Convergence of transformations - Exercise set 1
This exercise set contains some solved exercises on convergence of transformations. The theory needed to solve these exercises is introduced in the lecture entitled Convergence of transformations.
Exercise 1.1
Let
![[eq1]](s.gif)
be a sequence of
random vectors such
thatwhere
denotes convergence in distribution and
is a normal random vector with mean
and invertible covariance matrix
.
Let
be a sequence of
random matrices such
that:where
denotes convergence in probability and
is a constant matrix. Find the limit in distribution of the sequence of
products
.
By Slutski's
theoremwhereThe
random vector
has a multivariate normal distribution, because it is a linear transformation
of a multivariate normal random vector (see the lecture entitled
Linear combinations of
normal random variables). The expected value of
is:![[eq10]](http://images1.statlect.com/convergence_of_transformations_exercise_set_1__18.png)
and
its covariance matrix
is:![[eq11]](http://images2.statlect.com/convergence_of_transformations_exercise_set_1__19.png)
Therefore,
the sequence of products
converges in distribution to a multivariate normal random vector with mean
and covariance matrix
.
Exercise 1.2
Let
be a sequence of
random vectors such
thatwhere
denotes convergence in distribution and
is a normal random vector with mean
and invertible covariance matrix
.
Let
be a sequence of
random matrices such
that:where
denotes convergence in probability. Find the limit in distribution of the
sequence
By the Continuous mapping
theoremTherefore,
by Slutski's
theorem![[eq21]](http://images1.statlect.com/convergence_of_transformations_exercise_set_1__36.png)
Using
the Continuous mapping theorem
again:![[eq22]](http://images2.statlect.com/convergence_of_transformations_exercise_set_1__37.png)
Since
is an invertible covariance matrix, there exists an invertible matrix
such
that:Therefore![[eq24]](http://images1.statlect.com/convergence_of_transformations_exercise_set_1__41.png)
where
we have
definedThe
random vector
has a multivariate normal distribution, because it is a linear transformation
of a multivariate normal random vector (see the lecture entitled
Linear combinations of
normal random variables). The expected value of
is:![[eq26]](http://images2.statlect.com/convergence_of_transformations_exercise_set_1__45.png)
and
its covariance matrix
is:![[eq27]](http://images1.statlect.com/convergence_of_transformations_exercise_set_1__46.png)
Thus,
has a standard multivariate normal distribution (mean
and variance
)
and![[eq28]](http://images2.statlect.com/convergence_of_transformations_exercise_set_1__50.png)
is
a quadratic form in a
standard normal random vector. So,
has a Chi-square distribution with
degrees of freedom. Summing up, the sequence
converges in distribution to a Chi-square distribution with
degrees of freedom.
Exercise 1.3
Let everything be as in the previous exercise, except for the fact that now
has mean
.
Find the limit in distribution of the
sequence![[eq32]](http://images1.statlect.com/convergence_of_transformations_exercise_set_1__57.png)
where
is a sequence of
random vectors converging in probability to
.
DefineBy
Slutski's
theoremwhereis
a multivariate normal random variable with mean
and variance
.
Thus, we can use the results of the previous exercise on the
sequencewhich
is the same
asand
we find that it converges in distribution to a Chi-square distribution with
degrees of freedom.