1. XI. Estimation of linear model parameters
XI.A Linear models for designed experiments
XI.B Least squares estimation of the expectation parameters in simple linear models
XI.C Generalized least squares (GLS) estimation of the expectation parameters in general linear models
XI.D Maximum likelihood estimation of the expectation parameters
XI.E Estimating the variance
Statistical Modelling Chapter XI

2. XI.A Linear models for designed experiments
In the analysis of experiments, the models that we have considered are all of the form
where
y is an n1 vector of expected values for the observations,
Y is an n1 vector of random variables for the observations,
X is an nq matrix with nq,
q is a q1 vector of unknown parameters,
V is an nn matrix,
is the variance component for the ith term in the variance model and
Si is the summation matrix for the ith term in the variance model.
Note that generally the X matrices contain indicator variables, each indicating the observations that received a particular level of a (generalized) factor.
Statistical Modelling Chapter XI

3. Summation matrices
Si is the matrix of 1s and 0s that forms the sums for the generalized factor corresponding to the ith term.
Now,
where
Mi is the mean operator for the ith term in the variance model,
n is the number of observational units and
fi is the number of levels of the generalized factor for this term.
For equally-replicated, generalized factors n/fi = gi where gi is the no. of repeats of each of its levels.
Consequently, the Ss could be replaced with Ms.
Statistical Modelling Chapter XI

4. For example, variation model with Blocks random for RCBD with b = 3 and t = 4
Thus the model is of the form:
Statistical Modelling Chapter XI

5. Classes of linear models
Firstly, linear models split into
expectation model is of full rank
expectation model is of less than full rank.
The rank of an expectation model is given by the following definitions — 2nd is equivalent to that given in chapter I.
Definition XI.1: The rank of an expectation model is equal to the rank of its X matrix.
Definition XI.2: The rank of an n  q matrix X, with n  q, is the number of linearly independent columns in X.
Such a matrix is of full rank when its rank equal q and is of less than full rank when its rank is less than q.
It will be less than full rank if the linear combination of some columns is equal to the same linear combination as some other columns.
Statistical Modelling Chapter XI

6. Simple vs general models
Secondly, linear models split into those for which the variation model is
simple in that it is of the form s2In
more general form that involves more than one term.
For general linear models, S matrices in variation model can be written as direct products of I and J matrices.
Definition XI.3: The direct-product operator is denoted  and, if Ar and Bc are square matrices of order r and c, respectively, then
Statistical Modelling Chapter XI

7. Expressions for S matrices in general models
When all the variation terms correspond to only unrandomized, and not randomized, generalized factors, use the following rule to derive these expressions.
Rule XI.1: The Si for the ith generalized factor from the unrandomized structure that involves s factors, is the direct product of s matrices, provided the s factors are arranged in standard order.
Taking the s factors in the sequence specified by the standard order,
for each factor in the ith generalized factor include an I matrix in the direct product and
J matrices for those that are not.
The order of a matrix in the direct product is equal to the number of levels of the corresponding factor.
Statistical Modelling Chapter XI

8. Example IV.1 Penicillin yield (continued)
This experiment was an RCBD and its experimental structure is
Two possible models depending on whether Blends random or fixed.
For Blends fixed,
and taking the observations to be ordered in standard order for Blends then Flasks,
the only random term is BlendsFlasks and the maximal expectation and variation models are
Statistical Modelling Chapter XI

9. Example IV.1 Penicillin yield (continued)
Suppose observations are
in standard order for Blend then Treatment;
with Flask renumbered within Blend to correspond to Treatment as in the prerandomization layout.
Analysis for prerandomization and randomized layouts must be same as prerandomization layout is one of the possible randomized arrangements.
Then the vectors and matrices in the maximal expectation model
are:
Can write the Xs as direct products:
XB = I5  14
XT = 15  I4
Statistical Modelling Chapter XI

10. Example IV.1 Penicillin yield (continued)
Clearly, sum of indicator variables (columns) in XB and sum of those in XT both equal 120
So the expectation model involving the X matrix [XB XT] is of less than full rank — its rank is b + t -1.
The linear model is a simple linear model because the variance matrix is of the form s2In.
That is, the observations are assumed to have the same variance and to be uncorrelated.
Statistical Modelling Chapter XI

11. Example IV.1 Penicillin yield (continued)
On the other hand, for Blends random,
the random terms are Blends and BlendsFlasks and
the maximal expectation and variation models are
In this case the expectation is of full rank as XT consists of t linearly independent columns.
However, it is a general, not a simple, linear model as the variance matrix is not of the form s2In.
Statistical Modelling Chapter XI

12. Example IV.1 Penicillin yield (continued)
The form is illustrated for b = 3, t = 4 below, the extension to b = 5 being obvious.
Model allows for covariance, and hence correlation, between flasks from the same blend.
Clearly, occurs down the diagonal and occurs in diagonal 44 blocks.
Statistical Modelling Chapter XI

13. Example IV.1 Penicillin yield (continued)
The direct-product expressions for both Vs conform to rule XI.1.
Recall, the units are indexed with Blocks then Flasks.
Hence, the direct product for is I5  I4 and it involves two I matrices because both unrandomized factors, Blends and Flasks, occur in BlendsFlasks.
The direct product for is I5  J4 and it involves an I and a J because the factor Blends is in this term but Flasks is not.
Statistical Modelling Chapter XI

14. Example IX.1 Production rate experiment (revisited)
Layout:
The experimental structure for this experiment was:
Statistical Modelling Chapter XI

15. Example IX.1 Production rate experiment (revisited)
Suppose all the unrandomized factors are random
So the maximal expectation and variation models are, symbolically
y = E[Y] = MethodsSources
var[Y] = Factories + FactoriesAreas FactoriesAreasParts.
That is
Statistical Modelling Chapter XI

16. Example IX.1 Production rate experiment (revisited)
Now if the observations are arranged in standard order with respect to the factors Factories, Areas and then Parts, we can use rule XI.1 to obtain the following direct-product expression for V.
In this matrix
will occur down the diagonal,
will occur in 3  3 blocks down the diagonal and
will occur in 9  9 blocks down the diagonal.
So this model allows for
covariance between parts from the same area and
a different covariance between areas from the same factory.
Statistical Modelling Chapter XI

