4 ms·
It's not the ratio of interest to inflation that matters, it's the difference. The easiest way to conceptualize it is to consider an inflation rate of near-0,
by gibybo 14y ago
It's not the ratio of interest to inflation that matters, it's the difference.
The easiest way to conceptualize it is to consider an inflation rate of near-0, or even negative. The ratio would imply a months salary would be sufficient FU money if you could invest it at even 1%.
A slightly more mathematical example:
Consider $100 in today's money. Invested at 10% annually, it will become 100 * 1.1^25 = $1,083 in 25 years. If inflation is 1%, it will require $100 * 1.01^25 = $128 for every $100 in today's money, thus the $1,083 is worth $846 in today's money. If inflation is 2%, it will require $164, or $660.
Despite the F/I ratio doubling, the difference is only ~22% after 25 years.
If you use ((1+I) - F), you can achieve the same $846 and $660 numbers with a very straightforward calculation: S_0 * ((1+I)-F)^N
e.g.
$100 * (1.10 - 0.01) = $846,
$100 * (1.10 - 0.02) = $660
EDIT: Also, you mention the it is very hard to get r below 0.9, but it's not really. Or at least, it hasn't been for the last 100 years (I am not able to predict the next 100 years as accurately as the past). The average inflation rate of the US Dollar for the past 100 years is roughly 3% and the average annual return of the S&P 500 is 10%. That would produce an r = 0.3, well below 0.9.
- nileshtrivedi 14y agor = F/I = (1+f)/(1+i) according to what I have defined. So inflation of 3% and return of 10% means r = 1.03 / 1.1 = 0.9363, not 0.3.
- gibybo 14y agoSorry, I missed the F = (1+f)/R = (1+r) line. Pretty sure the rest of my argument holds though. Your formula merely approximates the correct answer for low values of i and f.
- nileshtrivedi 14y agoI also left out explanation for the edge cases. For low enough values of r, savings will never decrease and never reach zero. For practical values of r, the equation is correct.
- gibybo 14y agoOk I think I've got a counter example. First of all, I'm only going to consider values where N approaches infinity && r < 1 && r > 0 to make the math easier. lim [n-> inf, r < 1, r > 0 ] of S_0 = E_0 (1-r^n)/(1-r) -> S_0 = E_0 * 1 / (1-r) I propose that the actual relationship is: S_0 = E_0 / (i-f) E_0=1,f=.07,i=.10 - Original: S_0 = 36.666..., My proposal: S_0 = 33.333... E_0=1,f=.80,i=1.00 - Original: S_0 = 10, My proposal: S_0 = 5 So it seems your formula approximates mine at low values of i and f, but diverges strongly at higher values. I believe I can show mine is the correct one with a small amount of intuition when we let f=0: If I make 10% annually in interest, then $1mil will provide $100k/year forever. The same thing more concisely: i * S_0 = E_0 r = F/I = (1+0/1+i) After substitution and division, your formula is E_0 = S_0 / ((i+1)/i) In words: If I make 10% annually in interest, then $1mil will provide $90.9k/year forever, which is intuitively false.
- nileshtrivedi 14y agoIt is not. As I mentioned in the post, you need to set aside the money for the year's expenses at the beginning of the year. $1mil provides $90.9k/year because (1m - 90.9k)*0.1 = 90.9k. Interest is received at the end of the period but we need cash in hand throughout the year to pay for the expenses.
- gibybo 14y agoAlright, so I spent another hour playing with these numbers and I think I understand it correctly now and agree with your original math. A combination of mistakes (F/I vs f/i, taking money out at the start of the year) and my imperfect intuition of inflation led me astray. Thanks for teaching me something new, and for your patience :)
- nileshtrivedi 14y agoThank you for helping me find weak points in the post and clarify it. :)