But say when you reinvest you have to a tax 4% on the stock you already hold.

That should have an impact on when you invest, because the more frequently you buy, the more tax you have to paid. For example, if you buy after the 1st year and buy again after the second year, you have to pay roughly $8 in tax but if you wait to buy after the second year you only pay $4

I have done an excel simulation and it seems the best time to wait to reinvest is about 3 years.

The question is:

1. is there a way to solve the question analytically with a formula?

2. how about if the growth is continuous?

Continuous grow uses natural log.

Quote:deedubbsI'm a bit confused because stocks pay dividends, not interest. Semantics aside, ceteris paribus, the value of a stock will fall by the amount of the dividend. If $100 in stock pays $10 in dividends, then the stock will trade at $90, but it sounds like you are assuming that the stock pays 10% in dividends, and appreciates at 11.11% per year. The falling stock price lowers taxes in future years. You are also ignoring the cash drag caused by not holding the dividends in stock.

Continuous grow uses natural log.

Its a simulation. Let's call it "investment" instead of "stock" if you prefer.

And let's ignore the ex-dividend, script option blablabla. The investment pays 10% in cash, and if you want to reinvest you have to pay 4% tax on the book value of your holdings.

Anyway, I'd think if you had $100 and at the end of the year you got paid $10, wouldn't it just cost $0.40 to re-invest that $10. Then at the end of the second year, you'd have $110 + 10% of $110 = $110 + $11 = $121, so it'd cost 4% of $11 to reinvest that $11 which is $0.44?

Or are you saying at the end of the first year, you either have the option to keep $100 invested for free and pocket $10....or you can invest the entire $110 for a 4% of $110 fee (ie: You'd pay a $4.40 fee to invest $110, but it's be free to keep $100 invested)?

THE WHOLE THING CONFUSES ME

Quote:RS

Anyway, I'd think if you had $100 and at the end of the year you got paid $10, wouldn't it just cost $0.40 to re-invest that $10. Then at the end of the second year, you'd have $110 + 10% of $110 = $110 + $11 = $121, so it'd cost 4% of $11 to reinvest that $11 which is $0.44?

Can't blame you for being confused. What you said is what NORMALLY happens in the real world.

But no, thats not what happens in my question. In my question, you pay 4% on the book value of your HOLDINGS

ie in the end of first year if you reinvest, you pay 100 = 4, and if you reinvest again at the end of the 2nd yr that would be 4% * 106 = 4.24.

but if you don't reinvest in the first year and wait until the end of the 2nd year to reinvest, you would have received 20 in interest and you only have to pay 4 (because your HOLDINGS is still only 100)

Quote:andysifCan't blame you for being confused. What you said is what NORMALLY happens in the real world.

But no, thats not what happens in my question. In my question, you pay 4% on the book value of your HOLDINGS

ie in the end of first year if you reinvest, you pay 100 = 4, and if you reinvest again at the end of the 2nd yr that would be 4% * 106 = 4.24.

but if you don't reinvest in the first year and wait until the end of the 2nd year to reinvest, you would have received 20 in interest and you only have to pay 4 (because your HOLDINGS is still only 100)

it is confusing. But I am pretty sure you have something wrong.

>you pay 4% on the book value of your HOLDINGS

that's rare, at the moment I can only think of personal property and real estate taxes that do that [maybe in Communist countries they tax CD, stock, and bond holdings!]

>if you reinvest again at the end of the 2nd yr

you are also citing an uncommon tax trigger, usually you are taxed on the dividends when they are issued, and taxed on the capital gains when realized. Re-investment is not a tax trigger in stocks etc. Something issued to re-invest in other things? Rent? there's no delay in taxing that

so, I have to ask what this exercise is about? If you are anticipating a change in tax laws maybe you will be right to hold off on your re-investments in that future instance. Or are there some investments I havent thought of that can involve both taxing holdings and having an untaxed income stream too?

Quote:andysif

The question is:

1. is there a way to solve the question analytically with a formula?

If the holding period is N years:

P(0) = 100

P(N) = 100 * 0.96 * (1 + 0.1 * N)

P(2N) = (100 * 0.96 * (1 + 0.1 * N)) * 0.96 * (1 + 0.1 * N)..

P(KN) = 100 * 0.96

^{K}* (1 + 0.1 N)

^{K}

= 100 (0.96 + 0.096 N)

^{K}

Assume you want to know what is the strategy to maximize your value after T years:

P(T) = 100 (0.96 + 0.096 N)

^{(T/N)}

= 100 ((0.96 + 0.096 N)

^{(1/N)})

^{T}

Find N such that (0.96 + 0.096 N)^(1/N) is a maximum

d((0.96 + 0.096 N)

^{(1/N)})

= d(exp (1/N * ln (0.96 + 0.096 N)))

= exp (1/N * ln (0.96 + 0.096 N)) * d(1/N * ln (0.96 + 0.096 N))

= (0.96 + 0.096 N)^(1/N) * d(ln (0.96 + 0.096 N) / N)

= (0.96 + 0.096 N)^(1/N) * (N * 0.096/(0.96 + 0.096 N) - ln (0.96 + 0.096 N)) / N

^{2}

= (0.96 + 0.096 N)^(1/N) * (N / (10 + N) - ln (0.96 + 0.096 N)) / N

^{2}

This is 0 when N / (10 + N) = ln (0.96 + 0.096 N)

-> exp(N / (N + 10)) = 0.096 * (10 + N)

N = about 3.499

Whether 3 or 4 is better might depend on T, although you will notice that T is irrelevant when calculating the best N.

