are independent random variables each having a
distribution. Let 
, 
,
, 
and 
.
Give the names for the distributions of each of 
, U, V, X and Y
and use tables to find 
, 
, 
,
, 
, 
.
is 
, U is 
, V is 
, X is 
,
since 
and V are independent and 
, and 
has an 
distribution, using the fact that U and V are independent. Your answer
should specify the mean and variance for 
, the various degrees of
freedom and note the required independence for X and Y.
The required probabilities are 0.197, 0.05, 0.725, 0.025, 0.688, 0.034.
I used SPlus to compute these; if you used tables your answers will be less
accurate.
for 
;
the new process is used to measure the concentrations for these samples.
It is thought likely that the concentrations measured by the new process,
which we denote 
, will be related to the true concentrations via
where the 
are independent, have mean 0 and all have the
same variance 
which is unknown.
- If this model is fitted by least squares, (that is by minimizing
) show that the least squares estimate of
is
Differentiate
with respect to 
and
get 
which is 0 if and only if 
. The second derivative of the function being
minimized is 
so this is a minimum.
- Show that the estimator in part (a) is unbiased.
Let
. Then 
. Use
to see that 
- Compute (give a formula for) the standard error of
.
You have to compute
which is simply 
.
- The error sum of squares for this model is
which may be shown to have n-1 degrees of freedom.
If the 
are the numbers 1, 2, 3 and 4, 
and the
error sum of squares is 0.12 find a 95% confidence interval for 
and explain what further assumptions you must make to do so.
If we assume that the errors are independent
) random
variables then 
is independent of the usual estimate of
, samely 
in this case. The usual t
statistic then has a t distribution and the confidence interval
is 
which boils down to 
.
- Show that the estimator
is also unbiased.
Let
; then 
.
- Compute (give a formula for) the standard error of
. Which
is bigger, the standard error of 
or that of 
?
We have
.
The difference 
is then simply
The denominator is positive and the numerator is
times the usual
sample variance of the x's so this difference of variances is positive.
- Show that the mle of
in this model is 
, the least
squares estimate, if the 
have normal distributions.
In this case
is 
and the likelihood is
As usual you maximize the logarithm which is
Take the
derivative and get the same equation to solve for 
as in the first part of this problem.
for
and 
. We generally fit a so-called additive model
In the following questions consider the case I=2 and J=3.
- If we treat
, 
, 
, 
, 
and 
as the entries in the parameter vector 
what is the design matrix 
and what
is the rank of 
?
Writing the data as
the design matrix is
Letting
denote column i of 
we have 
and 
so that the rank of 
must be no more than
4. But if 
then from row 6
we get 
. Then from rows 4 and 5 get 
and 
. Finally use
row 3 to get 
. This shows that columns 1, 2 4 and 5 are linearly
independent so tha t
has rank at least 4 and so exactly 4.
- What is the determinant of the matrix
? Is this matrix invertible? How many
solutions do the normal equations have?
The matrix
is 6 by 6 but has rank only 4 so is singular and
must have determinant 0. The normal equations are
It may be seen that the second and third rows give equations which add up to the equation in the first row as do the fourth, fifth and sixth rows. Eliminate rows 3 and 6, say and solve. This leaves 4 linearly independent equations in 6 unknowns and so there are infinitely many solutions.
- Usually we impose the restrictions
and 
.
Use these restrictions to eliminate 
and 
from the model equation
and, for the parameter vector 
find the design
matrix 
.
The restrictions give
and 
. In each model equation which mentions either 
or
you replace that parameter by the equivalent formula. So, for
example,
The 6th row of the design matrix is obtained by reading off the coefficients of
which are 1, -1, -1 and
-1. This makes
- An alternate set of restrictions is called corner point coding where we assume
. With this restriction and the parameter vector
what is the design matrix 
?
This just makes the design matrix
just the corresponding columns, 1, 3, 5
and 6 of 
.
- Show that the three design matrices have the same column space by finding a matrix
A such that
and similarly for 
and 
and for 
and 
.
To write
just let A be the 
matrix which picks
out columns 1,3,5 and 6 of 
, namely,
To write
we just have to put back column 2 and 4 remembering that
col 2 is col 1 - col 3 and col 4 is col 1 - col 5 - col 6. Thus
Similarly
A vector in the column space of say
is of the form 
for a vector of
coefficients v. But such a vector is 
and so of the form 
for the vector 
and so in the column space of 
.
- Use the previous part to show that the vectors of fitted values
will be the
same for any solution of the normal equations for any of the three design matrices.
The easy way to do this is to say: the fitted vector
is the closest point
in the column space of the design matrix to the data vector Y. Since all three have the
same column space they all have the same closest point and so the same 
.
Algebra is an alternative tactic: The matrix
is invertible and we have
Plug in
for 
and get
Use
to see that all occurences of 
cancel
out to give
The algebraic approach makes it a bit more difficult to deal with the case of
because the normal equations have many solutions.
Suppose that 
is any solutions of the normal equations
. Then
The matrix
has rank 4. If any vector v satisfies 
then
so that
. This shows that
so that
.
Thus
Postponed to next assignment.