17. Example IX.1 Production rate experiment (revisited)
Actually, in ch. IX, Factories was assumed fixed and so
So the rules still apply as terms in V only involve unrandomized factors
This model only allows for
covariance between parts from the same area.
Statistical Modelling Chapter XI

18. XI.B Least squares estimation of the expectation parameters in simple linear models
Want to estimate the parameters q in the expectation model by establishing their estimators.
There are several different methods for doing this.
A common method is the method of least squares, as in the following definition originally given in chapter I.
Statistical Modelling Chapter XI

19. a) Ordinary least squares estimators for full-rank expectation models
Definition XI.4: Let Y = Xq + e where
X is an nq matrix of rank m  q with nq,
q is a q1 vector of unknown parameters,
e is an n1 vector of errors with mean 0 and variance s2In.
The least ordinary least squares (OLS) estimator of q is the value of q that minimizes
Note applies to the less-than-full rank case
Statistical Modelling Chapter XI

20. Least squares estimators of q
Theorem XI.1: Let Y = Xq + e where
Y is an n1 vector of random variables for the observations,
X is an nq matrix of full rank with nq,
q is a q1 vector of unknown parameters,
e is an n1 vector of errors with mean 0 and variance s2In.
The ordinary least squares estimator of q is given by
(The ‘^’ denotes estimator)
Proof: Need to set differential of e'e = (Y – Xq)'(Y – Xq) to zero (see notes for details);
Yields the normal equations whose solution is given above.
Statistical Modelling Chapter XI

21. Estimator of y
Using definition I.7 (expected values), an estimator of the expected values is
Now for experiments
the X matrix consists of indicator variables and
the maximal expectation model is generally the sum of terms each of which is an X matrix for a (generalized) factor and an associated parameter vector.
In cases where maximal expectation model involves a single term corresponding to just one generalized factor,
the model is necessarily of full rank and
the OLS estimator of its parameters is particularly simple.
Statistical Modelling Chapter XI

22. Example VII.4 22 Factorial experiment
To aid in proofs coming up consider a two-factor factorial experiment (as in chapter 7, Factorial experiments).
Suppose there are the two factors A and B with a = b = 2, r = 3 and n = 12.
Assume that data is ordered for A then B then replicates.
The maximal model involves a single term corresponding to the generalized factor AB — it is
That is, each column of XAB indicates which observations received one of the combinations of the levels of A and B.
Statistical Modelling Chapter XI

23. Example VII.4 22 Factorial experiment (continued)
Now, from theorem XI.1 (OLS estimator), the estimator of the expectation parameters is given by
So we require
Statistical Modelling Chapter XI

24. Example VII.4 22 Factorial experiment (continued) — obtaining inverse
Notice in the above that
rows of XAB are columns of XAB
each element in product is of the form
where xi and xj are the n-vectors for the ith and jth columns of X, respectively.
i.e. sum of pairwise products of elements of ith and jth columns of X.
Statistical Modelling Chapter XI

25. Example VII.4 22 Factorial experiment (continued) — obtaining product
where Yij. is the sum over the replicates for the ijth combination of A and B.
Statistical Modelling Chapter XI

26. Example VII.4 22 Factorial experiment (continued) — obtaining parameter estimates
where is the mean over the replicates for the ijth combination of A and B.
Statistical Modelling Chapter XI

27. Example VII.4 22 Factorial experiment (continued) — obtaining estimator of expected values
where is the mean operator producing the n-vector of means for the combinations of A and B.
Statistical Modelling Chapter XI

28. Proofs for expected values for a single-term model
First prove that model is of full rank and then derive the OLS estimator.
Lemma XI.1: Let X be an n  f matrix with n ≥ f.
Then rank(X) = rank(XX).
Proof: not given.
Statistical Modelling Chapter XI

29. Proofs for expected values for a single-term model (continued)
Lemma XI.2: Let Y = Xq + e where X is an n  f matrix corresponding to a generalized factor, F say, with f levels and n ≥ f.
Then, the rank of X is f so that X is of full rank.
Proof: Lemma XI.1 states that rank(X) = rank(XX) and it is easier to derive the rank of XX.
To do this we need to establish a general expression for XX which we do by considering the form of X.
Statistical Modelling Chapter XI

30. Proofs for expected values for a single-term model (continued)
Now
n rows of X correspond to observational units in the experiment and
f columns to the levels of the generalized factor.
All the elements of X are either 0 or 1
In the ith column, a 1 occurs for those units with the ith level of the generalized factor
so that, if the ith level of the generalized factor is replicated gi times, there will be gi 1s in the ith column.
In each row of X there will be a single 1 as only one level of the generalized factor can be observed with each unit
if it is the ith level of the generalized factor that occurs on the unit, then the 1 will occur in the ith column of X.
Statistical Modelling Chapter XI

31. Proofs for expected values for a single-term model (continued)
Considering the f  f matrix XX
its ijth element is the product of the ith column, as a row vector, with the jth column as given in the following expression (see example above):
When i = j, the 2 columns are same and the products are either of two 1s or two 0s, so
product is 1 for units that have ith level of generalized factor and
sum will be gi.
When i ≠ j, have product of vector for ith column, as a row vector, with that for jth column.
As each unit can receive only one level for the generalized factor, it cannot receive both ith and jth levels and the products are either two 0s or a 1 and a 0.
Consequently, sum is zero.
Statistical Modelling Chapter XI

32. Proofs for expected values for a single-term model (continued)
It is concluded that XX is a f  f diagonal matrix with diagonal elements gi.
Hence, in this case, the rank(X) = rank(XX) = f and so X is of full rank.
Statistical Modelling Chapter XI

33. Proofs for expected values for a single-term model (continued)
Theorem XI.2: Let Y = Xq + e
where
X is an n  f matrix corresponding to a generalized factor, F say, with f levels and n ≥ f.
q is a f 1 vector of unknown parameters,
e is an n1 vector of errors with mean 0 and variance s2In.
The ordinary least squares estimator of q is denoted by and is given by
double over-dots indicate not n-vector, but f-vector.
where is the f-vector of means for the levels of the generalized factor F.
Statistical Modelling Chapter XI

