Assignment 1 Solutions
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.
where the are independent, have mean 0 and all have the
same variance
which is unknown.
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.
Let . Then
. Use
to see that
You have to compute
which is simply
.
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
.
is also unbiased.
Let ; then
.
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.
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.
In the following questions consider the case I=2 and J=3.
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.
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.
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
This just makes the design matrix just the corresponding columns, 1, 3, 5
and 6 of
.
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
.
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.
DUE: Friday, 17 January.