Statistics 200b. Spring 2008 David Brillinger Statistical Models (2003) by A. C. Davison

1. Statistics 200b. Spring 2008 David Brillinger Statistical Models (2003) by A. C. Davison Chapter 1: Introduction Chapter 2: Variation Chapter 3: Uncertainty Chapter 4: Likelihood A tool for making statistical inferences (uncertain conclusions based on data) Midterm March 12, in class. Through Chapter 4.

3. Chapter 4: Likelihood. "value y of r.v. Y with p.d.f f(y;)  in  f known" "goal to make statements about distribution of Y based on observed data y" that are plausible, given y observed"

4. Likelihood, L. L() = f(y;)  in  regarded as function of  for fixed y "L will be relatively larger for  near that which generated the data" When Y's i.i.d., y = (y1,...,yn) sample L() = f(y;) =   f(yj;)

5. Normal. IN(,2)  = (,2) data: (y1,...,yn) L() =  f(yj;) =  [exp{-(yj- )2/2 2 } / {2} ] l() = log L() = - { log 2 }/2 - { (yj - )2}/2 2

6. An aside One sees that all one needs is l() is a maximum at

7. Poisson. L() = y exp{-} / y!>0, y = 0,1,2,3,... Goes to 0 as  0 or  Maximum at  = y sample y = (y1,...,yn)  f(yj;) = ^(y1+...+yn) exp{-n} / C Maximum at  =

8. Exponential distribution f(y;) = -1 exp{-y/} y > 0  > 0 L() =  f(yj;) = -n exp{-(y1+...+yn)/} Maximum at  =

9. Spring failure at 950 N/mm2 225 171 198 189 189 135 162 135 117 162 103 cycles 1 | 244 1 | 66799 2 | 03 decimal point is 2 digit(s) to the right of the | L() maximized at  approx 168 L(168) = 2.49E-27 L(150) = 2.32E-27 150 is .93 times less likely than 168 Might declare s with L() > .5 L(168) "plausible"

10. Spring failure times at stress 950 N/mm2 225 171 198 189 135 162 135 117 162 units of 1000 cycles stleaf L() maximized at theta approx 168 L(168) approx 2.49E-27 L(150) approx 2.32E-27 "150 is .93 times less likely that 168" Values with L() > .5 L(168) " plausible": (120,260)

11. Weibull. f(y;,) = y-1 exp{-(y/)}/ y>0 ,  > 0 L(,) =  f(yj;,) Contour plot of log L(,) Maximized at approx (181,6) Exponential for  = 1 dotted line

12. Richard Feynman

13. Challenger data. Space shuttle exploded after launch 28 January 1986 Presidential Commission - cause: O-rings not pliable in cold weather or holed in pressure test Thermal distress Data Plot proportions r/m vs temperature x1 and pressure x2 m=6 Statistical model, R binomial Bin(m,π), R=1,...,6 π = eu /(1+eu) u = β0 + β1 x1 + β2 x2

15. rj O-rings out of m with thermal distress at launch temperature xj Pr{distress} = π = eu /(1+eu) u = β0 + β1 x R ~ IBinomial (m,π), m = 6 L(β0 ,β1) =  Pr{Rj = rj; β0 ,β1} Contour plot maximum at (β0 ,β1) approx (5,-0.1) ridge

17. Summaries. "key idea is that in many cases log likelihood is approx quadratic in the parameter" {More later} Example exponential samples of sizes 5, 10, 20, 40, 80 each sample has = exp{-1} the maximum of the log likelihood is at l() = -n (log  + /n) l'( ) = 0 l"( ) = -n/ 2 maximum likelihood estimate,

20. Likelihood, L. L() = f(y;)  in  regarded as function of  for fixed y "L will be relatively larger for  near that which generated the data" When Y's i.i.d., y = (y1,...,yn) sample L() = f(y;) =   f(yj;) l() = log L() =  log f(yj;)

21. Maximum likelihood estimate likelihood equation. when things exist. score statistic/vector. observed information.

22. Local maximum? Computation and representation iterate to convergence

23. log RL() = l() - l( )  ( - )2l"( )/2 "In many cases the likelihood can be summarized in terms of the mle and the observed information -l"( ). " In a sense one can act as if the distribution is approximately normal There are large sample optimality results, but there are problems too, e.g. the uniform on (0,)

24. Observed information Fisher information I() = EJ()

25. Normal distribution

26. Properties of U(), u() E(U()) = nE(u()), E(u()) = 0 Remember

27. Var U() = n Var u() Var u() = i()

28. 0: true parameter value U(0), J(0) sums of i.i.d.'s LLN and CLT EU(0) = 0, Var(0) = I(0) = ni(0) EJ(0) = I(0) = ni(0) I(0)-1 J(0)  1 in probability I(0)-1/2 U(0)  Z in distribution, as n   Z ~ Np (0,Ip )

29. In normal case

31. Approximate 100(1-2)% CI for real 0 CR for vector 0