I'd like to ask the following question:
Mr. X has a certain capital (10000 Dollars). He buys one security (from am company, say) now. The present value of one security paper is 10000 Dollars, but after a while the value increases.
After 1 year the security has value 11000 Dollars. Mr. X gets 1000 Dollars paid, but still is the owner of the very security paper.
Mr. X is greedy (or just has enough / too much money to spend) and wants to re-invest the 1000 Dollars win into the same securities of the same company.
Now, since the value has increased, he can only buy 1/11 of the same / next security paper (1000 Dollars / 11 000 Dollars = 1/11).
Always when he wins 1000 Dollars, he re-invests again. Consequently, if the value of the securities increases continuously, and, in total, increases 1000 Dollars per year, the time until the next re-investment becomes shorter and shorter (since, for instance, (1+1/11) security papers win 1000 Dollars faster than 1 security paper).
When he has gained 2000 Dollars in total (so a bit less than 2 years after the beginning) he re-invests again and has now approximately (1 + 1000/11000 + 1000/12000) many security papers (indeed a bit more, because he doesn't have to wait the full second year until he has earned the second 1000 Dollars).
Question:
If Mr. Y does the same (all conditions the same), but starts with two security papers instead of one, and also re-invests after having gained 1000 Dollars in total (so the first time he re-invests is after half a year), how much more security papers does he own after x many years?
Thanks for the help.
Let $S_0$ be your starting capital. Let $S_n$ be your current capital after reinvesting the n-th payout. Let p be the (relative) gain per day.
Now, let $x_n$ be the time it takes till the next payout (in years) if your current capital is $S_n$:
(i.e. $x_n$ is the time from the $n$'th payout to the $n+1$'th payout) $$S_n\cdot p\cdot x_n = 1000 \implies x_n = \frac {1000} {p S_n}$$
Further we have $S_n = S_0 + n\cdot 1000$, and therefore $x_n = \frac {1000} {p(S_0 + n\cdot 1000)}$.
The total time till the $n$-th payout therefore is $$ \sum_{i=0}^{n-1} x_i = \sum_{i=0}^{n-1} \frac {1000} {p(S_0 + i\cdot 1000)} = \frac{1000}p \sum_{i=0}^{n-1} \frac {1} {S_0 + i\cdot 1000} = \frac{\psi\left(n+\frac{S_0}{1000}\right)}{p}-\frac{\psi \left(\frac{S_0}{1000}\right)}{p} $$ In the last line, $\psi$ denotes the digamma-function, for which holds $\sum_{i=1}^n \frac 1i = \psi(n+1)+\gamma$
We would now have to solve this for $t$, where $t$ is the time in years you're interested in, solve it for $n$, floor the result and then plug that value into the formula for $S_n$.