With a Little Help from My Friends

Is collective investing superior to individual investing? Let's use mathematics to find out!

To begin with: what do we mean by "collective investing"? In short, it is when multiple investment accounts pool their resources to incrementally help one account to grow at a time.

So for collective investing, accounts #2 to #5 give all proceeds to account #1 until it has reached a threshold. All accounts other than #2 then collectively give proceeds to #2 until it has reached the same threshold. And so on and so forth, until all 5 accounts have reached the threshold.

This contrasts with individual investing, whereby each account is on its own, growing singularly and contributing proceeds to none of the other accounts.

We will first mathematically investigate the collective scenario, determining how long it takes, before using that same timespan in the individual scenario to calculate what if any differences there are.

There are several mathematical principles we must first address before the calculations:

    - an equation f(x) is the output of the variable x. So for instance, f(x) = x is a straight line, while f(x) = 1/x is a curving line that asymptotically approaches zero over time.

    - a derivative f'(x) is the pace at which f(x) changes over the course of the variable x. The derivative of f(x) = x is f'(x) = 1, because the straight line changes by a fixed value over time.

    - a differential equation is the relationship between f(x) and f'(x), which can be used to arrive at the formula for f(x). So f'(x) = f(x)² results in f(x) = -1/x as a solution.

With that out of the way, we can begin our collective investment scenario.

There are 5 accounts, each of which start with $1000. Accounts 2, 3, 4 and 5 give their proceeds to account 1 until it reaches $2000. Then accounts 1, 3, 4 and 5 give their proceeds to account 2 until it too reaches $2000. And so on and so forth until all accounts have reached $2000.

Whatever the balance of an account, it will generate 10% return on its value per unit of time. This implies exponential growth, with compounding value over time.

So how do we represent such a relationship? To begin with, account 1's own growth is represented by the differential equation: f₁'(t) = 0.1 * f₁(t) + 4*100. The "4*100" represents the 10% contributions of accounts 2 through 5.

The solution to that differential equation is f₁(t) = e^{((t+(10*ln(500)))/10)} * 10 - 4000. This represents the growth function of account 1.

Now we must figure out how long it takes from account 1 to reach $2000. We do this by setting f₁(t) to 2000, and solving the equation for "t", which we will denote as T₁.

2000 = e^{((T₁+(10*ln(500)))/10)} * 10 - 4000 is the equation, and solving it results in T₁ = 1.8232155679395462621171802515451.

One account down, and 4 more to go.

Account 2 has a slightly different differential equation, owing to the fact that account 1 now has $2000 in it. Its equation is f₂'(t) = 0.1 * f₂(t) + 0.1*2000 + 300. The "0.1*2000" represents account 1's 10% contribution, while the "300" are accounts 3 through 5.

The solution is f₂(t) = e^{((t+(10*ln(600)))/10)} * 10 - 5000. This represents the growth function of account 2.

Again, we calculate how long it takes for account 2 to reach $2000. This is done by solving the equation 2000 = e^{((T₂+(10*ln(600)))/10)} * 10 - 5000. It reveals that T₂ = 1.5415067982725830429287538506248.

Hopefully you are getting the hang of it. Account 3's differential equation is f₃'(t) = 0.1 * f₃(t) + 0.1*4000 + 200. The "0.1*4000" represents accounts 1 and 2, while the "200" represents accounts 4 and 5.

Solving results in f₃(t) = e^{((t+(10*ln(700)))/10)} * 10 - 6000. Setting it equal to 2000 and substituting in T₃ before solving reveals that T₃ = 1.3353139262452262314634362093135.

Account 4 is f₄'(t) = 0.1 * f₄(t) + 0.1*6000 + 100, where "0.1*6000" are accounts 1 through 3, and "100" is account 5.

Solving results in f₄(t) = f₄(t) = e^{((t+(10*ln(800)))/10)} * 10 - 7000, and setting it equal to 2000 before substituting in T₄ and solving results in T₄ = 1.1778303565638345453879410947052.

Finally, account 5's differential equation is f₅'(t) = 0.1 * f₄(t) + 0.1*8000, where "0.1*8000" are the previous four accounts.

Solving it, we arrive at f₅(t) = e^{((t+(10*ln(900)))/10)} * 10 - 8000. Setting it equal to 2000 and substituting in T₅ we obtain T₅ = 1.0536051565782630122750098083931.

The hardest part is now over. The total time of the entire procedure is obtained by adding all the times together.

T = T₁ + T₂ + T₃ + T₄ + T₅ = 6.9314718055994530941723212145817

This is how long it takes for our collective investment scenario.

But how long for individual investment? Thankfully, this is less complicated to figure out. We must simply come up with the growth equations, and plug in our value for T.

Each individual account can be represented by the differential equation g(t)'= 0.1*g(t). Because they all grow at the same pace, we only need concern ourselves with this equation.

Solving it results in g(t) = e^{(0.1*t + ln(1000))}. Again, setting it equal to 2000 and solving for T, we obtain T = 6.9314718055994530941723212145818

You may have noticed that this is precisely the same value as our collective investing scenario. This tells us that collective investment yields no advantage whatsoever over individual investment.

And that is it! If you have made it to the end, it is my hope that you learned something about mathematics. Have a good one!