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So now that we've seen LaGrange method

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let's take another look at minimum
variance portfolios,

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and just see what we have to do to solve
one of these optimization problems.

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So a minimum variance portfolio.

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This was the quantity I was calling risk,

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which is just the variance of my
portfolio.

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And I want to minimize my risk, so I

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want to minimize the variance, subject to
two conditions.

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One, that I hit my target rate of return.

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And two because this is sort of a more
realistic like there's

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no way around this condition.

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That my money and my portfolio has to all
be sort of used up.

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So the proportions of my portfolio, my
weights, have to all sum up to one.

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So to set this up.
For Lagrange method.

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The objective function, the f, the thing
that I'm trying to minimize

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or maximize, is going to be the variance.
The w transpose sigma w.

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And the constraints.

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So g is going to be a function of this
weights vector.

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So the weights of my portfolio.

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It's a vector-valued function with two
pieces.

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The first one is just this condition.

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So I've, here I haven't written them equal
to 0.

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So when I write this out in the, and the g
of x, I

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have to oh, I guess I swapped them too,
but that won't make any difference.

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So instead of saying expected the return
of the

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portfolio should be equal to my target
expected return I'm

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going to say the expected return of my
portfolio minus

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my target expected return has to be equal
to zero.

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And that the sum of the weights minus 1
has to be equal to 0.

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And then I'll check my necessary
condition, so If I take the gradient

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of this, I'm just going to end up with mew
transposed as the first row.

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And the second row being all one, and so
the only way I could get in to trouble is

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if I manage to pick n assets that all had
exactly the same expected return.

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So if one, the vector one, so one in every
element,

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that's proportional to any vector.
That has the same

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number for every element, so as long as I
have at least two distinct expected

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returns among my assets, then this matrix
is going to be full rank.

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It's going to have two pivots for two
rows.

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And I need a indermediate result.

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Which is something called the derivative
of a quadratic form.

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So I'm going to let A be a symmetric
matrix,

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so this is just a two by two example.

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So I have diagonal entries A and C,

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and then the off-diagonal elements, they
have to match.

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So, for a two by two case, there's only
one off-diagonal element, B.

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And so, so I guess it's two, but they have
to be the same.

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So the 1, 2 and the 2,1 element are the
same value.

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And then I'll define, f of x to be equal
to, so x is going

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to be a vector of, of two elements.
X transpose a x.

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And when I do that arithmetic what I end
up with is a x1 squared

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plus 2b x1x2 plus c x2 squared.
If I take the

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derivative of that, so the partial
derivatives

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of this, I end up getting 2 x transpose A.

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So you can kind of see what's, you know,

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it makes sense because this x transpose A
x, this is sort of the matrix equivalent

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of just, what a single variable function
would have as a x squared.

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If I take the derivative of a function
that's equal to a x squared, I get 2 a x.

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And so, that's pretty much what I've got
here, except I just

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have to respect the order of operations,
so that the, I get

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the right answer that I want to.
So the gradient has to be a row vector.

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So I need to do this as x transpose A so

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that the answer here is going to be a row
vector.

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And then I pick up this factor of 2
because I, it's

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sort of the same thing as taking a
derivative of x squared.

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[COUGH]

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And it turns out that in general if a is
an n by n symmetric

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matrix,so that's what makes this thing a
quadratic form.

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Is if I write this here.
So x is my vector.

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It's my, has my variables in it, x1 up to
xn.

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And a is a symmetric matrix.

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Then this thing here is called a quadratic
form.

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So the derivative of a quadratic form, so
that the gradient.

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Of, of the corresponding function here.

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You can just write that as two x transpose
a.

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So it works for the two by two case like
this, and if you actually

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did the same thing but for larger cases,
three by three or n by n.

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You'd be doing a whole lot of
calculations, but in

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the end you're always going to get two x
transpose a.

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That's it, okay.
So

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now let's look at the Lagrangian I have
for my, my minimum variance problem.

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So I put the variance here, so this is the
little f.

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Then I have lambda 1 times constraint 1
plus lambda 2 times constraint 2.