34. Proofs for expected values for a single-term model (continued)
Proof: From theorem XI.1(OLS estimator) we have that the OLS estimator is
Need expressions for XX and XY in this special case .
In the proof of lemma XI.2 it was shown that XX is a f  f diagonal matrix with diagonal elements gi.
Consequently, (XX)-1 is a diagonal matrix with diagonal elements
Statistical Modelling Chapter XI

35. Proofs for expected values for a single-term model (continued)
Now XY is an f-vector whose ith element is the total for the ith level of the generalized factor of the elements of Y.
This can be seen when it is realized that the ith element of XY is the product of the ith column of X with Y.
Clearly, the result will be the sum of the elements of Y corresponding to elements of the ith column of X that are one:
those units with the ith level of the generalized factor.
Finally, taking the product of (XX)-1 with XY divides each total by its replication forming the f-vector of means as stated.
Statistical Modelling Chapter XI

36. Proofs for expected values for a single-term model (continued)
Corollary XI.1: With the model for Y as in theorem XI.2, the estimator of the expected values are given by
where
is the n-vector an element of which is the mean for the level of the generalized factor F for the corresponding unit and
MF is the mean operator that replaces each observation in Y with the mean for the corresponding level of the generalized factor F.
Statistical Modelling Chapter XI

37. Proofs for expected values for a single-term model (continued)
Proof: Now and, as stated in proof of lemma XI.2 (rank X), each row of X contains a single 1
if it is the ith level of the generalized factor that occurs on the unit, then the 1 will occur in the ith column of X.
So corresponding element of will be the ith element of which is the mean for the ith level of the generalized factor F and
To show that , we have from theorem XI.1 (OLS estimator) that
Letting yields
Now, given also, MF must be the mean operator that replaces each observation in Y with the mean for the corresponding level of the F.
Statistical Modelling Chapter XI

38. Example XI.1 Rat experiment
See lecture notes for this example that involves a single, original factor whose levels are unequally replicated.
Statistical Modelling Chapter XI

39. b) Ordinary least squares estimators for less-than-full-rank expectation models
The model for the expectation is still of the form E[Y] = Xq but in this case the rank of X is less than the number of columns.
Now rank(X) = rank(XX) = m < q and so there is no inverse of XX to use in solving the normal equations
Our aim is to show that, in spite of this and the difficulties that ensue, the estimators are functions of means.
For example, the maximal, less-than-full-rank, expectation model for an RCBD is:
yB+T = E[Y] = XBb + XTt
Will show, with some effort, the estimator of the expected values is:
Statistical Modelling Chapter XI

40. Less-than-full-rank vs full-rank model
In the full rank model it is assumed that the parameters specified in the model are unique.
There exists exactly one set of real numbers {q1, q2, …, qq} that describes the system.
In the less-than-full-rank model there are infinitely many sets of real numbers that describe the system.
For our example, there are infinitely many choices for {b1, …, bb, t1, …, tt}. The model parameters are said to be nonidentifiable.
In the full rank model XX is nonsingular and so (XX)-1 exists and the normal equations have the one solution:
In a less-than-full-rank model there are infinitely many solutions to the normal equations.
In the full rank model all linear functions of {q1, q2, …, qq} can be estimated unbiasedly.
In the less-than-full-rank model some functions can while others cannot. Want to identify functions with invariant estimated values.
Statistical Modelling Chapter XI

41. yB+T = E[Y] = XBb + XTt and var[Y] = s2In
Example XI.2 The RCBD
Simplest model that is less than full rank is the maximal expectation model for the RCBD,
E[Y] = Blocks + Treats.
Generally, RCBD involves b blocks with t treatments so that there are n = b  t observations.
The maximal model used for an RCBD, in matrix terms, is:
yB+T = E[Y] = XBb + XTt and var[Y] = s2In
However, as previously mentioned, the model is not of full rank because the sums of the columns of both XB and XT are equal to 1n.
Consequently, for X = [XB XT], rank(X) = rank(XX) = b + t - 1
Statistical Modelling Chapter XI

42. Illustration for b=t=2 that the parameters are nonidentifiable
Suppose following information is known about the parameters:
Then the parameters are not identifiable for:
if b1 = 5, then t1 = 5, t2 = 10 and b2 = 7;
if b1 = 6, then t1 = 4, t2 = 9 and b2 = 8.
Clearly can pick a value for any one of the parameters and then find values of the others that satisfy the above equations.
Infinitely many possible values for the parameters.
However, no matter what values are taken for b1, b2, t1 and t2, the value of b2 - b1 = 2 and of t2 - t1 = 5;
these functions are invariant.
Statistical Modelling Chapter XI

43. Coping with less-than-full-rank models
Extend estimation theory presented in section a), Ordinary least squares estimators for full-rank expectation models.
Involves obtaining solutions to the normal equations using generalized inverses.
Statistical Modelling Chapter XI

44. Introduction to generalized inverses
Suppose that we have a system of n linear equations in q unknowns such as Ax = y
where
A is an n  q matrix of real numbers,
x is a q-vector of unknowns and
y is a n-vector of real numbers.
There are 3 possibilities:
the system is inconsistent and has no solution;
the system is consistent and has exactly one solution;
the system is consistent and has many solutions.
Statistical Modelling Chapter XI

45. Consistent vs inconsistent equations
Consider the following equations:
x1 + 2x2 = 7
3x1 + 6x2 = 21
They are consistent in that the solution of one will also satisfy the second because the second equation is just 3 times the first.
However, the following equations are inconsistent in that it is impossible to satisfy both at the same time:
3x1 + 6x2 = 24
Generally, to be consistent, any linear relations on the left of the equations must also be satisfied by the righthand side of the equation.
Theorem XI.3: Let E[Y] = Xq be a linear model for the expectation. Then the system of normal equations
is consistent.
Proof: not given.
Statistical Modelling Chapter XI

46. Generalized inverses
Definition XI.5: Let A be an n  q matrix. A q  n matrix A- such that
AA-A = A
is called a generalised inverse (g-inverse for short) for A.
Any matrix A has a generalized inverse but it is not unique unless A is nonsingular, in which case A- = A-1.
Statistical Modelling Chapter XI

47. Example XI.3 Generalized inverse of a 22 matrix
It is easy to see that the following matrices are also generalized inverse for A:
This illustrates that generalized inverses are not necessarily unique.
Statistical Modelling Chapter XI

