WEBVTT
00:00:00.760 --> 00:00:03.020
In this lecture sequence,
we will do a lot
00:00:03.020 --> 00:00:05.130
with normal random variables.
00:00:05.130 --> 00:00:07.460
And for this reason,
it is useful to start
00:00:07.460 --> 00:00:10.140
with a simple observation
that will allow us
00:00:10.140 --> 00:00:12.890
later on to move much faster.
00:00:12.890 --> 00:00:16.090
Recall that a normal random
variable with a certain mean
00:00:16.090 --> 00:00:19.160
and variance has a PDF
of this particular form.
00:00:19.160 --> 00:00:22.040
What if somebody gives
you a PDF of this form
00:00:22.040 --> 00:00:24.220
and asks you what
it corresponds to?
00:00:24.220 --> 00:00:27.180
You can answer that
this is exactly
00:00:27.180 --> 00:00:32.110
of this form provided that you
make the identification that 3
00:00:32.110 --> 00:00:34.140
is equal to mu.
00:00:34.140 --> 00:00:39.210
So this is a normal PDF
with a mean equal to 3
00:00:39.210 --> 00:00:41.860
and whose variance
can also be found
00:00:41.860 --> 00:00:45.820
by matching this constant that
appears here with the number 8.
00:00:45.820 --> 00:00:48.470
This constant here is
in the denominator.
00:00:48.470 --> 00:00:51.140
So we have a term
1/2 sigma squared.
00:00:51.140 --> 00:00:53.110
This must be equal to 8.
00:00:53.110 --> 00:00:56.690
And from this, we can
infer that the variance
00:00:56.690 --> 00:01:00.690
of this random variable
is equal to 1/16.
00:01:00.690 --> 00:01:04.220
And if you also want to find out
the value of this constant c,
00:01:04.220 --> 00:01:07.160
you check the formula
for normal PDFs,
00:01:07.160 --> 00:01:11.370
the constant c is 1 over
the standard deviation,
00:01:11.370 --> 00:01:13.050
which is 1/4 in our case.
00:01:13.050 --> 00:01:16.940
The square root of this number
times the square root of 2 pi.
00:01:19.600 --> 00:01:23.850
Now suppose that somebody
gives you a PDF of this form.
00:01:23.850 --> 00:01:26.140
It's a constant times
a negative exponential
00:01:26.140 --> 00:01:28.850
of a quadratic function in x.
00:01:28.850 --> 00:01:33.430
We will argue that this
PDF is also a normal PDF.
00:01:33.430 --> 00:01:36.700
And identify the
parameters of that normal.
00:01:36.700 --> 00:01:39.200
First, let's start with
a certain observation.
00:01:39.200 --> 00:01:44.460
A PDF must integrate to 1 so
it cannot blow up as x goes
00:01:44.460 --> 00:01:49.500
to infinity, which means that
this exponential needs to die
00:01:49.500 --> 00:01:52.190
out as x goes to infinity.
00:01:52.190 --> 00:01:55.630
And that will only happen if
this coefficient alpha here
00:01:55.630 --> 00:01:56.830
is positive.
00:01:56.830 --> 00:02:00.300
So that we have e to the minus
something positive and large
00:02:00.300 --> 00:02:02.180
which is going to die out.
00:02:02.180 --> 00:02:07.280
Therefore, we must have alpha
being a positive constant.
00:02:07.280 --> 00:02:11.190
And let us assume from now
on that this is the case.
00:02:11.190 --> 00:02:12.870
What we will do
next is we will try
00:02:12.870 --> 00:02:16.220
to write this PDF in this form.
00:02:16.220 --> 00:02:20.310
And the trick that we're
going to use is the following.
00:02:20.310 --> 00:02:26.070
We will focus on the term
in the exponent, which
00:02:26.070 --> 00:02:28.180
we rewrite this way.
00:02:28.180 --> 00:02:30.185
We take out a factor of alpha.
00:02:37.500 --> 00:02:42.470
And then we will try to
make this expression here
00:02:42.470 --> 00:02:47.340
appear like a square of this
kind, like a perfect square.
00:02:47.340 --> 00:02:50.480
So what is involved
is a certain method,
00:02:50.480 --> 00:02:54.250
a certain trick called
completing the square.
00:02:54.250 --> 00:03:00.110
That is, we write this term here
in the form x plus something
00:03:00.110 --> 00:03:01.430
squared.
00:03:01.430 --> 00:03:04.310
And then we may need
some additional terms.
00:03:04.310 --> 00:03:06.830
What should that something be?
00:03:06.830 --> 00:03:08.900
We would like that
something be such
00:03:08.900 --> 00:03:11.830
that when we expand
this quadratic,
00:03:11.830 --> 00:03:14.420
we get this term and that term.
00:03:14.420 --> 00:03:16.470
Well we get an x
squared and then
00:03:16.470 --> 00:03:18.370
there's going to
be a cross term.
00:03:18.370 --> 00:03:22.410
What do we need here so that
the cross term is equal to this?
