WEBVTT

00:00:01.460 --> 00:00:03.880
For part E and F of the problem,
we'll be introducing

00:00:03.880 --> 00:00:07.160
a new notion of convergence,
so-called the convergence E

00:00:07.160 --> 00:00:09.260
mean squared sense.

00:00:09.260 --> 00:00:16.820
We say that xn converges to a
number c in mean squared, if

00:00:16.820 --> 00:00:24.090
as we take and go to infinity,
the expected value of xn minus

00:00:24.090 --> 00:00:29.200
c squared goes to 0.

00:00:29.200 --> 00:00:33.130
To get a sense of what this
looks like, let's say we let c

00:00:33.130 --> 00:00:38.740
equal to the expected value of
xn, and let's say the expected

00:00:38.740 --> 00:00:40.940
value of xn is always
the same.

00:00:40.940 --> 00:00:44.060
So the sequence of random
variables has the same mean.

00:00:44.060 --> 00:00:48.010
Well, if that is true, then mean
square convergence simply

00:00:48.010 --> 00:00:54.170
says the limit of the
variance of xn is 0.

00:00:54.170 --> 00:00:58.820
So as you can imagine, somehow
as xn becomes big, the

00:00:58.820 --> 00:01:03.130
variance of xn is very small,
so xn is basically highly

00:01:03.130 --> 00:01:05.300
concentrated around c.

00:01:05.300 --> 00:01:10.080
And by this I mean, the density
function for xn.

00:01:10.080 --> 00:01:11.400
So that's the notion
of convergence

00:01:11.400 --> 00:01:12.940
we'll be working with.

00:01:12.940 --> 00:01:16.440
Our first task here is to show
that the mean square

00:01:16.440 --> 00:01:20.000
convergence is in some sense
stronger than the convergence

00:01:20.000 --> 00:01:23.410
in probability that we have been
working with from part A

00:01:23.410 --> 00:01:28.290
to part D. That is, if I know
that xn converged to some

00:01:28.290 --> 00:01:33.620
number c in mean squared, then
this must imply that xn

00:01:33.620 --> 00:01:37.660
converges to c in probability.

00:01:37.660 --> 00:01:42.560
And now, we'll go show
that for part E.

00:01:42.560 --> 00:01:45.400
Well, let's start with a
definition of convergence in

00:01:45.400 --> 00:01:46.620
probability.

00:01:46.620 --> 00:01:49.990
We want to show that for a fixed
constant epsilon the

00:01:49.990 --> 00:01:55.650
probability that xn minus
c, greater than epsilon,

00:01:55.650 --> 00:02:00.990
essentially goes to 0 as
n goes to infinity.

00:02:00.990 --> 00:02:03.830
To do so, we look at the
value of this term.

00:02:03.830 --> 00:02:07.470
Well, the probability of
absolute value xn minus c

00:02:07.470 --> 00:02:11.700
greater than epsilon is equal
to the case if we were to

00:02:11.700 --> 00:02:15.150
square both sides of
the inequality.

00:02:15.150 --> 00:02:19.000
So that is equal to the
probability that xn minus c

00:02:19.000 --> 00:02:22.320
squared greater than
epsilon squared.

00:02:22.320 --> 00:02:26.170
We can do this because both
sides are positive, hence this

00:02:26.170 --> 00:02:28.590
goes through.

00:02:28.590 --> 00:02:33.630
Now, to bound this equality,
we'll invoke the Markov's

00:02:33.630 --> 00:02:46.970
Inequality, which it says this
probability of xn, some random

00:02:46.970 --> 00:02:51.420
variable greater than epsilon
squared, is no more than is

00:02:51.420 --> 00:02:56.950
less equal to the expected value
of the random variable.

00:02:56.950 --> 00:03:00.850
In this case, the expected value
of x minus c squared

00:03:00.850 --> 00:03:04.890
divided by the threshold that
we're trying to cross.

00:03:04.890 --> 00:03:08.530
So that is Markov's
Inequality.

00:03:08.530 --> 00:03:12.430
Now, since we know xn converges
to c in mean

00:03:12.430 --> 00:03:15.590
squared, and by definition,
mean square we know this

00:03:15.590 --> 00:03:19.710
precise expectation right
here goes to 0.

00:03:19.710 --> 00:03:25.040
And therefore, the whole
expression goes to 0 as n goes

00:03:25.040 --> 00:03:26.140
to infinity.

00:03:26.140 --> 00:03:30.380
Because the denominator here is
a constant and the top, the

00:03:30.380 --> 00:03:32.920
numerator here, goes to 0.

00:03:32.920 --> 00:03:34.190
So now we have it.

00:03:34.190 --> 00:03:39.760
We know that the probability of
xn minus c absolute value

00:03:39.760 --> 00:03:43.630
greater than epsilon goes to 0
as n goes to infinity, for all

00:03:43.630 --> 00:03:47.660
fixed value of epsilons and
this is the definition of

00:03:47.660 --> 00:03:48.910
convergence in probability.

00:03:55.310 --> 00:03:58.740
Now that we know if xn converges
to c mean squared,

00:03:58.740 --> 00:04:02.720
it implies that xn converges
to c in probability.

00:04:02.720 --> 00:04:06.160
One might wonder whether
the reverse is true.

00:04:06.160 --> 00:04:09.430
Namely, if we know something
converges in probability to a

00:04:09.430 --> 00:04:12.080
constant, does the same sequence
of random variables

00:04:12.080 --> 00:04:14.820
converge to the same constant
in mean squared?

00:04:14.820 --> 00:04:17.279
It turns out that is
not quite the case.

00:04:17.279 --> 00:04:19.980
The notion of probability
converges in probability is

00:04:19.980 --> 00:04:25.030
not as strong as a notion of
convergence in mean squared.

00:04:25.030 --> 00:04:27.810
Again, to look for a counter
example, we do not have to go

00:04:27.810 --> 00:04:31.410
further than the yn's we
have been working with.

00:04:31.410 --> 00:04:38.210
So here we know that yn
converges to 0 in probability.

00:04:38.210 --> 00:04:40.160
But it turns out it does
not converge to

00:04:40.160 --> 00:04:42.030
0 in the mean squared.

00:04:42.030 --> 00:04:45.520
And to see why this is the case,
we can take the expected

00:04:45.520 --> 00:04:50.820
value of yn minus 0 squared,
and see how that goes.

00:04:50.820 --> 00:04:54.780
Well, the value of this can be
computed easily, which is

00:04:54.780 --> 00:05:01.090
simply 0, if yn is equal to 0,
with probability 1 minus n

00:05:01.090 --> 00:05:06.800
plus n squared when yn takes a
value of n, and this happens

00:05:06.800 --> 00:05:09.140
with probability 1 over n.

00:05:09.140 --> 00:05:13.490
The whole expression evaluates
to n, which blows up to

00:05:13.490 --> 00:05:18.400
infinity as n going
to infinity.

00:05:18.400 --> 00:05:26.490
As a result, the limit n going
to infinity of E of yn minus 0

00:05:26.490 --> 00:05:32.200
squared is infinity and
is not equal to 0.

00:05:32.200 --> 00:05:35.580
And there we have it, even
though yn converges to 0 in

00:05:35.580 --> 00:05:39.290
probability, because the
variance of yn, in some sense,

00:05:39.290 --> 00:05:42.660
is too big, it does
not converge in a

00:05:42.660 --> 00:05:43.910
mean squared sense.