STA3043/7/8 Generalised Linear Models Exam Summary

EXPONENTIAL FAMILY & GLMS
Three components required for GLM
Response variables Y, Yn assumed be from a distribution
1. . .
.,
to
belonging to the

exponential family
. A
2 set of parameters B and explanatory variables si for i= 1 ,2
, ...,
N

. A
3 monotone link function g) such that g(mi)-ociTB for i= 1, 2
, ...,
N



Exponential family parameter of interest


f(y : 2) =
exp(a(y)b(z) + c(q) +
d(y)
·

aly) y - = canonical form

·
bla) is the natural parameter
mean and Variance
(2)
E(a(y))
-


=
D'(8)
C'(G)b"(a) -

C"(2)b'(G)
var(a(y)) =
(b'(2))3

Likelihood function

e(y ; 8) =
a(y)b(a) + c(a) +
d(y)
e'(y : 2) =
a(y)b'(z) +
c'(2)

e"(yiz) =
a(y)b"(2) + c"(2)
Score function

U(aiy) G = =
a(y)b'(a) + C'(2) function

U =
a(Y)b' (2) + c' (2) statistic

E(U) =
b'(G)E(a(y)] + c'(z)
=-
D'Ac
E(U] =
0

var(u] =
E(u2) -


E(u) =
E(Un
var(u) =
-

f(u) = Y
-

E(U) =
b"(8)f(a(Y)] -

c"(2)
-
Cable
Y =
(b'(a)) Var(a(y)]

,