00:03:22.410 --> 00:03:29.250
What we need is a term
equal to beta over 2 alpha.
00:03:29.250 --> 00:03:31.970
Because in that
case, the cross term
00:03:31.970 --> 00:03:37.700
is going to be 2 times x
times beta divided by 2 alpha.
00:03:37.700 --> 00:03:40.930
The 2 in the beginning
and that 2 cancel out,
00:03:40.930 --> 00:03:43.320
so we're left with
x beta over alpha
00:03:43.320 --> 00:03:45.600
which is exactly
what we got here.
00:03:45.600 --> 00:03:48.480
However, this quadratic is
going to have an additional term
00:03:48.480 --> 00:03:51.240
which is going to be the square
of this, which is not present
00:03:51.240 --> 00:03:51.880
here.
00:03:51.880 --> 00:03:56.250
So to keep the two sides equal,
we need to subtract that term.
00:04:00.780 --> 00:04:04.035
And finally, we have the
last term involving gamma.
00:04:07.440 --> 00:04:12.780
Therefore, the PDF
of X is of the form.
00:04:12.780 --> 00:04:16.060
We have a certain
constant from here.
00:04:16.060 --> 00:04:22.780
Then, we have the negative
exponential of this term,
00:04:22.780 --> 00:04:31.720
e to the minus alpha x plus
beta over 2 alpha squared.
00:04:31.720 --> 00:04:34.150
And then there's the
negative exponential
00:04:34.150 --> 00:04:39.700
of the rest, which is going
to be a term of the form
00:04:39.700 --> 00:04:49.050
e to the minus alpha times beta
squared over 4 alpha squared
00:04:49.050 --> 00:04:53.510
plus gamma over alpha.
00:04:53.510 --> 00:04:57.540
Now this term here does
not involve any x's.
00:04:57.540 --> 00:05:01.550
So it can be absorbed
into the constant c.
00:05:01.550 --> 00:05:05.420
The dependence on x is
only through this term.
00:05:05.420 --> 00:05:10.430
And now this term looks exactly
like what we've got up here,
00:05:10.430 --> 00:05:14.030
provided that you make the
following identifications.
00:05:14.030 --> 00:05:16.690
Mu has to be equal
to what we have here,
00:05:16.690 --> 00:05:19.920
but here there's a minus sign,
here there's no minus sign.
00:05:19.920 --> 00:05:22.540
And so mu is going to
be the negative of what
00:05:22.540 --> 00:05:24.010
we have up here.
00:05:24.010 --> 00:05:28.420
It's minus beta over 2 alpha.
00:05:28.420 --> 00:05:31.370
And as for sigma
squared, we match
00:05:31.370 --> 00:05:39.080
and say that 1/2 sigma squared
must be equal to the constant
00:05:39.080 --> 00:05:41.700
that we have up
here, which is alpha.
00:05:41.700 --> 00:05:47.300
And from this, we conclude
that sigma squared is equal
00:05:47.300 --> 00:05:49.020
to 1/2 alpha.
00:05:51.909 --> 00:05:55.340
So we have concluded
that a PDF of this type
00:05:55.340 --> 00:05:57.890
is indeed a normal PDF.
00:05:57.890 --> 00:06:01.090
It has a mean equal
to that value.
00:06:01.090 --> 00:06:03.860
And a variance
equal to that value.
00:06:03.860 --> 00:06:07.320
Actually, to figure out what
the mean of this PDF is,
00:06:07.320 --> 00:06:10.650
we do not need to go
through this whole exercise.
00:06:10.650 --> 00:06:14.440
Once we're convinced that
this is a normal PDF,
00:06:14.440 --> 00:06:19.750
then we know that the mean is
equal to the peak of the PDF.
00:06:19.750 --> 00:06:23.700
To find the peak, we want
to maximize this over all
00:06:23.700 --> 00:06:27.450
x's, which is the same as
minimizing this quadratic
00:06:27.450 --> 00:06:29.680
function over all x's.
00:06:29.680 --> 00:06:32.610
Where is this quadratic
function minimized?
00:06:32.610 --> 00:06:35.830
To find that place, we
can look at the exponent,
00:06:35.830 --> 00:06:39.570
take its derivative,
and set it to 0.
00:06:39.570 --> 00:06:42.350
So setting the derivative
of the exponent to 0
00:06:42.350 --> 00:06:48.680
gives us the equation 2
alpha x plus beta equal to 0.
00:06:48.680 --> 00:06:51.870
And from this, we solve for x.
00:06:51.870 --> 00:06:56.210
And we can tell that the
peak of the distribution
00:06:56.210 --> 00:07:00.020
is going to be when x takes
this particular value.
00:07:00.020 --> 00:07:03.240
This value must also
be equal to the mean.
00:07:03.240 --> 00:07:06.090
So this is a very
useful fact to know.
00:07:06.090 --> 00:07:08.680
And we will use
it over and over.
00:07:08.680 --> 00:07:11.940
Negative exponential of
a quadratic function of x
00:07:11.940 --> 00:07:14.840
is always a normal PDF.