48. An algorithm for finding generalized inverses
To find a generalized inverse A- for an matrix A of rank m:
Find any m  m minor H of A i.e. the matrix obtained by selecting any m rows and any m columns of A.
Find H-1.
Replace H in A with (H-1)'.
Replace all other entries in A with zeros.
Transpose the resulting matrix.
Statistical Modelling Chapter XI

49. Properties of generalized inverses
Let A be an n  q matrix of rank m with n  q  m.
Then
A-A and AA- are idempotent.
rank(A-A) = rank(AA-) = m.
If A- is a generalized inverse of A, then (A-)' is a generalized inverse of A'; that is, (A-)' = (A')-.
(A'A)-A' is a generalized inverse of A so that A = A(A'A)-(A'A) and A' = (A'A)(A'A)-A'.
A(A'A)-A' is unique, symmetric and idempotent as it is invariant to the choice of a generalized inverse. Furthermore rank(A(A'A)-A') = m.
Statistical Modelling Chapter XI

50. Generalized inverses and the normal equations
Lemma XI.3: Let Ax = y be consistent.
Then x = A-y is a solution to the system where A- is any generalized inverse for A.
Proof: not given.
Statistical Modelling Chapter XI

51. Generalized inverses & the normal equations (cont’d)
Theorem XI.4: Let Y = Xq + e where X is an n  q matrix of rank m  q with n  q, q is a q  1 vector of unknown parameters, e is an n  1 vector of errors with mean 0 and variance s2In.
The ordinary least squares estimator of q is denoted by and is given by
Proof: In our proof for the full-rank case, showed that the LS estimator is the solution of the normal equations.
Nothing in our derivation of this result depended on the rank of X.
So the least squares estimator in the less-than-full-rank case is still a solution to the normal equations
Applying theorem XI.3 (normal eqns consistent) and lemma XI.3 (g-inverse gives soln) to the normal equations, yields as a solution of the
normal equations as claimed. (not unique — see notes)
Statistical Modelling Chapter XI

52. A useful lemma where a and b are scalars.
Next theorem derives the OLS estimators for the less-than-full-rank model for the RCBD.
But, first a useful lemma.
Lemma XI.4: The inverse of is
where a and b are scalars.
Proof: First show that
Think of J matrices as “sum-everything” matrices.
Consequently,
so that is the inverse of
Statistical Modelling Chapter XI

53. OLS estimators for the less-than-full-rank model for the RCBD
Theorem XI.6: Let Y be a n-vector of jointly-distributed random variables with
y = E[Y] = XBb + XTt and var[Y] = s2In
where
b, XB, t and XT are the usual vectors and matrices.
Then a nonunique estimator of q, obtained by deleting the last row and column of X in computing the g-inverse, is
Triple over-dots indicates a modified, double over-dots vector
In this case, t-vector modified
where
is the b-vector of block means of Y,
is the (t -1)-vector of the first t -1 treatment means of Y,
is the mean of Y for treatment t and
is the grand mean of Y.
Statistical Modelling Chapter XI

54. Proof of theorem XI.6 According to theorem XI.4 (OLS for LTFR),
and so we require a generalized inverse of X'X where X = [XB XT], so that
no. times each treatment occurs in each block
Statistical Modelling Chapter XI

55. An algorithm for finding generalized inverses (from before)
To find a generalized inverse A- for an matrix A of rank m:
Find any m  m minor H of A i.e. the matrix obtained by selecting any m rows and any m columns of A.
Find H-1.
Replace H in A with (H-1)'.
Replace all other entries in A with zeros.
Transpose the resulting matrix.
Statistical Modelling Chapter XI

56. Proof of theorem XI.6
Our algorithm tells us that we must take a minor of order b+t-1 and obtain its inverse.
So delete any row and column and find its inverse
say last row and column.
To remove the last row and column of X'X,
let Z = [XB XT-1] where XT-1 is the n(t -1) matrix formed by taking the first t -1 columns of XT.
Then,
is X'X with its last row and column removed.
Now Z'Z is clearly symmetric so that its inverse will be also. Our generalized inverse will be with a row and column of zeros added. (no need to transpose)
Statistical Modelling Chapter XI

57. Proof of theorem XI.6 (continued)
Now to find the inverse of Z'Z we note that
for a partitioned, symmetric matrix H where
then
So first we need to find W, then V and lastly U
In our case
Statistical Modelling Chapter XI

58. Proof of theorem XI.6 (continued)
Note that lemma XI.4 (special inverse) is used to obtain the inverse in this derivation.
Statistical Modelling Chapter XI

59. Proof of theorem XI.6 (continued)
Consequently,
and
Statistical Modelling Chapter XI

60. Proof of theorem XI.6 (continued)
Now want to partition X'Y similarly:
X'Y is a vector of totals
Total = mean  n.
Express as means for notational economy
where
XT-1 is the n(t -1) matrix formed by taking the first t -1 columns of XT,
xt is column t of XT,
is the b-vector of block means of Y,
is the (t -1)-vector of the first t -1 treatment means of Y,
is the mean of Y for treatment t and
Statistical Modelling Chapter XI

61. Proof of theorem XI.6 (continued)
Hence the nonunique estimators of q are
Now to simplify this expression further we look to simplify
Statistical Modelling Chapter XI

62. Proof of theorem XI.6 (continued)
Firstly, each element of is the sum of the b Block means and
where is the grand mean of all the observations.
Hence,
Similarly,
Next, each element of is the sum of the first (t -1) Treatment means and
Hence,
Similarly,
Statistical Modelling Chapter XI

63. Proof of theorem XI.6 (continued)
Summary:
Therefore
So the estimator is a function of block, treatment and grand means, albeit a rather messy one.
Statistical Modelling Chapter XI

64. c) Estimable functions
In the less-than-full-rank model the problem we face is that q cannot be estimated uniquely as it changes for the many different choice of (X'X)-.
To overcome consider those linear functions of q that are invariant to the choice of (X'X)- and hence to the particular obtained.
That is, consider functions whose estimators are invariant to different s in that their values remain the same regardless of which solution to the normal equations is used.
This property is true for all estimable functions.
Statistical Modelling Chapter XI