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And now I want to take the gradient of the
lagrangian, so I can do

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this again doing sort of the x part first
and then the lambda part second.

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So what I end up with is the gradient.
Of the objective function, so

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the little f.
Ans so that's, I've just mentioned how I

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can take the derivative of this quadratic
form so that's

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two w transpose sigma and then I'm going
to have lambda

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to one times the gradient of g, so the
gradient of g is

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going to just be.

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If I look at the, the gradient here it
will just be a vector of ones.

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So it's one times w, and I'm taking the

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derivatives with respect to each of these
w's, so each

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time only one of those terms is going to
pop out and it's always going to be a one.

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And then similarly, when I do the same
thing

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here, oh and I guess this one is constant.

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This mu p is constant, so they're going to
disappear when I take derivatives.

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And the same thing's going to happen here.

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When I take the gradient here, I'm just
going to

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get mu one, mu two, mu three in a row.

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Because everything else I, I'm taking the

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partial derivative with respect to a
different variable.

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So I can write that, just as lambda 1

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times e transpose plus lambda 2 times mu
transpose.

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And then, the partial derivatives with
respect to the lambda here.

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They are just going to be my constraints
again.

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So I end up with e transpose w minus 1,
and mu transpose w minus mu p.

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And so I guess we should check that all of
the dimensions work out here.

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So w W transposed.

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W's a column vector with n elements in it.

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And this is an n by n matrix, so when

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I take w transpose, that has dimensions 1
by n.

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Sigma has dimensions n by n, so when I do
that product, I'm going to get 1 by n.

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E transposed, this is this column vector
of m, sorry n

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ones, so when I take the transpose of that
I'm going to get.

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A vector here that has, it's a row vector
with n elements and

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similarly mu transposed is also going to
be a row vector with n elements.

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So I can do this sum, and I end up with a
row vector with n elements.

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And then these are each scalar values.

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So I have n elements here and then I have
two lambdas so I end up with two more.

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So m more terms here, but in this
particular case m is equal to 2.

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And I need to find values now of w.

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Lambda one and lambda two, that are going
to make this equal to 0.

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And so it turns out, you can write

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that problem down just as solving a linear
system.

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So I have 2 Sigma, so that's your 2 Sigma.
Times w, so this is just the

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transpose of this so it's kind of more
convenient to work, with this as a column

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vector when I want to solve it.
lambda 1 times e transpose.

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So that's going to be, if I just take the
transpose

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of that, lambda 1 a constant, so That'll
just become e.

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And then lambda 2 times u transposed.
So I end up with mu here.

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And now if I do the dot product of this
row and this column,

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I end up with, 2 sigma w e plus e lambda 1
plus mu.

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Lambda 2, which is just the transpose of
this here.

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So if I can make this thing equal to

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zero, then I'm also making this thing
equal to zero.

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Then I need e transposed w has to equal 1.

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So here I have e transposed times w.
Plus 0 times lambda

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00:10:25,291 --> 00:10:30,440
one plus 0 times lambda 2 equals 1.
So that gets

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me this constraint here and then u
transposed

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00:10:35,190 --> 00:10:39,940
times w, plus 0 times lambda 1 plus 0
times

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lambda 2 equals mu p.
So that gets me this constraint here.

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So it turns out in the, in the case of
this minimum variance portfolio, I'm

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able to find, there's going to just be one
critical point because this is a

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linear system, and I'm able to find that
just by solving this ax equals b

146
00:10:59,572 --> 00:11:04,740
problem that we worked on in the, in the
linear algebra section of the lectures.

147
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And so that's it for, for the lecture
today, there's still

148
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one more quiz, but I just wanted to
mention some further reading.

149
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So there is also a, a

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00:11:17,772 --> 00:11:21,648
reasonably complicated second order
condition for whether

151
00:11:21,648 --> 00:11:26,750
a point you find using Lagrange
multipliers is a maximum or a minimum.

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00:11:26,750 --> 00:11:33,638
And that's described in the course
textbook, so Theorem 9.2 and Corollary 9.1