Imagine you have a large sum of money $M$. You decide to put this large sum of money into the stock market and, as the careful investor that you are, you decide to invest it in something low risk for a small return rate of $r$.

Of course, one of your life goals is never to have to work this crappy job again. You ask to yourself "would it be possible for me to live off $M$, and still grow my money?" In other words, you might take out some fixed amount at regular intervals, say $m$, and hope that even if your cash pile gets smaller, compound interest would have you covered.

Disclaimer. This is not financial advice. However I do think that this question, and its answers can be insightful (or at the very least, interesting.) There are still many things that this model does not consider, and it's important to be aware of these things too (see the section on shortcomings.)

All models are wrong, but some are useful. George E. P. Box

Thinking in Periods

Until this point, I've been trying really hard not to use the phrase "per month", or "per year." This is very intentional, because it doesn't matter at all! All that matters is that we work in terms of some regular period (every month, every year, every 10 days, every 6 minutes, you decide!) As long as $r$ and $m$ are both expressed in terms of that period, the math will work out.

Modeling the problem

Let's make the scenario abundantly clear: you have an initial sum $M$, every period, you take out $m$ from it and the result accrues by $r$, also every period.

Sidenote. Let's turn our attention to $r$ for a minute. Understanding its value will be very important throughout. Suppose that $M$ accrues by 5% each period. Then $r$ should be $1.05$ (not $5$, not $0.05$, but $1.05$ !) Why? Well suppose that you put in $100, and those $100 accrue 5% per period. Then, at the end of one period, you'll have $105, but that's equal to $100 \times 1.05$. If our rate were 12%, $r$ would be $1.12$. If it were 100%, $r$ would be $2$. If it were 1337%, $r$ would be $14.37$. In general, $r$ your percentage number divided by 100, plus one. I really recommend making sure that this makes sense before moving on.

Ok, now we're ready to model the problem. Let's take a look at what happens each period.

• At first, you have $M - m$. This accrues by $r$, so we get $(M - m)r$. This is how much you have at the end of the first period.

• Of course, when the second period rolls around, you come back to take $m$ out, so you have $(M - m)r$ (from the first period) minus $m$

  $$\underbrace{(M - m)r}_{\text{Result of the first period}} - m$$

  This all accrues by $r$ of course, so by the end of the second period, you have $((M - m)r - m)r$

Now we can see the pattern: each period, you take out $m$, and the result accrues by $r$. In general, we have

$$\cdots (((M - m)r - m)r - m)r \cdots$$

This kind of nesting is a pain to write, so let's rewrite it differently. First, notice that the $M$ on the very inside is multiplied by all the $r$s. Secondly, notice that the most nested $m$ is also multiplied by all the $r$s, and that the second most nested $m$ is multiplied by all the $r$ except one, the third by all the $r$s except two, and so on...

Now suppose that we do this whole process for $k$ periods. Then, we can rewrite the expression in this form

$$\begin{align*} & (M \times \underbrace{r \times r \times \cdots \times r}_{k \text{ times}}) - (m \underbrace{\times r \times r \times \cdots \times r}_{k \text{ times}}) - (m \times \underbrace{r \times r \times \cdots \times r}_{k - 1 \text{ times}}) - \cdots \\ = & Mr^k - mr^k - mr^{k - 1} - mr^{k - 2} - \cdots - mr^1 - mr^0 \\ = & Mr^k - m(r^k + r^{k - 1} + \cdots + r^1 + r^0) \end{align*}$$

This is already much easier to write! But where to go from here? If you've never seen it before I don't blame you, but we can actually simplify $r^k + \cdots + r^0$. This is called a geometric sum and in general, we have

$$r^k + r^{k - 1} + \cdots + r^0 = \frac{1 - r^{k + 1}}{1 - r}$$

Why this is equation is true is not obvious and beyond the scope of this article, but you can just trust me that it is (or try it for, say $k = 3$ and $r = 2$ !)

With this newfound knowledge descended to us from the heavens above, we can rewrite our form from earlier!

$$\begin{align*} = & Mr^k - m(r^k + r^{k - 1} + \cdots + r^1 + r^0) \\ = & Mr^k - m \left(\frac{1 - r^{k + 1}}{1 - r}\right) \end{align*}$$

That's it! We've now modeled the problem and massaged it into a form which is not even much of a pain to write out. At this point, it's worth stopping to mention that, you could stop right here, plug in your numbers, and figure out how much money you'd have. If that's you, let me at least reiterate what everything above is saying, so that you can be sure you did it right.

• $M$ is your initial large sum of money that you decide to invest. Maybe this is $50,000. Maybe it's $100,000. Maybe it's $500,000. You can decide this for yourself.

• $r$ is the accruing rate over a period. For example, you might have heard that the S&P 500 returns on average, around 8% per year. It's important for $r$ to be written in terms of the period you are going for! In the example of the S&P 500 above, 8% per year does not mean 8% per month (I mean could you imagine??) Instead, to get the monthly interest rate, you need to divide it equally into each month (12 months in a year) and get a month interest rate of $8/12 \approx 0.67\%$ per month. Already, a lot more believable.

