Quantitative methods for finance - Appunti completi (ENG)

Quantitative methods for finance – lecture notes

In the first module we are going to study pricing problems, more specifically pricing of assets with no-arbitrage opportunities. We will define what an arbitrage-free market is and the conditions that need to be satisfied.

There are two types of assets:

• Existing traded securities: liquid assets;
• Unreplicable assets: illiquid assets that cannot be traded in the market.

These pricing problems will be interesting if you’ll work in a trading desk of an investment bank (like Morgan Stanley). In the second part of the first module the theme is portfolio allocation problems (the optimal allocation of wealth) and the tools useful for building an optimal portfolio.

The second part of the course, held by another professor, will focus on equilibrium valuation of assets in continuous time. Excel (and MatLab) is not part of the exam but it’s useful.

Linear algebra (review)

1 × × ) 2 × 2

The ( identity matrix has 1s down the main diagonal and 0s elsewhere. The identity matrix, for example, is:

Remember that is the neutral element of the matrix multiplication. If we want to multiply two matrices we need to remember that multiplication is possible only with:

[ [× ] ∙ × ] = [ × ]

Let’s try an example:

We proceed rows by columns, so the calculus (for the first row of the final matrix) is:

(1 (0 (1 (0 (1 (0∙ 3) + ∙ 5) = 3; ∙ 4) + ∙ 2) = 4; ∙ 0) + ∙ 2) = 0

An identity matrix, multiplied for any conformable matrix, does not affect the multiplication.

1 An identity matrix is a diagonal matrix (with non 0 elements only on main diagonal) with all elements equal to 1.

Transposition and multiplication

To transpose a matrix ( ) we simply switch the two dimensions (rows and columns) so the first row becomes the first column, the second row becomes the second column, etc. A very important property of transposition is that if you transpose a matrix product you multiply the transposed matrices in reverse ( ∙ ) = ∙ order: . Let’s see an example:

As we can see, the result is identical.

Linear independence

1 2
[ ] [ ]

The vectors and are linearly independent if and only if (iff) the linear relationship:

2 3

Is true only for the trivial weights: = = 0

1 2

We need to multiply each vector to a scalar quantity unidentified for the moment. Then we impose this linear combination to be equal to 0. We can also write it as: + 2 = 0

1 2
{2 + 3 = 0
1 2 = = 0)

When this relation has only one solution that satisfies the condition ( then the two vectors are 1 2 linearly independent. Instead, if there are other solutions, the vectors are non-linearly independent (they are dependent). Stating that one of the vectors is a scaled version of the other is impossible:

−

We transport the second member to the right side and then we divide both sides for . Remember that 2 every number of the vector needs to be multiplied for the entire fraction.

−

Inverse of a matrix ( )

If a matrix is not square you cannot even talk about inversion. If you multiply the inverse of a matrix for the original one, you’ll obtain the identity matrix: −1 ∙ =

Linear independence and inverse of a matrix

−1

There is linear independence because there exists the inverse matrix of their coefficients ( ). If exists the inverse matrix then the vectors are linearly independent. In fact, if you premultiply both sides of the original condition with the inverse matrix you get the following expression:

−1 −1 ∙ ∙ = ∙ 0 ⇒ ∙ = 0 ℂ = 0
2×1

Let’s see an example:

That is equal to:

How to calculate the inverse of a matrix:

1−1 [()] = ×det ()det ()

Remember that need to be different from 0 to exist the inverse of a matrix (so it needs to be a non-singular matrix):

−1∃ ⇔ det () ≠ 0
2 × 2

Let’s try to invert a generic matrix on the notebook. Some key aspects:

• Cofactors matrix has the same size of the original matrix;
• () × The of a matrix is equal to its only element;
• × For a matrix there is a simple formula to calculate the determinant and this is: main diagonal multiplied the other diagonal, the second one with negative sign.

To calculate the determinant with the cofactors matrix (useful for every generic matrix) we simply need to choose one random row/column and multiply that for the correspondent row/column of the cofactors matrix (if there are some 0s, choose that row/column):

Do not forget the transposition of the matrix of cofactors. Inversion and transposition are interchangeable: −1 −1 ( ) ( )=

Let’s do the example as homework.

Linear independence, invertibility and full rank

× ⇔ × Now let’s consider this statement: the columns of an matrix are linearly independent the ⇔ × matrix is invertible the matrix has full rank (). The rank of a matrix is the highest order of non-singular square submatrices. The rank is the dimension of the largest possible square submatrix that you det () ≠ 0 3 × 3 can construct (non-singular so and so it’s invertible). Let’s practice on a matrix. To 3 × 3 compute the determinant of a matrix (only for them), you can use the Sarrus’ rule:

We are attaching the first two columns to the original matrix and calculating the determinant by diagonals.

