This was an exercise I completed in the second half of 2017, as a building block for a bigger project, which ultimately failed to bear fruit.
My intention at the time had been to build a viable model for predicting asset prices, which I would then use to provide a discrete probability distribution for a new type of option pricing model that would use a discrete set of trading prices as opposed to the currently dominant Black-Scholes-Merton model, which assumes a log-normal distribution for prices and hence uses a continuous set of trading prices.
I began by attempting to construct a geometric pricing model – a model that would allow one to move randomly from an initial price to the initial price multiplied by a randomly selected factor from a given set and thence to proceed in a similar fashion, using the initial set of factors, for as long as one pleased. I discovered, of course, that that geometric model was excluding perfectly valid trading prices in quite a significant way.
In light of the issues I had been having with geometric pricing models (both bounded [i.e. using a fixed set of specific random factors] and unbounded), I switched to looking instead at a linear pricing model – a model that would allow one to move randomly from an initial price to the initial price plus or minus a randomly selected factor from a given set and thence to proceed in a similar fashion, using the initial set of factors, for as long as one pleased. While the exclusion of perfectly valid trading prices was less of an issue with the bounded linear model, it did still prove to be a major problem, so I had to switch to an unbounded linear model, with infinitely many adjustment factors. Confirming a viable distribution has been keeping me occupied on and off ever since – I just need to get around to testing.
But I see I have digressed. This is my point: I meant to use the formula below to help me determine the probability of a given path of prices, which I had thought might be useful in the option pricing. It turned out not to be (because the relevant probability is the path-independent probability of reaching a given price). Nonetheless, I still have the formula below, which I now present for your reading pleasure. Enjoy!
Hello there. And welcome. I trust this finds you well.
Now that you are here, let’s get started with some definitions.
- An integer is a whole number; a natural number is a positive integer.
- The prime factorisation of a natural number N is the one way in which to write N as a product of powers of prime numbers (numbers with only two factors: one (1) and themselves) – for any natural number N, only one such factorisation exists and re-arranging its constituents makes absolutely no difference whatsoever.
- For the purposes of this text, a “Q-length factorisation” is a way in which to write N as a product of Q natural numbers, without considering multiplicity. So the prime factorisation will ordinarily not be included in this, because in its ordinary form it is compactly written as products of powers of prime numbers – but an expanded form that repeats the prime number every time it appears will be included.
- The factorial of N is only defined for non-negative integers and is the product of every natural number less than or equal to N and is denoted herein by N! Due to the mathematical convention for an empty product (multiplying over zero factors), the factorial of zero equals one.
- The capital pi notation indicates an iterated product, just like the capital sigma notation indicates an iterated sum.
Statement
Let N be a natural number that is the product of V prime numbers, whose prime factorisation yields T factors.
Then the number of Q-length (not necessarily prime) factorisations of N depends on T and V and is given by the following formula:
[Note: T0 = 1; Ti is greater than 1 for every i greater than 0; V is a natural number; T is a natural number; Q is a natural number; Ti gives the number of factors of N that come from the ith prime number in the prime factorisation of N; T1 * T2 * … * TV-1 * TV = T.]
Proof
The proof will be by induction on V, Q and T.
For the sake of completeness, we begin with the trivial case. Let V equal zero.
If V equals zero, then N has no primes in its factorisation. As a result, N is neither prime nor composite because N only has one factor, which means that N equals one. Therefore N only has one factorisation of any length Q.
Now let V equal 1, which is equivalent to letting N have one prime in its factorisation.
Now that V = 1, we want to prove that the number of Q-length factorisations N has depends on T and is given by the formula
We begin with the trivial case, i.e. T = 2. N is prime, in this case. As such, it has Q factorisations of length Q.
Let the base case be that T = 4.
As an aside, it is easily proven that when T = 3, N is the square of a prime number and – more importantly –
Given V = 1 and T = 4, N has one 1-length factorisation, four 2-length factorisations, ten 3-length factorisations and twenty 4-length factorisations. The emerging pattern is clearly the sequence of sums of the first K triangular numbers (the Kth term is the sum of the first K triangular numbers [K is a natural number]). Therefore,
Our inductive hypothesis is therefore the following: there exists a W > 4 such that, given V = 1, when T = W + 1,
Assume the restricted statement is true for T = W; i.e.
Let T = W + 1. The following numbers give the number of 1-length, 2-length, 3-length and 4-length factorisations of N when N has one prime in its factorisation and W + 1 factors.
Suppose that Q= G > 4. When Q increases by one, the number of factorisations increases by a scale factor of
When Q= G, the number of G-length factorisations of N equals
We have proven that when V = 1, the number of Q-length factorisations of N, which has T factors, is given by
Now let V = 2. There are two distinct primes in the factorisation of N, so N has at least four factors. (Also note, the number of factors T now features in the form T1T2 where T1T2 = T.)
We wish to prove that when V = 2,
Let the base case be that T = 4. Then N = P1P2.
A quick pen-and-paper test tells us the following.
The emergent pattern here is clearly the sequence of square numbers (the Kth term is simply K2 [K is a natural number]). Therefore,
Our inductive hypothesis is that there exists an X > 4 such that, given V = 2, when T = X + 1 = x1x2,
Assume the restricted statement is true for T = X; i.e.
Let T = X + 1. The following numbers give the number of 1-length, 2-length, 3-length and 4-length factorisations of N when N has two primes in its factorisation and X + 1 factors.
Suppose that Q= H > 4. When Q increases by one, the number of factorisations increases by a scale factor of
When Q= H, the number of H-length factorisations of N equals
We have proven the following with respect to V = 1 and V = 2:
Assume now that the main statement is true for V = Y > 2, i.e.
Now let V = Y + 1. We wish to prove that when V = Y + 1,
The following numbers give the number of 1-length, 2-length, 3-length and 4-length factorisations of N when N has Y + 1 primes in its factorisation and T factors.
Suppose that Q = I > 4. When Q increases by one, the number of factorisations increases by a scale factor of
When Q = I, the number of I-length factorisations of N equals
QED