Program Flow and Logicals

Preliminaries

Very often in your life you have to repeat the same operation many times (move your right and left leg sequentially while walking/running) or behave differently depending on external conditions (there is or there isn’t a bus on the bus stop). Quite often these two are combined together. Say, if there is a bus on the bus stop, then you run trying to catch it, otherwise walk or stop and enjoy the usual Manchester weather. The same is true for programming. Quite often you want to repeat the same operation many times, or you want to change the way you process your data depending on some conditions. We start with conditional statements. They execute different pieces of the code depending on whether condition is true or false. There are several ways you can formulate it. The shortest

if condition
statement1;
statement2;
...
end

, executes statement1;statement2,... only if condition is satisfied. condition can be anything that generate non-zero or 0 (True or False), say i>0, size(y,1)\sim=40, or 5-i. The last condition is True always but for i=5. A slightly longer version

  if condition
  statement1;
  statement2;
  ...
  else
  statement1a;
  statement2a;
  ...
  end

runs statement1;statement2;..., if condition is true and statement1a;statement2a;... otherwise. The most general specification is

  if condition1
  statement1;
  statement2;
  ...
  elseif condition2
  statement1a;
  statement2a;
  ...
  ...
  ...
  elseif conditionN
  statement1b;
  statement2b;
  ...
  else
  statement1c;
  statement2c;
  ...
  end

In this case, however, you have to ensure that condition1, condition2, …, conditionN are mutually disjoint. As an example, you might think about different actions depending on your final grade. Condition1: grade<30; Condition 2: (grade>=30)&&(grade<40); Condition 3: (grade>=40)&&(grade<50); etc.

MATLAB has two statements that create a loop. First, it is unconditional loop:

for CounterVariable=[range of values]
statement1;
statement2;
...
end

It repeats at most as many times as many elements it has in the [range of values]. If the range of values is empty, this loop does not run. Say, if you define a range 10:1, MATLAB creates an empty range. Thus, this loop will not be executed. If you define a range 1:3:10, MATLAB creates a range of four values [1 4 7 10], and the loop runs four times. During the first iteration, CounterVariable=1, during the second CounterVariable=4, etc. After the end of the loop CounterVariable=10. Please note, it is very unwise to modify the counter inside the loop. All modifications will disappear after the next iteration. Please also note, that the values in the range could be anything, including filenames or matrices from a cell vector. There are two commands that can modify the execution of the loop. continue breaks the current iteration of the loop. Once it is executed, the loop continues skipping current iteration. break stops the execution of the loop and your program continues after this point. These commands are used inside if statements. For example, if CounterVariable == 10 continue;end skips the loop iteration for CounterVariable = 10.

Second, it is the conditional loop

while condition
statement1;
statement2;
...
end

This version of the loop executes statements as long as condition is true. If condition is always true, your loop runs forever.

for ... end loop

A standard application for for ... end loop is the reconstruction of AR(p) series once AR(p) coefficients and the vector of error terms is known. [math]y_t=\phi_0+\sum_{i=1}^p \phi_i y_{t-i}+e_t.[/math] For simplicity, we assume that [math]p=1[/math]. Also, to be able to compute [math]y_1[/math], we need to provide [math]y_0[/math]. Since we don’t know [math]y_0[/math],the best guess for [math]y_0[/math] is [math]E(y_0)[/math]. For stationary AR(1) process, that is for the case [math]|\phi_1|\lt 1[/math], [math]E(y_0)=\phi_0/(1-\phi_1)[/math]. Thus, knowing [math]y_0[/math] and [math]e_t[/math] for [math]t=1,\ldots,T[/math], we can reconstruct [math]y_t,\ t=1\ldots,T[/math]:

[math]\begin{aligned} y_1=&\phi_0+\phi_1 y_0+e_1\\ y_2=&\phi_0+\phi_1 y_1+e_2\\ &\ldots\\ y_t=&\phi_0+\phi_1 y_{t-1}+e_t\\ &\ldots\\ y_T=&\phi_0+\phi_1 y_{T-1}+e_T\end{aligned}[/math]

Definitely, if you are patient enough and [math]T[/math] is not very large, you can create your m file with [math]T[/math] lines in it. However, once [math]T[/math] is unknown, this approach would not work. Fortunately, there is a better alternative for this type of operations. All these computations can be summarized using the following algorithm, assuming vector e is already created:

1. Find the length of a vector of error terms e: T=size(e,1)
2. Initialize a vector y of the same length as vector e: y=zeros(T,1)
3. Compute y(1)=phi0+phi1*(phi0/(1-phi1))+e(1). Please remember, we assume that [math]y_0=E(y)=\phi_0/(1-\phi_1)[/math]
4. Compute y(i)=phi0+phi1*y(i-1)+e(i) for [math]i=2[/math]
5. Repeat line 4 for [math]i=3,...,T[/math]

