 
For part E and F of the problem,
we'll be introducing
a new notion of convergence,
so-called the convergence E
mean squared sense.
We say that xn converges to a
number c in mean squared, if
as we take and go to infinity,
the expected value of xn minus
c squared goes to 0.
To get a sense of what this
looks like, let's say we let c
equal to the expected value of
xn, and let's say the expected
value of xn is always
the same.
So the sequence of random
variables has the same mean.
Well, if that is true, then mean
square convergence simply
says the limit of the
variance of xn is 0.
So as you can imagine, somehow
as xn becomes big, the
variance of xn is very small,
so xn is basically highly
concentrated around c.
And by this I mean, the density
function for xn.
So that's the notion
of convergence
we'll be working with.
Our first task here is to show
that the mean square
convergence is in some sense
stronger than the convergence
in probability that we have been
working with from part A
to part D. That is, if I know
that xn converged to some
number c in mean squared, then
this must imply that xn
converges to c in probability.
And now, we'll go show
that for part E.
Well, let's start with a
definition of convergence in
probability.
We want to show that for a fixed
constant epsilon the
probability that xn minus
c, greater than epsilon,
essentially goes to 0 as
n goes to infinity.
To do so, we look at the
value of this term.
Well, the probability of
absolute value xn minus c
greater than epsilon is equal
to the case if we were to
square both sides of
the inequality.
So that is equal to the
probability that xn minus c
squared greater than
epsilon squared.
We can do this because both
sides are positive, hence this
goes through.
Now, to bound this equality,
we'll invoke the Markov's
Inequality, which it says this
probability of xn, some random
variable greater than epsilon
squared, is no more than is
less equal to the expected value
of the random variable.
In this case, the expected value
of x minus c squared
divided by the threshold that
we're trying to cross.
So that is Markov's
Inequality.
Now, since we know xn converges
to c in mean
squared, and by definition,
mean square we know this
precise expectation right
here goes to 0.
And therefore, the whole
expression goes to 0 as n goes
to infinity.
Because the denominator here is
a constant and the top, the
numerator here, goes to 0.
So now we have it.
We know that the probability of
xn minus c absolute value
greater than epsilon goes to 0
as n goes to infinity, for all
fixed value of epsilons and
this is the definition of
convergence in probability.
 
Now that we know if xn converges
to c mean squared,
it implies that xn converges
to c in probability.
One might wonder whether
the reverse is true.
Namely, if we know something
converges in probability to a
constant, does the same sequence
of random variables
converge to the same constant
in mean squared?
It turns out that is
not quite the case.
The notion of probability
converges in probability is
not as strong as a notion of
convergence in mean squared.
Again, to look for a counter
example, we do not have to go
further than the yn's we
have been working with.
So here we know that yn
converges to 0 in probability.
But it turns out it does
not converge to
0 in the mean squared.
And to see why this is the case,
we can take the expected
value of yn minus 0 squared,
and see how that goes.
Well, the value of this can be
computed easily, which is
simply 0, if yn is equal to 0,
with probability 1 minus n
plus n squared when yn takes a
value of n, and this happens
with probability 1 over n.
The whole expression evaluates
to n, which blows up to
infinity as n going
to infinity.
As a result, the limit n going
to infinity of E of yn minus 0
squared is infinity and
is not equal to 0.
And there we have it, even
though yn converges to 0 in
probability, because the
variance of yn, in some sense,
is too big, it does
not converge in a
mean squared sense.
 
