function based of the first n elements of a matrix

I have a matrix, A, with 40000 binomial random elements. I am trying to complete the following code and would appreciate help:

I need to create 40000 arrays X_n, where X_n represents the array containing the first n elements of the matrix A for each n:[1,40000]

then for each X_n I need to calculate the value of the function, O_n = (the number of elements of X_n that are less than 25)/n

(This is all based around trying to approximate the proportion of overbookings of a flight. The matrix A is 40000 simulations of my situation.)

[Cedric]

It is unclear to me. So X_1 will have one element, X_2 two, and so on?

[Dst]

Yes. X_1 represents the array that contains the first element of A. X_2 represents the array with the first two elements of A and so on.

Answer 1

[Cedric]

Based on what I understand, i.e. that you don't really need to build the X_n arrays, but just the O_n results, I propose that you compute directly O_n for all n. Here is a demo with a 10x1 numeric array A (for sake of simplicity), that we populate with random integers between 1 and e.g. 60:

 >> A = randi( 60, 10, 1 )
 A =
     8
    30
    58
    21
    36
    14
    46
    16
    31
    42

We start by creating a vector of logicals testing for the conditon "is less than 25" ( <25) :

 >> results = A < 25
 results =

     1
     0
     0
     1
     0
     1
     0
     1
     0
     0

We perform a cumulative sum of these 1's taken as integers, and we get, for each index, the count of elements lower than 25 from the first element to this index:

 >> results = cumsum( results )
 results =
     1
     1
     1
     2
     2
     3
     3
     4
     4
     4

Finally, we divide each count by its index ( n in your statement):

 >> results = results ./ (1:numel(A)).'
 results =
    1.0000
    0.5000
    0.3333
    0.5000
    0.4000
    0.5000
    0.4286
    0.5000
    0.4444
    0.4000

At this stage, results is a vector of what the function O_n of your statement would output for each n.

Putting everything together, you get the following one liner:

 >> results = cumsum( A < 25 ) ./ (1:numel( A )).'
 results =
    1.0000
    0.5000
    0.3333
    0.5000
    0.4000
    0.5000
    0.4286
    0.5000
    0.4444
    0.4000