65. Definition of estimable functions
Definition XI.6: Let E[Y] = Xq and var[Y] = s2In where X is n  q of rank m  q.
A function 'q is said to be estimable if there exists a vector c such that E[c'Y] = 'q for all possible values of q.
This definition is saying that, to be estimable, there must be some linear combination of the random variables Y1, Y2, …, Yn that is a linear unbiased estimator of 'q.
In the case of several linear functions, we can extend the above definition as follows:
If L is a k  q matrix of constants, then Lq is a set of k estimable functions if and only if there is a matrix C of constants so that E[CY] = Lq for all possible values of q.
However, how to identify estimable functions?
The following theorems give us some important cases that will cover the quantities of most interest.
Statistical Modelling Chapter XI

66. Expected value results
Lemma XI.5: Let
a be a k  1 vector of constants,
A an s  k matrix of constants and
Y a k  1 vector of random variables or random vector.
Statistical Modelling Chapter XI

67. Estimable if linear combination of X
Theorem XI.7: Let E[Y] = Xq and var[Y] = s2In where X is n  q of rank m  q.
Then 'q is an estimable function if and only if there is a vector of constants c such that ' = c'X.
(Note that ' = c'X implies that  is a linear combination of the rows of X.)
Moreover, for a k  q matrix L, Lq is a set of k estimable functions if and only if there is a matrix C of constants such that L = CX.
Statistical Modelling Chapter XI

68. Proof First suppose that 'q is estimable.
Then, by definition XI.6 (estimable function), there is a vector c so that E[c'Y] = 'q.
Now, using lemma XI.5 (E rules) , E[c'Y] = c'E[Y] = c'Xq.
But also E[c'Y] = 'q and so c'Xq = 'q.
Since this last equation is true for all q, follows that c'X = '.
Second, suppose that ' = c'X.
Then 'q = c'Xq = E[c'Y] and, by definition XI.6 (estimable function), 'q is estimable.
Similarly, if E[CY] = Lq, CXq = Lq for all q and thus CX = L.
Likewise, if L = CX, then Lq = CXq = E[CY] and hence Lq is estimable.
Statistical Modelling Chapter XI

69. Expected values are estimable
Theorem XI.8: Let y = E[Y] = Xq and var[Y] = s2In where X is n  q of rank m  q.
Each of the elements of y = Xq is estimable.
Proof: From theorem XI.7, Lq is a set of estimable functions if L = CX.
We want to estimate Xq, so that L = X
and so to have L = CX we need to find a CX = X
Clearly, setting C = In we have that CX = X = L and as a consequence y = Xq is estimable.
Statistical Modelling Chapter XI

70. A linear combination of estimable functions is itself estimable
Theorem XI.9: Let be a collection of k estimable functions.
Let be a linear combination of these functions.
Then z is estimable.
Proof: If each of is estimable then, by definition XI.6 (estimable function), there exists a vector of constants ci such that
Then
where
and so z is estimable.
Statistical Modelling Chapter XI

71. Finding all the estimable functions
First find y = Xq, all of whose rows are estimable functions.
It can be shown that all estimable functions can be expressed as a linear combination of the entries in y = Xq.
So find all linear combinations of y = Xq and you have all the estimable functions.
Statistical Modelling Chapter XI

72. Example XI.2 The RCBD (continued)
For the RCBD, the maximal model is
y = E[Y] = XBb + XTt.
So for b = 5, t = 4,
From this we identify the 20 basic estimable functions, bi + tk.
Any linear combination of these 20 is estimable and any estimable function can be expressed as a linear combination of these.
Hence bi – bi' and tk – tk' are estimable.
On the other hand, neither bi nor tk + tk' are estimable.
Statistical Modelling Chapter XI

73. d) Properties of estimable functions
So we have seen what estimable functions are and how to identify them, but why are they important?
Turns out that they can be unbiasedly and uniquely estimated using any solution to the normal equations and that they are, in some sense, best.
It is these properties that we establish in this section.
But first a definition.
Statistical Modelling Chapter XI

74. The LS estimator of an estimable function
Definition XI.7: Let 'q be an estimable function and let be a least squares estimator of q.
Then is called the least squares estimator of 'q.
Now the implication of this definition is that is the unique estimator of 'q.
This is not at all obvious because there are infinitely many solutions to the normal equations when they are less than full rank.
The next theorem establishes that the least squares estimator of an estimable function is unique.
Statistical Modelling Chapter XI

75. Recall properties of g-inverses
Let A be an n  q matrix of rank m with n  q  m.
Then
A-A and AA- are idempotent.
rank(A-A) = rank(AA-) = m.
If A- is a generalized inverse of A, then (A-)' is a generalized inverse of A'; that is, (A-)' = (A')-.
(A'A)-A' is a generalized inverse of A so that A = A(A'A)-(A'A) and A' = (A'A)(A'A)-A'.
A(A'A)-A' is unique, symmetric and idempotent as it is invariant to the choice of a generalized inverse. Furthermore rank(A(A'A)-A') = m.
Statistical Modelling Chapter XI

76. LS estimator of estimable function is unique
Theorem XI.10: Let 'q be an estimable function.
Then least squares estimator is invariant to the least squares estimator .
That is, the expression for is the same for every expression for
Proof: From theorem XI.7 (estimable if ' = c'X), we have ' = c'X and, from theorem XI.4 (LS/LTFR),
Using these results we get that
Since, from property 5 of generalized inverses, is invariant to choice of (X'X)-, it follows that has the same expression for any choice of (X'X)-.
Hence is invariant to (X'X)- and hence .
Statistical Modelling Chapter XI

77. Unbiased and variance
Next we show that the least squares estimator of an estimable function is unbiased and obtain its variance.
Theorem XI.11: Let E[Y] = Xq and var[Y] = s2In where X is n  q of rank m  q.
Also let 'q be an estimable function and be its least square estimator.
Then is unbiased for 'q.
Moreover, , which is invariant to choice of generalized inverse.
Statistical Modelling Chapter XI

78. Proof of theorem XI.11 Since and ' = c'X we have
This verifies that is unbiased for 'q.
Statistical Modelling Chapter XI