When vector e is known in advance (for the sake of certainty, let e=randn(1000,1)), the MATLAB code is

  
  T=size(e,1);
  y=zeros(T,1);
  y0=phi0/(1-phi1);
  y(1)=phi0+phi1*y0+e(1);
  for i=2:T
    y(i)=phi0+phi1*y(i-1)+e(i);
  end

However, if phi1=1, [math]E(y_t)[/math] is not constant, then the formula we use in the code does not work and will create either a series y of [math]\pm\infty[/math], if phi0 \ne 0 or a series of not a numbers NaN, if phi0=0.

if else end or if end

To avoid these inconveniences, we have to consider separately two cases:

1. AR(1) process is stationary, i.e. [math]|\phi_1|\lt 1[/math]
2. AR(1) process is nonstationary, i.e.  [math]|\phi_1|\ge 1[/math]

For the latter, we have to acknowledge the fact that [math]E(y_t)=\mu_t[/math], i.e. unconditional expectation is a function of time. In this case we have to set [math]E(y_0)[/math] to some value. A standard assumption for non-stationary series is to assume that [math]E(y_0)=0[/math].

The algorithm in this case would look like:

1. Find a length of a vector of error terms e: T=size(e,1)
2. Initialize a vector y of the same length as vector e: y=zeros(T,1)
3. Check whether abs(phi1)<1. If this statement is true, then y0=phi0/(1-phi1). Else, y0=0. Please remember, we set [math]y_0=E(y_0)[/math].
4. Compute y(1)=phi0+phi1*y0+e(1).
5. Compute y(i)=phi0+phi1*y(i-1)+e(i) for [math]i=2[/math]
6. Repeat line 5 for [math]i=3,...,T[/math]

Assuming vector e is known in advance (for the sake of certainty, let e=randn(1000,1)), the MATLAB code is

  T=size(e,1);
  y=zeros(T,1);
  if abs(phi1)<1
  y0=phi0/(1-phi1);
  else
  y0=0;
  end
  y(1)=phi0+phi1*y0+e(1)
  for i=2:T
    y(i)=phi0+phi1*y(i-1)+e(i);
  end

If you don’t like the word else, you can skip it:

  T=size(e,1);
  y=zeros(T,1);
  y0=0;
  if abs(phi1)<1
  y0=phi0/(1-phi1);
  end
  y(1)=phi0+phi1*y0+e(1)
  for i=2:T
    y(i)=phi0+phi1*y(i-1)+e(i);
  end

while end loop

An alternative way of running the same code is to use a conditional loop (purely for demonstration purposes). Usually the conditional loop is used when the number of iterations is not known in advance.

1. Find the length of a vector of error terms e: T=size(e,1)
2. Initialize a vector y of the same length as vector e: y=zeros(T,1)
3. Check whether abs(phi1)<1. If this statement is true, then y0=phi0/(1-phi1). Else, y0=0. Please remember, we set [math]y_0=E(y_0)[/math].
4. Set $i=1$.
5. Compute y(i)=phi0+phi1*y0+e(i).
6. Increase $i$ by one, i.e. $i=i+1$ (please note, in programming this is a valid statement)
7. Check whether [math]i\lt =T[/math]. If yes, proceed. Else go to line 11.
8. Compute y(i)=phi0+phi1*y(i-1)+e(i)
9. Increase i by 1, i.e. [math]i=i+1[/math]
10. Repeat lines 7 $-$ 9
11. ... (continuing the script)

Assuming vector e is known in advance (for the sake of certainty, let e=randn(1000,1)), the MATLAB code is

  T=size(e,1);
  y=zeros(T,1);
  if abs(phi1)<1
    y0=phi0/(1-phi1);
  else
    y0=0;
  end
  y(1)=phi0+phi1*y0+e(1)
  i=2;
  while i<=T
    y(i)=phi0+phi1*y(i-1)+e(i);
    i=i+1;
  end

Imperfect substitutes of the above

MATLAB has two powerful tools that make programmer’s life much easier and utilization of loops/if less frequent. In addition, quite often it makes the code run faster. In particular,

1. Logical expressions work not only on scalars, but also on vectors, matrices and, in general, on n-dimensional arrays.
2. Subvectors/submatrices can be extracted using logical 0-1 arrays.