1 2 3,
[ ] [ ] [ ]

Linear independence also applies to 3 vectors. The vectors and are linearly independent if and only if (iff) the linear relationship:

2 3 2
3 3 2

Is true only for the trivial weights: = = = 0

1 2 3

Stating that one of the vectors is a linear combination of the others is impossible:

3 1 2 = = [ ]

It’s not possible to find and to get so there’s linear independence (otherwise linear 2− −3 3 2 dependence).

Systems of linear equations (review)

= 3 = 3

Consider the following linear system in equations and unknowns:

+ 2 + 3 = 6
1 2 3
2 + 3 + 2 = −1
{ 1 2 3
3 + 3 + 2 = 0
1 2 3

∙ = )

We can rewrite it in matrix form ( or as a constrained linear combination:

Systems of linear equations (review)

The system admits a solution iff the vector can be seen as a linear combination of the vectors:

1∗ 2∗ 3∗

With proper weights , and . This can happen iff (Rouché-Capelli Theorem) the number of independent columns of equals the number of independent columns of which is obtained by adding () = (|) = 3. to its right the vector of constants. In other words, the rank of a matrix measures how many independent columns the matrix has. The rank of a matrix is the highest order of non-singular square submatrices:

det() = −5 ≠ 0

Moreover, so the rank is 3 (full rank). If these ranks are equal to each other, then there exists at least one solution to the system:

∗ ∗
() = (|) = ⇔ ∃ . . ∙ = :

Let’s call the rank

• = ∃! ; If (unknowns): then the system admits only one solution:
• − < : ∃∞ . If then the system admits an infinite number of solutions:

1∗ 2∗ 3∗

The rank cannot be bigger than the number of unknowns. We can work out the solution , and by means of the Cramer's Theorem (only if the system admits one solution):

Another method (the easiest in our situation) is using means of inversion (given non-singularity) because, in −1 −1 ∗ −1 ∙ ∙ = ∙ ⇓ = ∙ that way, we can write: simply premultiplying both sides for the inverse matrix. For example:

General resolution method for linear systems

Consider the following linear system of equations and unknowns (in matrix notation):

( × ) × ( × 1) = ( × 1)

() = (|) = , As we said, if the system admits solutions. The resolution scheme (particularly < < ) useful when and/or is:

• ̃, × : Within the matrix identify a non-singular square submatrix of dimension this is; typically the matrix you have identified to calculate the rank of
• ̃ − − Discard the rows of not belonging to : so you discard redundant equations;
• ̃ Keep on the left-hand side of the remaining equations of the system all terms belonging to : ̃ − while bring to the right-hand side all terms not belonging to . The unknowns brought to the right-hand side become free parameters.
• − × < , ∞ − Solve the resulting system: when the solutions will be a function of the ̃ unknowns that have been brought to the right-hand side. The matrix of coefficients will be .

= 3 = 4) , , Consider the following linear system ( and in the unknowns ( and ):

1 2 3 4

∙ = Let’s express it as (see the notebook) or we can also write is as:

The maximum rank of the matrix of the coefficients is 3 so if the system admits a solution, it won’t be > . unique because and we don’t have enough information. Let’s calculate the rank of The first three det() ≠ 0 → () = 3: vectors are independent so

Now let’s evaluate the rank of the complete matrix (it’s a three by five matrix that also contains the matrix of coefficients). It’s obvious that its rank is 3. The system admits solutions (Rouché-Capelli Theorem):

() = 3 = (|) = = 3
4−3, < ∃∞ .

Because of (one degree of freedom). To solve the system, we follow the previous ̃ scheme so we start from our matrix and we see that there are no additional rows (point 2). The fourth ̃ column is not loaded in so coefficients of will be brought to the right-hand side of the system, 4 considering them part of our vector of constants:

− + 2 = 1 +
1 2 3 4
̃
̃ 2 − − = −2
∙ ̃ = → { 1 2 3 4
− + 2 + 2 = 1 −
1 2 3 4

Exercise (part 2)

We can also write it as:

Now we can solve the system using means of inversion (given non-singularity) simply multiplying the ̃ ̃ ̃−1 ∙ ̃ = inverse matrix to both sides of this equation: . The solution (with one degree of freedom) can be written as:

We obtained a solution which depends on so we can attribute any value to in order to obtain a 4 4 1 2 4 = 0 = , = , = = 0. different solution. For example, if the solutions are and 4 1 2 3 4 3 9 9

Third example

We have a system in four equations and three unknowns (this system is at risk of not having solutions):

Let's state it as follows:

4 × 4

The maximum rank of is 3 (it’s a three by four matrix). The complete matrix is a so the maximum rank could be four. If the complete matrix is a non-singular, then the system has no solution. We create the ̃ matrix and we evaluate its rank. The first three vectors are independent because:

2 7 1
̃ det(̃)= → = −99 ≠ 0 → () = 3
[ ]
5 6 8
0 9 0 |.