79. Proof of theorem XI.11(continued)
Next
Statistical Modelling Chapter XI

80. Unbiased only for estimable functions
The above theorem is most important.
This follows from that fact that is not unbiased for any 'q.
In general,
which is not necessarily equal to 'q.
Thus the case when 'q is estimable, and hence is an unbiased estimator of it, is a very special case.
Statistical Modelling Chapter XI

81. BLUE = Best Linear Unbiased Estimator
Definition XI.8: Linear estimators are estimators of the form AY, where A is a matrix of real numbers.
That is, each element of the estimator is a linear combination of the elements of Y.
The unbiased least squares estimator developed here is an example of a linear estimator with A = '(X'X)-X'.
Unfortunately, unbiasedness does not guarantee uniqueness.
There might be more than one set of unbiased estimators of q.
Another desirable property of an estimator is that it be minimum variance.
Theorem XI.12, the Gauss-Markoff theorem, guarantees that, among the class of linear unbiased estimators of 'q, the least squares estimator is the best in the sense that the variances of the estimator, , is minimized.
For this reason the least squares estimator is called the BLUE (best linear unbiased estimator).
Statistical Modelling Chapter XI

82. A result that will be required in the proof of the Gauss-Markoff theorem
Lemma XI.6: Let U and V be random variables and a and b be constants. Then
Proof:
Note that last step uses definitions I.5 and I.6 of the variance and covariance.
Statistical Modelling Chapter XI

83. Gauss-Markoff theorem
Theorem XI.12 [Gauss-Markoff]: Let E[Y] = Xq and var[Y] = s2In where X is n  q of rank m  q. Let 'q be estimable.
Then the best linear unbiased estimator of 'q is where is any solution to the normal equations.
This estimator is invariant to the choice of .
Proof: Suppose that a'Y is some other linear unbiased estimator of 'q so that E[a'Y] = 'q.
The proof proceeds as follows:
Establish a result that will be used subsequently.
Prove that is the minimum variance estimator by proving that
Prove that is the unique minimum variance estimator.
Statistical Modelling Chapter XI

84. Proof of theorem XI.12 Firstly, E[a'Y] = a'E[Y] = a'Xq, and
since also E[a'Y] = 'q,
a'Xq = 'q for all q.
Consequently, a'X = '
this is the result that will be used subsequently.
Secondly, we proceed to show that
Now, from lemma XI.6 (var[aU + bV]), as and a'Y are scalar random variables and is their linear combination with coefficients 1 and –1, we have
Statistical Modelling Chapter XI

85. Proof of theorem XI.12 (continued)
(The last step uses theorem XI.11 — LS and var of est. func.)
Statistical Modelling Chapter XI

86. Proof of theorem XI.12 (continued)
So,
Since,
Hence, has minimum variance amongst all linear unbiased estimators.
Statistical Modelling Chapter XI

87. Proof of theorem XI.12 (continued)
Finally, to demonstrate that is the only linear unbiased estimator with minimum variance, we suppose that
Then and hence is a constant, say c.
Since
so that
But so that
Now we have that c = 0 and so
Thus any linear unbiased estimator of 'q which has minimum variance must equal
So, provided restrict our attention to estimable functions, our estimators will be unique and BLU.
Statistical Modelling Chapter XI

88. e) Properties of the estimators in the full rank case
The above results apply generally to models either of less than full rank or of full rank.
However, in the full rank case, some additional results apply.
In particular, so that in this particular case q itself is estimable.
So are BLUE for q.
Further, since 'q is a linear combination of q, it follows that 'q is estimable for all constant vectors .
Statistical Modelling Chapter XI

89. f) Estimation for the maximal model for the RCBD
In the next theorem, because each of the parameters in q' = [b' t'] is not estimable, we consider the estimators of y.
theorem XI.8 tells us that these are estimable,
theorem XI.10 that they are invariant to the solution to the normal equation chosen and
theorem XI.12 that they are BLUE.
Statistical Modelling Chapter XI

90. Estimator of expected values for RCBD
Theorem XI.13: Let Y be a n-vector of jointly-distributed random variables with
y = E[Y] = XBb + XTt and var[Y] = s2In
where b, XB, t and XT are the usual vectors and matrices.
Then where , and are the n-vectors of block, treatment and grand means, respectively.
Statistical Modelling Chapter XI

91. Proof of theorem XI.13
Proof: Using the estimator of q, established in theorem XI.6 (LS estimator of q in RCBD), the estimator of the expected values is given by
Statistical Modelling Chapter XI

92. Proof of theorem XI.13 (continued)
Now each row of [XB XT-1] corresponds to 1 obs: say obs. in ith block that received jth treat.
In the row for obs. in ith block that received jth treat:
the element in the ith column of XB will be 1,
as will the element in the jth column of XT-1;
all other elements in the row will be zero.
Note if treatment is t, only nonzero element will be ith column of XB.
Put B-bar-.. and T-bar … vectors on board and consider what get for say observations 7 and 8 when evaluate expected value expression
Statistical Modelling Chapter XI

93. Proof of theorem XI.13 (continued)
So from find estimator of expected value for obs. in ith block with jth treat is the sum of:
ith element of and
jth element of ,
unless treat is t (j = t) when it is ith element of
Estimator of expected value for obs.
with treatment t in ith block:
with jth treatment (j ≠ t) in the ith block:
That is, for all i and j, and clearly
Statistical Modelling Chapter XI

94. Notes on estimator for RCBD
Now,
That is the estimator of the expected values can be written as the sum of the grand mean plus main effects.
Note that the estimator is a function of means with
where Xi = XB, XT or XG.
That is, MB, MT and MG are the block, treatment and grand mean operators, respectively.
Further, if the data in the vector Y has been arranged in prerandomized order with all the observations for a block placed together, the operators are:
Statistical Modelling Chapter XI

95. Example XI.2 The RCBD (continued)
For the estimable function y23 = b2 + t3,
In this case, the mean operators are as follows:
Statistical Modelling Chapter XI

96. Example XI.2 The RCBD (continued)
Statistical Modelling Chapter XI

97. Example XI.2 The RCBD (continued)
Statistical Modelling Chapter XI

98. XI.C Generalized least squares (GLS) estimation of the expectation parameters in general linear models
Define a GLS estimator.
Definition XI.9: Let Y be a random vector with y = E[Y] = Xq and var[Y] = V where X is an n  q matrix of rank m  q, q is a q  1 vector of unknown parameters, V is an n  n positive-definite matrix and n  q.
Then the generalized least squares estimator of q is the estimator that minimizes the "sum of squares"
Statistical Modelling Chapter XI

