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#1 2022-06-17 01:25:58

Kongus
Member
Registered: 2022-06-17
Posts: 1

Geometric sequence - finding the formula(s) and convergence

Hi All,

I have a problem as follows, and was looking for some help:


There are two assets A and B, of unit price

and
respectively, which are traded on the open market with a price ratio of:

The instantaneous price ratio at any time is determined by the market but is always between these bounds. i.e. we always have:

Let's suppose I start by depositing an initial number of units of asset A into an account,

, having a value of 

Every time I deposit a number of units of asset A into the account, I am gifted an equal number of units of asset B,

, having a value of:

Where:

I am free to withdraw asset B ONLY, and may trade it on the market in return for an equal-valued amount of asset A. In other words, I can sell all of my asset

, and receive more asset A, of the same value:

I may then deposit the new amount of asset A,

, back into the account, where I will again be gifted an equal amount of units of asset B, which again I may withdraw trade, and re-deposit. I may do this as many times as I like, let's say N times, to increase my amount of asset A in the account from what I initially deposited.

For this problem, to simplify, let's suppose we have a price ratio:

that is CONSTANT and does not change each time we do this.

Note also here that unless

, then we always have
and therefore
. So to simplify again let's suppose
, then we can see that if I repeat this process an infinite number of times (N tends to infinity), the amount of asset A in the account will converge to some fixed amount,
, which is greater than the initial amount deposited, and the amount of asset B gifted to me each time I deposit will tend to 0. So Asset A and B will basically each have a convergent geometric sequence in terms of N.


What I want is a formula for each asset, where I can input:

, and output


Additionally I'd like to find a formula for

, given
, and I know that to get this I would have to take the limit as N tends to infinity of the formula for asset A.

I've been messing around for a while on a spreadsheet and paper, but my college maths is a bit rusty, and I can't seem to arrive at a solution that works for any arbitrary N. (Also, apologies for the formatting error in the post - LaTeX seems to add a new-line each time I insert something, and I'm not sure how to prevent it doing this?)

Many Thanks


EDIT:

I think I've worked out that the general formulas for Asset A and asset B will be:


Where:

is initial deposit of asset A
is total amount of asset A in the account after n repeats of the cycle
is amount of asset B gifted upon each deposit in the account after n repeats of the cycle

I still can't find a formula where I can input e.g. n=10, R_P =0.5, A_0 = 100, and find the amount of A a the end. (seems to require some recursive calculations - maybe will write an excel macro to calculate)

Also cannot find a formula for the value that asset A converges to when n tends to infinity. Think there is some algebraic solution to this kind of problem?

Last edited by Kongus (2022-06-17 03:46:39)

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#2 2022-06-18 02:03:42

Bob
Administrator
Registered: 2010-06-20
Posts: 10,052

Re: Geometric sequence - finding the formula(s) and convergence

hi Kongus

Welcome to the forum.

My brain has struggled a bit but I think I've got it now.  I'm calling Vn the total value of asset A after n iterations.  So it looks like

Vo = Ao.Pa
V1 = Ao.Pa  + Ao.R.Pa
V2 = Ao.Pa + Ao.R.Pa + Ao.R^2 .Pa

and so on.

So Vn = Ao.Pa .(1 + R + R^2 + R^3 + ... + R^n)

When R < 1 the bracket is a converging geometric sequence with sum to infinity (Ao.Pa)/(1-R)

The formula for this is on this page https://www.mathsisfun.com/algebra/sequ … etric.html with the sum to infinity right down the bottom of the page.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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