• $k$ is the number of periods. For instance, if you're working in months and want to calculate for $2$ years, that would be $k = 2 \times 12 = 24$. If you're working in years and want to find out for $10$ years, $k = 10$. Just make sure your $r$ value is correctly adjusted for the period you're working in (see bullet above.)

• Finally, what the whole expression is telling you is how much money you will have at the end of $k$ periods by following this process.

How do I know I'm not slowly going bankrupt?

In the previous section, I told you that we had completely modeled the problem. While this is true, there are still some questions we haven't answered. We know at the end of each period how much money we'll have, but what if that number is going down over more and more periods? (Spoiler alert: that does mean you're going bankrupt.)

It's probably important for us to figure out exactly when it does go down so we can avoid it.

A crude way of doing this is to just plug in your numbers for larger and larger values of $k$. If you see the number going down as $k$ gets larger, that probably means you're taking too much money out each period.

Of course, we can be a lot more precise with the right math. This rate of change problem calls for some calculus! (Please don't be scared)

Remember our expression from the previous section? Let's call it $f$, its a function (of a couple variables) that tells us how much money we'll have after $k$ periods.

$$f(M, m, r, k) = Mr^k - m \left(\frac{1 - r^{k + 1}}{1 - r}\right)$$

The question we're interested in is: does $f$ go up as $k$ goes up? That is, if we chose values for everything else ($M$, $m$, and $r$) and we change $k$, does our money go up or down?

Some of you might recognize this as the derivative of $f$, with respect to $k$. If you don't recognize this, it just means what I said above, and we write it like this

$$\frac{df}{dk} = \text{The change of $f$ {\it with respect} to $k$}$$

Our goal of course is for $f$ to go up as $k$ goes up. If you've never taken a partial derivative before, don't worry, I'll do it for you.

First, we massage $f$ a little bit

$$\begin{align*} f(M, m, r, k) &= Mr^k - m \left(\frac{1 - r^{k + 1}}{1 - r}\right) \\ &= Mr^k - \frac{m}{1 - r} + \left(\frac{m}{1 - r}\right) r^{k + 1} \end{align*}$$

Now, we take the derivative with respect to $k$.

$$\frac{df}{dk} = M r^k \ln(r) + \left(\frac{m}{1 - r}\right)r^{k + 1} \ln(r)$$

If this rate of change is positive, we're making money! But where to go from here? Well, we know that we have no control over $r$ (we could make riskier investments of course, but let's not do that) Instead, let's ask the following question: "How big can I make $m$ in proportion with $M$, without losing money?" In other words, how much of this snowball can I steal from and still have my snowball grow? What we're really looking for here is a ratio of $M$ and $m$, so let's solve for that!

$$\begin{align*} M r^k \ln(r) + \left(\frac{m}{1 - r}\right)r^{k + 1} \ln(r) &> 0 \\ M r^k \ln(r) &> - \left(\frac{m}{1 - r}\right)r^{k + 1} \ln(r) \\ \frac{M}{m} &> - \left(\frac{1}{1 - r}\right)\left(\frac{r^{k + 1} \ln(r)}{r^k \ln(r)}\right) \\ \frac{M}{m} &> - \left(\frac{1}{1 - r}\right)\left(\frac{r \cdot \cancel{r^k \ln(r)}}{\cancel{r^k \ln(r)}}\right) \\ \frac{M}{m} &> \frac{r}{r - 1} \end{align*}$$

Wow! Look at how simple that result is. I have to admit that when I first solved this system, I was definitely a bit surprised myself. Notice how $k$ is completely gone! This should make sense, since changing the number of periods we do this over isn't going to make us lose more or less money: it's just going to tell us how much we've made (or lost.)

This result tells us that our ratio should be at least $r / (r - 1)$, but what does that say about $m$? If we solve for it, we can know the answer

$$\begin{align*} \frac{M}{m} &> \frac{r}{r - 1} \\ m &< M \left(\frac{r - 1}{r}\right) \\ \end{align*}$$

If $m$ is less than this quantity, we're not losing money!

Shortcomings

It would be disingenuous not to mention certain things that this model fails to consider. Firstly, interest rates are hardly constant. For instance, I mentioned earlier that the S&P 500 returns on average 8% per year, but this is an average. Some years, it might be less (maybe even negative), and some it might be more. The second thing that this model fails to consider is that normal people often have unpredictable expenses. While this model assumes that a constant amount $m$ is withdrawn each period, one might not find it difficult to imagine that an unforeseen expense may blow out this budget.

There are solutions to these problems of course, one could consider overestimating for $m$, or underestimating for $r$. However none of these changes would ever truly capture the actual complexity of the problem we attempt to solve. It is up to the reader to decide how to weigh these options.