99. GLS estimator
Theorem XI.14: Let Y be a random vector with y = E[Y] = Xq and var[Y] = V where X is an n  q matrix of rank m  q, q is a q  1 vector of unknown parameters, V is an n  n positive-definite matrix and n  q.
Then the generalized least squares estimator of q is denoted by and is given by
Proof: similar to proof for theorem XI.1.
Unsurprisingly, properties of GLS estimators parallel those for OLS estimators.
In particular, they are BLUE as the next theorem states.
Statistical Modelling Chapter XI

100. y = E[Y] = Xq and var[Y] = V
Gauss-Markoff for GLS
Theorem XI.15 [Gauss-Markoff]: Let Y be a random vector with
y = E[Y] = Xq and var[Y] = V
where X is an n  q matrix of rank m  q, q is a q  1 vector of unknown parameters, V is an n  n positive-definite matrix and n  q.
Let 'q be an estimable function.
Then the generalized least squares estimator
is the best linear unbiased estimator of 'q, with
Proof: not given
Statistical Modelling Chapter XI

101. Estimation of fixed effects for the RCBD when Blocks are random
We now derive the estimator of t in the model for the RCBD where Blocks are random.
Symbolically, the model is
E[Y] = Treats and var[Y] = Blocks + BlocksUnits.
One effect of making Blocks random is that the maximal, marginality-compliant model for the experiment is of full rank and so the estimator of t will be BLUE.
Before deriving estimators under this model, we establish the properties of a complete set of mutually-orthogonal idempotents as they will be useful in proving the results.
Statistical Modelling Chapter XI

102. Complete set of mutually-orthogonal idempotents
Definition XI.10: A set of idempotents {QF: F  } forms a complete set of mutually-orthogonal idempotents (CSMOI) if
where and otherwise.
Of particular interest is that the set of Q matrices from a structure formula based on crossing and nesting relationships forms a CSMOI.
e.g. RCBD
Statistical Modelling Chapter XI

103. Inverse of CSMOI
Lemma XI.7: For a matrix that is the linear combination of a complete set of mutually-orthogonal matrices, such as ,
its inverse is
Proof:
Statistical Modelling Chapter XI

104. Estimators for only Treatments fixed
In the following theorem we show that the GLS estimator for the expectation model when only Treatments is fixed is the same as the OLS estimator for a model involving only fixed Treatments, such as would occur with a CRD.
However, first a useful lemma.
Lemma XI.8: Let A, B, C and D be square matrices and a, b and c be scalar constants. Then,
provided A and C, as well as B and D, are conformable.
Statistical Modelling Chapter XI

105. Estimators for only Treatments fixed (cont’d)
Theorem XI.16: Let Y be an random vector for the observations from an RCBD with the Yijk arranged such that those from the same block occur consecutively in treatment order as in a prerandomized layout.
Also, let
Then the generalized least squares estimator of t is given by
and the GLS estimator of the expected values is
where
Statistical Modelling Chapter XI

106. Proof of theorem XI.16
From theorem XI.14 (GLS estimator), the GLS estimator of t is given by
the generalized inverse being replaced by an inverse here because XT is of full rank.
To prove the second part of the result, we
derive an expressions for V-1,
use this to obtain ,
invert the last result, and
show that
Statistical Modelling Chapter XI

107. Proof of theorem XI.16 (continued)
We start by obtaining an expression for V in terms of QG, QB and QBU as these form a complete set of mutually orthogonal idempotents.
Now,
But, from section IV.C, Hypothesis testing using the ANOVA method for an RCBD, we know that
so that
Statistical Modelling Chapter XI

108. Proof of theorem XI.16 (continued)
Consequently
where
Since QG, QB and QBU form a complete set of mutually orthogonal idempotents, we use lemma XI.7 (CSMOI inverse) to obtain
we have our expression for V-1.
Statistical Modelling Chapter XI

109. Proof of theorem XI.16 (continued)
Next to derive an expression for , note that
so that
We then see that
Statistical Modelling Chapter XI

110. Proof of theorem XI.16 (continued)
Thirdly, to find the inverse of this:
note that and are two symmetric idempotent matrices whose sum is I and whose product is zero
so that they form a complete set of mutually orthogonal idempotents — see also lemma XI.4 (special inverse).
Now
Statistical Modelling Chapter XI

111. Proof of theorem XI.16 (continued)
Finally,
Statistical Modelling Chapter XI

112. Proof of theorem XI.16 (continued)
The GLS estimator of the expected values is then given by
But, as
and, from corollary XI.1 (expected values for single-term model),
So the estimator of the expected values under the model
are just the treatment means, the same as for the simple linear model
This result is generally true for orthogonal experiments, as stated in the next theorem that we give without proof.
Statistical Modelling Chapter XI

113. Orthogonal experiments
Definition XI.11: An orthogonal experiment is one for which QFQF* equals either QF, QF* or 0 for all F, F* in the experiment.
All the experiments in these notes are orthogonal.
Theorem XI.17: For orthogonal experiments with a variance matrix that can be expressed as a linear combination of a complete set of mutually-orthogonal idempotents, the GLS estimator is the same as the OLS estimator for the same expectation model.
Proof: not given
This theorem applies to all orthogonal experiments for which all the variance terms correspond to unrandomized generalized factors
Statistical Modelling Chapter XI

114. XI.D Maximum likelihood estimation of the expectation parameters
Maximum likelihood estimation: based on obtaining the values of the parameters that make the data most 'likely'.
Least squares estimation minimizes the sum of squares of the differences between the data and the expected values.
Need likelihood function to maximize.
Statistical Modelling Chapter XI

115. The likelihood function
Definition XI.12: The likelihood function is the joint distribution function, f(y; ), of n random variables Y1, Y2, …, Yn evaluated at y = (y1, y2, …, yn).
For fixed y the likelihood function is a function of the k parameters,  = (1, 2, …, n), and is denoted by L(; y).
If Y1, Y2, …, Yn represents a random sample from f(y; ) then L(; y) = f(y; )
Notation f(y; ) indicates that
y is the vector of variables for the distribution function
the values of  are fixed for the population under consideration.
This is read ‘the function f of y given ’.
The likelihood function reverses the roles:
the variables are and y is considered fixed
so we have the likelihood of given y.
Statistical Modelling Chapter XI

