - 2  ) CI for  2 ( (n-1) S 2 / c n-1 (1-  ), (n-1) S 2 / c n-1 (  ) ) Cross-fertilized maize. n 1 = 15, s 1 2 = 837.3,  =.025 ( 14  / , 14  / ) (449,2082) eighths of inches squared

Left: chi-squared Right: students t

Student's t distribution. Maize data, differences n = 15 = s 2 = % CI  (1424.6/15) 1/2  2.14 (0.03,41.84) Is H 0 :  = 0 plausible? Not in the 95% confidence interval

F distribution. F = (W / )/(W' /  ) W's independent ~ F,  F,  ~  2 / F 1,  ~ T  2 Maize. n 1, n 2 =15 s 1 2 = 837.3, s 2 2 = Variances  2,   2 CI for  ( F -1,  -1 (  ) s 2 2 /s 1 2, F -1,  -1 (1-  ) s 2 2 /s 1 2 )  =.025 (0.108,.958) H 0 :  = 1

Normal random sample. =  +n -1/2  Z S 2 = (n-1) -1  2 W Z ~ N(0,1) W ~  n-1 2 independently T = Z/{W/(n-1)} 1/2 is students t with n-1 df T is a pivotal quantity for  100  % CI  n -1/2 s t n-1 (  )

Bivariate data

Bivariate distribution. cov(Y 1,Y 2 ) = E[(Y 1 -  1 )(Y 2 -  2 )] =  12 = cov(Y 2,Y 1 ) Collect into a square array cov(Y,Y) =  covariance matrix 2 by 2 variances,  11 and  22, on diagonal covariances,  12 and  21, off diagonal correlation  =  12 /  (  11  22 )

correlations -0.7, 0, 0.7

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Multivariate normal. p-variate Y = (Y 1,..., Y p ) T p linear combinations of IN(0,1) Linear combinations of normals are normal If it exists, density function f(y; ,  ) E(Y) =  cov(Y,Y) =  These are vectors and matrices Curves of constant density - ellipses

Properties. Marginals also (multivariate) normal Conditionals are (multivariate) normal Bivariate. E(Y 1 ), E(Y 2 ) = 0; var(Y 1 ), var(Y 2 ) = 1; cov(Y 1, Y 2 ) =  Y 1, Y 2 are N(0,1) Conditional distribution: Y 1 given Y 2 is N(  Y 2, 1 -  2 ) If Y 1 and Y 2 are uncorrelated they are independent

If Y is N p ( ,  ) then a + B T Y ~ N q (a + B T , B T  B) A surprise (Y -  ) T  -1 (Y -  ) ~  p 2 Another surprise and S 2 are statistically independent

Proof. S 2 is based on Y i - These are uncorrelated with and all are normal, hence the Y i - are independent of Use. Suppose have samples size n i from IN(  i,  i 2 ) is normal mean:  1 -  2, variance:  1 2 /n 1 +  2 2 /n 12

Pooled estimate of  2 S 2 = {(n 1 -1)S (n 2 -1)S 2 2 }/(n 1 + n 2 -2)  2  2 / independ of  confidence interval ( )  {S 2 (n n 2 -1 )} 1/2  t (  ) = n 1 + n 2 -2 Maize  /2 (1/15+1/15) 1/2  % CI (3.34,38.53) Doesn't include 0

Simulation. Computer generation of artificial data How much variability to expect Adequacy of approximation Sensitivity of conclusions To provide insight How variable are normal probability plots? What does bivariate normal data look like?  Based on pseudo-random, e.g. approx IN(0,1)

Tiger Woods, 20% Lance Armstrong, 30% Serena Willians, 50% Pictures in cereal boxes with these percents How many boxes do you expect to have to buy to get all 3? X = 3, 4, 5, …

Assume pictures distributed randomly R.v. Pr{X=Tig} =.2, Pr{X=Lan}=.3, Pr{X=Ser}=.5 Simulate times summary() Min. 1st Qu. Median Mean 3rd Qu. Max

Linear congruential generator. X j+1 = (aX j +c) mod M U j = X j /M M = 2 48, a = 5 17, c = 1 Study by simulation!

Other distributions. Continuous cdf F, inverse F -1 Y = F -1 (U) ~ F(y) N(0,1). Z =  -1 (U), Y =  +  Z Exponential. - log(1-U)/ qnorm qgamma qchisq qt, qf Discrete - layout segments, lengths p i, along [0,1]

Birth data. Poisson arrivals, = 12.9/day N ~ y e - /y!, y=0,1,2,3,... (2.6) Arrival times uniform during the day V 1,..., V N 1/24, 0<y<24 Women remain for gamma, shape  = 3.15, mean  = 7.93 hours G 1,...,G N  y  -1 exp{- y}/  (  ) y > 0,  =  / (2.7) V 1 + G 1,..., V N + G N Record how many women present at each arrival/departure