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