116. Maximum likelihood estimates
Definition XI.13: Let L(; y) = f(y; ) be the likelihood function for Y1, Y2, …, Yn.
For a given set of observations, y, the value that maximizes L(; y) is called the maximum likelihood estimate (MLE) of .
An expression for this estimate as a function of y can be derived.
The maximum likelihood estimators are defined to be the same function as the estimate, with Y substituted for y.
Statistical Modelling Chapter XI

117. Procedure for obtaining the maximum likelihood estimators
Write the likelihood function L(; y) as the joint distribution function for the n observations.
Find  = ln[L(; y)].
Maximize  with respect to  to derive the maximum likelihood estimates.
Obtain the maximum likelihood estimators.
The quantity  is referred to as the log likelihood.
Maximizing  is equivalent to maximizing L(; y) since ln is a monotonic function.
It will turn out that the expression for  is often simpler than that for L(; y).
Statistical Modelling Chapter XI

118. Probability distribution function
Now the general linear model for a response variable is
We have not previously specified a probability distribution function to be used as part of the model.
However, need one for maximum likelihood estimation. (least squares estimation does not).
A distribution function commonly used for continuous variables is the multivariate normal distribution.
Definition XI.14: The multivariate normal distribution function for is
where
Y is a random vector, y = E[Y] = Xq,
X is an n  q matrix of rank m  q with n ≥ q,
q is a q  1 vector of unknown parameters, and var[Y] = V.
Statistical Modelling Chapter XI

119. Deriving the maximum likelihood estimators of q
Theorem XI.18: Let Y be a random vector whose distribution is multivariate normal.
Then, the maximum likelihood estimator of q is denoted by and is given by
Statistical Modelling Chapter XI

120. Proof of theorem XI.18 The likelihood function L(; y)
Given the distribution function for Y, the likelihood function L(; y) is
Find  = ln[L(; y)]
Given the distribution function for Y, the likelihood function L(; y) is
Statistical Modelling Chapter XI

121. Proof of theorem XI.18 (continued)
Maximize  with respect to q
The maximum likelihood estimates of are then obtained by maximizing  with respect to , that is, by differentiating  with respect to and setting the result equal to zero.
Clearly, the maximum likelihood estimates will be the solution of
Now,
Statistical Modelling Chapter XI

122. Proof of theorem XI.18 (continued)
and applying the rules for differentiation given in the proof of theorem XI.1, we obtain
Setting this derivative to zero we obtain
Obtain the maximum likelihood estimators
Hence, as claimed.
So for general linear models the maximum likelihood and generalized least squares estimators coincide.
Statistical Modelling Chapter XI

123. XI.E Estimating the variance
a) Estimating the variance for the simple linear model
As stated in theorem XI.11, and so to determine the variance of our OLS estimates need to know s2.
Generally, it is unknown and has to be estimated.
Now the definition of the variance is
var[Y] = E[(Y – E[Y])2].
The logical estimator is one that parallels this definition:
put in our estimate for E[Y] and
replace the first expectation operator by the mean.
Statistical Modelling Chapter XI

124. Fitted values and residuals
Definition XI.15: The fitted values are:
the estimated expected values for each observation,
obtained by substituting the x-values for each observation into the estimated equation.
given by
The residuals are:
the deviations of the observed values of the response variable from the fitted values.
denoted by
the estimates of
Statistical Modelling Chapter XI

125. Estimator of s2 The logical estimator of s2 is:
and our estimate would be ee/n.
Notice that the numerator of this expression measures the spread of observed Y-values from the fitted equation.
We expect this to be random variation.
Turns out that this is the ML estimator as claimed in the next theorem.
Theorem XI.19: Let Y be a normally distributed random vector representing a random sample with E[Y] = Xq and var[Y] = VY = s2In where X is an n  q matrix of rank m  q with n ≥ q, and q is a q  1 vector of unknown parameters.
The maximum likelihood estimator of s2 is denoted by and is given by
Proof: left as an exercise
Statistical Modelling Chapter XI

126. Is the estimator unbiased?
Theorem XI.20: Let
with
Y the n  1 random vector of sample random variables,
X is an n  q matrix of rank m  q with n ≥ q,
is a q  1 vector of estimators of q.
Then
so an unbiased estimator of s2 is
Proof: The proof of this theorem is postponed.
Latter estimator indicates how we might go more generally,
once it is realized that this estimator is one that would be obtained using the Residual MSQ from an ANOVA based on simple linear models.
Statistical Modelling Chapter XI

127. b) Estimating the variance for the general linear model
Generally, the variance components (s2s) that are the parameters of the variance matrix, V, can be estimated using the ANOVA method.
Definition XI.16: The ANOVA method for estimating the variance components consists of equating the expected mean squares to the observed values of the mean squares and solving for the variance components.
Statistical Modelling Chapter XI

128. Example IV.1 Penicillin yield
This example involves 4 treatments applied using an RCBD employing 5 Blends as blocks.
Suppose it was decided that the Blends are random.
The ANOVA table, with E[MSq]s, for this case is
The estimates of and are obtained as follows:
so that
That is, the magnitude of the two components of variation is similar.
Statistical Modelling Chapter XI

129. Example IV.1 Penicillin yield
(see variance matrix in section XI.A, Linear models for designed experiments, above),
Now
so that
Clearly, the two components make similar contributions to the variance of an observation.
Statistical Modelling Chapter XI

130. XI.G Exercises
Ex asks you to provide matrix expressions for models and the estimator of the expected values.
Ex involves a generalized inverse and its use.
Ex asks you to derive the variance of the estimator of the expected values.
Ex involves identifying estimable parameters and the estimator of expected values for a factorial experiment.
Ex asks for the E[Msq], matrix expressions for V and estimates of variance components.
Ex involves max. likelihood estimation (not covered).
Ex requires the proofs of some properties of Q and M matrices for the RCBD.
Statistical Modelling Chapter XI