Now we evaluate the rank of the complete matrix The last vector linearly depends on the first three vectors as:

Because of this determinant is equal to zero (to calculate use cofactor method with only third row because () = (|) = = 3 there are zeros), obviously so the system admits solutions, specifically we ̃ = → ∃! . have that To solve the system we know that the submatrix doesn’t load on the fourth row so we can delete that equation.

Third example (part 2)

So we have that:

2 7 1 151
̃
̃ ∙ ̃ = → ∙ =
[ ] [ ] [ ]
5 6 8 320
9 0 273

We can work out the solution by focusing on these equations and using means of inversion (given non-singularity):

̃ The same result is obtained if we chose another matrix. To conclude, the last equation is met by construction:

On the notebook we have another example of a complete and useful exercise.

One-period financial markets

We are going to work in discrete time where the evolution of time will be described by time periods of finite length (one day, one month, etc.). In one-period financial markets there is only one period (initial date and terminal date ). The uncertainty will be described through a finite number of possible 0 1 scenarios. is the moment just before a decision (fed funds level decision for example) and is just after 0 1 ±= 25 it. We have different scenarios (like bp) and we don’t know what is going to happen.

Timing (notations)

= 0 = 1 = 0,

Investors face two trading dates only, (current time) and (future time). At time investors = 1 choose their investment strategy to allocate money in available assets. At time they receive the liquidation value of their strategy (they close the position). The assets available (investor can choose among + 1 = 0, … , .them) at time 0 are securities for which we use the index with

The riskless security

= 0 = 1, … ,

There is one risk-free asset corresponding to and risky assets corresponding to (that’s + 1). = 1 why the total is For the risk-free asset we already know the value at because there is no (0) ≡ 1 uncertainty. denotes the time-0 price of the riskless security (where stands for bond issued (1) ≡ 1 + from a reliable institution not subject to default). denotes its time-1 price. The quantity is ≥ 0. the risk-free rate, with The return of the riskless asset is:

(1) − (0) 1 + − 1
(1) = = =
(0) 1

The bond does not pay coupons.

The risky securities

̃ (1) > 0, (0)

For the securities with denotes their time-0 price (where stands for stock). is a = 1, … , . = 0 random variable that denotes their time-1 price, with At we don’t know what is going to happen so the return on the risky security is:

̃ (1) − (0)
̃ (1) =
(0)

This risky security doesn’t pay dividend, so its return is only given by capital gain. We use “Tilda” because it’s a random variable.

Uncertainty

We assume that there are possible states of the world at time 1 so by time 1 the market uncertainty will -th resolve in one of possible scenarios. indicates the generic state of the world at time 1. The ’s Ω are relevant economic/financial scenarios. indicates the set of all states of the world:

Ω = { , … , }
1

(1)( ) -th indicates the time-1 price of the security in scenario . We are going to represent all × 1): possible prices for a given security in a vector (

(1)( ) 1
̃ …
(1) = [ ]
(1)( )

The payoff matrix

+ 1 + 1

The payoff matrix has rows and columns. Each column of represents the cashflows -th = 0,1, … , ): from buying (row 1) or having bought (the other rows) 1 unit of the security ( + 1 , The columns represent the assets (so there are columns, the first column is the second one is 1 and the last is ). In the dimension of the rows, we need do describe both time and uncertainty dimensions. Information on the first row corresponds to time 0 and time 1 on the other rows. In the other rows we have the information corresponding to all possible states. In the second row we have and in 1 the last we have . Inside the matrix we basically have cashflows assuming buy one unit of the asset. At time one when you are liquidating the position, you are going to receive a positive cashflow. In the state 1 we jointly know all the prices of the assets so if this state realizes we know all of these. So, the matrix represents also their joint distributions and give information on their correlation. Other important ̃ (0) (1).notations are for the first row of the matrix and the remaining rows are

Exemplifying a market with 3 securities and 3 states

= 2 = 3. (1) = 1.00 = 0%.

Consider a one-period market with and Assume that so that The time- (0) (0) = 0.5 = 2.5.0 prices of the risky securities are and The time-1 prices of the risky securities are:

1 20 0
̃ ̃