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In this lecture sequence,
we will do a lot

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with normal random variables.

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And for this reason,
it is useful to start

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with a simple observation
that will allow us

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later on to move much faster.

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Recall that a normal random
variable with a certain mean

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and variance has a PDF
of this particular form.

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What if somebody gives
you a PDF of this form

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and asks you what
it corresponds to?

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You can answer that
this is exactly

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of this form provided that you
make the identification that 3

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is equal to mu.

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So this is a normal PDF
with a mean equal to 3

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and whose variance
can also be found

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by matching this constant that
appears here with the number 8.

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This constant here is
in the denominator.

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So we have a term
1/2 sigma squared.

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This must be equal to 8.

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And from this, we can
infer that the variance

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of this random variable
is equal to 1/16.

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And if you also want to find out
the value of this constant c,

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you check the formula
for normal PDFs,

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the constant c is 1 over
the standard deviation,

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which is 1/4 in our case.

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The square root of this number
times the square root of 2 pi.

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Now suppose that somebody
gives you a PDF of this form.

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It's a constant times
a negative exponential

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of a quadratic function in x.

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We will argue that this
PDF is also a normal PDF.

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And identify the
parameters of that normal.

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First, let's start with
a certain observation.

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A PDF must integrate to 1 so
it cannot blow up as x goes

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to infinity, which means that
this exponential needs to die

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out as x goes to infinity.

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And that will only happen if
this coefficient alpha here

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is positive.

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So that we have e to the minus
something positive and large

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which is going to die out.

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Therefore, we must have alpha
being a positive constant.

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And let us assume from now
on that this is the case.

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What we will do
next is we will try

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to write this PDF in this form.

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And the trick that we're
going to use is the following.

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We will focus on the term
in the exponent, which

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we rewrite this way.

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We take out a factor of alpha.

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And then we will try to
make this expression here

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appear like a square of this
kind, like a perfect square.

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So what is involved
is a certain method,

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a certain trick called
completing the square.

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That is, we write this term here
in the form x plus something

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squared.

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And then we may need
some additional terms.

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What should that something be?

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We would like that
something be such

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that when we expand
this quadratic,

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we get this term and that term.

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Well we get an x
squared and then

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there's going to
be a cross term.

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What do we need here so that
the cross term is equal to this?

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What we need is a term
equal to beta over 2 alpha.

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Because in that
case, the cross term

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is going to be 2 times x
times beta divided by 2 alpha.

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The 2 in the beginning
and that 2 cancel out,

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so we're left with
x beta over alpha

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which is exactly
what we got here.

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However, this quadratic is
going to have an additional term

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which is going to be the square
of this, which is not present

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here.

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So to keep the two sides equal,
we need to subtract that term.

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And finally, we have the
last term involving gamma.

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Therefore, the PDF
of X is of the form.

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We have a certain
constant from here.

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Then, we have the negative
exponential of this term,

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e to the minus alpha x plus
beta over 2 alpha squared.

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And then there's the
negative exponential

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of the rest, which is going
to be a term of the form

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e to the minus alpha times beta
squared over 4 alpha squared

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plus gamma over alpha.

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Now this term here does
not involve any x's.

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So it can be absorbed
into the constant c.

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The dependence on x is
only through this term.

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And now this term looks exactly
like what we've got up here,

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provided that you make the
following identifications.

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Mu has to be equal
to what we have here,

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but here there's a minus sign,
here there's no minus sign.

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And so mu is going to
be the negative of what

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we have up here.

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It's minus beta over 2 alpha.

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And as for sigma
squared, we match

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and say that 1/2 sigma squared
must be equal to the constant

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that we have up
here, which is alpha.

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And from this, we conclude
that sigma squared is equal

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to 1/2 alpha.

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So we have concluded
that a PDF of this type

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is indeed a normal PDF.

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It has a mean equal
to that value.

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And a variance
equal to that value.

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Actually, to figure out what
the mean of this PDF is,

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we do not need to go
through this whole exercise.

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Once we're convinced that
this is a normal PDF,

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then we know that the mean is
equal to the peak of the PDF.

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To find the peak, we want
to maximize this over all

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x's, which is the same as
minimizing this quadratic

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function over all x's.

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Where is this quadratic
function minimized?

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To find that place, we
can look at the exponent,

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take its derivative,
and set it to 0.

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So setting the derivative
of the exponent to 0

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gives us the equation 2
alpha x plus beta equal to 0.

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And from this, we solve for x.

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And we can tell that the
peak of the distribution

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is going to be when x takes
this particular value.

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This value must also
be equal to the mean.

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So this is a very
useful fact to know.

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And we will use
it over and over.

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Negative exponential of
a quadratic function of x

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is always a normal PDF.