Irrelevant but useful example

typing a=1:5 in MATLAB command window creates a [math]1\times5[/math] row-vector a with values [math][1\ 2\ 3\ 4\ 5][/math]. Logical expression ind=(a>3.5) will create a so called logical vector ind with values [math][0\ 0\ 0\ 1\ 1][/math], i.e. it is 1 if the according element is greater than 3.5 and 0 otherwise. Now, typing b=a(ind) will generate a [math]2\times1[/math] subvector b with values [math][4\ 5][/math]. You can also create some vectors or matrices with specific values changed: the command a(ind)=a(ind)*2 replace the last two values of the original vector a. As a result, the vector a becomes [math][1 \ 2\ 3\ 8\ 10][/math].

Slightly less irrelevant example

In some occasions you would like to modify the matrix of interest. Say, in some surveys “no answer” is coded as 999. Once you import the whole dataset in X, you might want to replace these with, say, NaN. It can be done for the whole matrix of interest: X(X==999)=NaN.

Relevant example

To demonstrate these capabilities in a more relevant environment, let’s run a very simple example. Assume that we have [math]T\times1[/math] vector of returns r and we want to

1. Compute number of positive, negative and zero returns
2. Compute means of positive and negative returns

The algorithm for this is quite straightforward:

1. Find out the length of vector r, T
2. Initiate three counter variables, Tplus=0, Tzero=0, Tminus=0, and vectors rplus=zeros(T,1), rminus=zeros(T,1) (since we don’t know how many negative and positive returns we will observe
3. Check whether r(i) is greater, smaller or equal to 0 for i=1
4. If r(i)>0, add 1 to Tplus, set rplus(Tplus)=r(i);
5. Else if r(i)<0 add 1 to Tminus, set rminus(Tminus)=r(i);
6. Else add 1 to Tzero
7. Repeat 3-6 for [math]i=2,\ldots,T[/math]
8. Remove excessive zeros from rplus and rminus: rplus=rplus(1:Tplus);
   

rminus=rminus(1:Tminus);

1. Compute means of rminus and rplus. Number of positive, negative and zero returns are stored in Tplus,Tminus,Tzero

MATLAB translation:

T=size(r,1);
Tplus=0;Tminus=0;Tzero=0;
rplus=zeros(T,1);rminus=zeros(T,1);
for i=1:T
    if r(i)>0
        Tplus=Tplus+1;%increasing Tplus by one if return is positive
        rplus(Tplus)=r(i);%storing positive return in the proper subvector
    elseif r(i)<0
        Tminus=Tminus+1;%increasing Tminus by one if return is negative
        rminus(Tminus)=r(i);%storing negative return in the proper subvector
    else
        Tzero=Tzero+1;%increasing Tzero by one if return is neither positive nor negative
    end
end
rplus=rplus(1:Tplus);%removing excessive zeros from a subvector of positive returns
rminus=rminus(1:Tminus);%removing excessive zeros from a subvector of negative returns
meanplus=mean(rplus);%computing mean of positive returns
meanminus=sum(rminus)/Tminus;%computing mean of negative returns

Using MATLAB capabilities mentioned in this section, the algorithm can be reduced to:

1. Construct a vector indplus that has 1 for positive returns and 0 for negative returns
2. Construct a vector indminus that has 1 for negative returns and 0 for positive returns
3. Assign to Tplus a sum of elements of indplus. This is the number of positive returns
4. Assign to Tminus a sum of elements of indminus. This is the number of negative returns
5. Compute Tzero which is T-Tplus-Tminus
6. Construct a vector of positive returns rplus=r(indplus) and compute its mean
7. Construct a vector of negative returns rminus=r(indminus) and compute its mean

MATLAB implementation:

  T=size(r,1);
  indplus  = r>0;%constructing an indicator vector with 1s if r(i)>0, 0 otherwise
  indminus = r<0;%constructing an indicator vector with 1s if r(i)<0, 0 otherwise
  Tplus=sum(indplus);%computing a number of positive returns
  Tminus=sum(indminus);%computing a number of negative returns
  Tzero=T-Tplus-Tminus;%computing a number of zero returns
  rplus=r(indplus);%constructing a vector of positive returns
  rminus=r(indminus);%constructing a vector of negative returns
  meanplus=sum(rplus)/Tplus; %computing mean of positive returns
  meanminus=mean(rminus); %computing mean of negative returns

Or, a slightly shorter version of the same thing

  T=size(r,1);
  rplus  = r(r>0);%constructing a vector of positive returns
  rminus = r(r<0);%%constructing a vector of negative returns
  Tplus=size(rplus,1);%computing a number of positive returns
  Tminus=size(rminus,1);%computing a number of negative returns
  Tzero=T-Tplus-Tminus;%computing a number of zero returns
  meanplus=sum(rplus)/Tplus; %computing mean of positive returns
  meanminus=mean(rminus); %computing mean of negative returns

A shorter code is less exposed to errors and easier to read (at least after some practice).