So now that we've seen LaGrange method
let's take another look at minimum
variance portfolios,
and just see what we have to do to solve
one of these optimization problems.
So a minimum variance portfolio.
This was the quantity I was calling risk,
which is just the variance of my
portfolio.
And I want to minimize my risk, so I
want to minimize the variance, subject to
two conditions.
One, that I hit my target rate of return.
And two because this is sort of a more
realistic like there's
no way around this condition.
That my money and my portfolio has to all
be sort of used up.
So the proportions of my portfolio, my
weights, have to all sum up to one.
So to set this up.
For Lagrange method.
The objective function, the f, the thing
that I'm trying to minimize
or maximize, is going to be the variance.
The w transpose sigma w.
And the constraints.
So g is going to be a function of this
weights vector.
So the weights of my portfolio.
It's a vector-valued function with two
pieces.
The first one is just this condition.
So I've, here I haven't written them equal
to 0.
So when I write this out in the, and the g
of x, I
have to oh, I guess I swapped them too,
but that won't make any difference.
So instead of saying expected the return
of the
portfolio should be equal to my target
expected return I'm
going to say the expected return of my
portfolio minus
my target expected return has to be equal
to zero.
And that the sum of the weights minus 1
has to be equal to 0.
And then I'll check my necessary
condition, so If I take the gradient
of this, I'm just going to end up with mew
transposed as the first row.
And the second row being all one, and so
the only way I could get in to trouble is
if I manage to pick n assets that all had
exactly the same expected return.
So if one, the vector one, so one in every
element,
that's proportional to any vector.
That has the same
number for every element, so as long as I
have at least two distinct expected
returns among my assets, then this matrix
is going to be full rank.
It's going to have two pivots for two
rows.
And I need a indermediate result.
Which is something called the derivative
of a quadratic form.
So I'm going to let A be a symmetric
matrix,
so this is just a two by two example.
So I have diagonal entries A and C,
and then the off-diagonal elements, they
have to match.
So, for a two by two case, there's only
one off-diagonal element, B.
And so, so I guess it's two, but they have
to be the same.
So the 1, 2 and the 2,1 element are the
same value.
And then I'll define, f of x to be equal
to, so x is going
to be a vector of, of two elements.
X transpose a x.
And when I do that arithmetic what I end
up with is a x1 squared
plus 2b x1x2 plus c x2 squared.
If I take the
derivative of that, so the partial
derivatives
of this, I end up getting 2 x transpose A.
So you can kind of see what's, you know,
it makes sense because this x transpose A
x, this is sort of the matrix equivalent
of just, what a single variable function
would have as a x squared.
If I take the derivative of a function
that's equal to a x squared, I get 2 a x.
And so, that's pretty much what I've got
here, except I just
have to respect the order of operations,
so that the, I get
the right answer that I want to.
So the gradient has to be a row vector.
So I need to do this as x transpose A so
that the answer here is going to be a row
vector.
And then I pick up this factor of 2
because I, it's
sort of the same thing as taking a
derivative of x squared.
[COUGH]
And it turns out that in general if a is
an n by n symmetric
matrix,so that's what makes this thing a
quadratic form.
Is if I write this here.
So x is my vector.
It's my, has my variables in it, x1 up to
xn.
And a is a symmetric matrix.
Then this thing here is called a quadratic
form.
So the derivative of a quadratic form, so
that the gradient.
Of, of the corresponding function here.
You can just write that as two x transpose
a.
So it works for the two by two case like
this, and if you actually
did the same thing but for larger cases,
three by three or n by n.
You'd be doing a whole lot of
calculations, but in
the end you're always going to get two x
transpose a.
That's it, okay.
So
now let's look at the Lagrangian I have
for my, my minimum variance problem.
So I put the variance here, so this is the
little f.
Then I have lambda 1 times constraint 1
plus lambda 2 times constraint 2.
And now I want to take the gradient of the
lagrangian, so I can do
this again doing sort of the x part first
and then the lambda part second.
So what I end up with is the gradient.
Of the objective function, so
the little f.
Ans so that's, I've just mentioned how I
can take the derivative of this quadratic
form so that's
two w transpose sigma and then I'm going
to have lambda
to one times the gradient of g, so the
gradient of g is
going to just be.
If I look at the, the gradient here it
will just be a vector of ones.
So it's one times w, and I'm taking the
derivatives with respect to each of these
w's, so each
time only one of those terms is going to
pop out and it's always going to be a one.
And then similarly, when I do the same
thing
here, oh and I guess this one is constant.
This mu p is constant, so they're going to
disappear when I take derivatives.
And the same thing's going to happen here.
When I take the gradient here, I'm just
going to
get mu one, mu two, mu three in a row.
Because everything else I, I'm taking the
partial derivative with respect to a
different variable.
So I can write that, just as lambda 1
times e transpose plus lambda 2 times mu
transpose.
And then, the partial derivatives with
respect to the lambda here.
They are just going to be my constraints
again.
So I end up with e transpose w minus 1,
and mu transpose w minus mu p.
And so I guess we should check that all of
the dimensions work out here.
So w W transposed.
W's a column vector with n elements in it.
And this is an n by n matrix, so when
I take w transpose, that has dimensions 1
by n.
Sigma has dimensions n by n, so when I do
that product, I'm going to get 1 by n.
E transposed, this is this column vector
of m, sorry n
ones, so when I take the transpose of that
I'm going to get.
A vector here that has, it's a row vector
with n elements and
similarly mu transposed is also going to
be a row vector with n elements.
So I can do this sum, and I end up with a
row vector with n elements.
And then these are each scalar values.
So I have n elements here and then I have
two lambdas so I end up with two more.
So m more terms here, but in this
particular case m is equal to 2.
And I need to find values now of w.
Lambda one and lambda two, that are going
to make this equal to 0.
And so it turns out, you can write
that problem down just as solving a linear
system.
So I have 2 Sigma, so that's your 2 Sigma.
Times w, so this is just the
transpose of this so it's kind of more
convenient to work, with this as a column
vector when I want to solve it.
lambda 1 times e transpose.
So that's going to be, if I just take the
transpose
of that, lambda 1 a constant, so That'll
just become e.
And then lambda 2 times u transposed.
So I end up with mu here.
And now if I do the dot product of this
row and this column,
I end up with, 2 sigma w e plus e lambda 1
plus mu.
Lambda 2, which is just the transpose of
this here.
So if I can make this thing equal to
zero, then I'm also making this thing
equal to zero.
Then I need e transposed w has to equal 1.
So here I have e transposed times w.
Plus 0 times lambda
one plus 0 times lambda 2 equals 1.
So that gets
me this constraint here and then u
transposed
times w, plus 0 times lambda 1 plus 0
times
lambda 2 equals mu p.
So that gets me this constraint here.
So it turns out in the, in the case of
this minimum variance portfolio, I'm
able to find, there's going to just be one
critical point because this is a
linear system, and I'm able to find that
just by solving this ax equals b
problem that we worked on in the, in the
linear algebra section of the lectures.
And so that's it for, for the lecture
today, there's still
one more quiz, but I just wanted to
mention some further reading.
So there is also a, a
reasonably complicated second order
condition for whether
a point you find using Lagrange
multipliers is a maximum or a minimum.
And that's described in the course
textbook, so Theorem 9.2 and Corollary 9.1