This reminds me of the Googleplex Minus 1 (aka Googol Digits of Pi) problem: you want to print a number with 10

^{100}digits (Googleplex - 1 has this many digits, all of which are 9s), but printer speeds increase exponentially with time (I use a variation of Moore's Law, and assume that if you wait time T, the speed will increase by a factor of 2

^{T / (5x10^7) seconds}(so speeds will double about every 19 months)); however, if you want too long, the Law of Diminishing Returns takes effect (if the speed is up to 3 years, you don't wait 19 months to save 18 months), so how long do you wait before starting?

(Related problem: suppose you want to save the number to a quantum hard drive, in binary, where each bit is represented by an atom; if the hard drive is a sphere, what is its minimum possible radius?)

Quote:odiousgambitit is confusing. But I am pretty sure you have something wrong.

>you pay 4% on the book value of your HOLDINGS

that's rare, at the moment I can only think of personal property and real estate taxes that do that [maybe in Communist countries they tax CD, stock, and bond holdings!]

>if you reinvest again at the end of the 2nd yr

you are also citing an uncommon tax trigger, usually you are taxed on the dividends when they are issued, and taxed on the capital gains when realized. Re-investment is not a tax trigger in stocks etc. Something issued to re-invest in other things? Rent? there's no delay in taxing that

so, I have to ask what this exercise is about? If you are anticipating a change in tax laws maybe you will be right to hold off on your re-investments in that future instance. Or are there some investments I havent thought of that can involve both taxing holdings and having an untaxed income stream too?

OK, don't kill me over this, but it's from a game so they can make up any funny rules, not related to anything real world. and apparently their rule is x% return periodically, and y% tax on your HOLDINGS when you decide to reinvest.

but nonetheless this makes an interesting math problem. if the tax is on the part of the reinvestment only, it wouldn't be anything interesting because whatever the tax rate is, you would still want to reinvest as soon as you got your money. Only when it is on the holdings the question become interesting. wait too long and you lose the power of compounding, buy too often and you pay too much tax.

Quote:ThatDonGuy

If the holding period is N years:

P(0) = 100

P(N) = 100 * 0.96 * (1 + 0.1 * N)

P(2N) = (100 * 0.96 * (1 + 0.1 * N)) * 0.96 * (1 + 0.1 * N)..

P(KN) = 100 * 0.96^{K}* (1 + 0.1 N)^{K}

= 100 (0.96 + 0.096 N)^{K}

Assume you want to know what is the strategy to maximize your value after T years:

P(T) = 100 (0.96 + 0.096 N)^{(T/N)}

= 100 ((0.96 + 0.096 N)^{(1/N)})^{T}

Find N such that (0.96 + 0.096 N)^(1/N) is a maximum

d((0.96 + 0.096 N)^{(1/N)})

= d(exp (1/N * ln (0.96 + 0.096 N)))

= exp (1/N * ln (0.96 + 0.096 N)) * d(1/N * ln (0.96 + 0.096 N))

= (0.96 + 0.096 N)^(1/N) * d(ln (0.96 + 0.096 N) / N)

= (0.96 + 0.096 N)^(1/N) * (N * 0.096/(0.96 + 0.096 N) - ln (0.96 + 0.096 N)) / N^{2}

= (0.96 + 0.096 N)^(1/N) * (N / (10 + N) - ln (0.96 + 0.096 N)) / N^{2}

This is 0 when N / (10 + N) = ln (0.96 + 0.096 N)

-> exp(N / (N + 10)) = 0.096 * (10 + N)

N = about 3.499

Whether 3 or 4 is better might depend on T, although you will notice that T is irrelevant when calculating the best N.

This reminds me of the Googleplex Minus 1 (aka Googol Digits of Pi) problem: you want to print a number with 10^{100}digits (Googleplex - 1 has this many digits, all of which are 9s), but printer speeds increase exponentially with time (I use a variation of Moore's Law, and assume that if you wait time T, the speed will increase by a factor of 2^{T / (5x10^7) seconds}(so speeds will double about every 19 months)); however, if you want too long, the Law of Diminishing Returns takes effect (if the speed is up to 3 years, you don't wait 19 months to save 18 months), so how long do you wait before starting?

(Related problem: suppose you want to save the number to a quantum hard drive, in binary, where each bit is represented by an atom; if the hard drive is a sphere, what is its minimum possible radius?)

thanks. this looks like the solution but i will have to take some time to digest it.

PS. i have absolutely no idea how your "related problem" is related to the original question. LOL

Quote:andysif

PS. i have absolutely no idea how your "related problem" is related to the original question. LOL

That's because it's not - it's related to the Googleplex Minus 1 problem, which in turn is related to the original problem in that both are "how long do you wait before waiting takes longer/costs more than any time saved/money earned by waiting".

It's like how, Alec Baldwin's brother Stephen is related to Alec and Kim Basinger's daughter Ireland (of "that voicemail message" fame), who is also related to Kim Basinger's sister Ashley, but Stephen and Ashley are not related to each other.