Monevator is one of the personal finance sites that I read fairly frequently. They have a fairly good weekly column that includes a number of interesting links, both about personal finance as well as on other topics like psychology or recent news events. The article I read yesterday begins as follows:
Unlike every other personal finance blogger around, I’ve started 2020 wondering whether it’s time I spent more money.
I would not call myself a personal finance blogger, but I had thoughts along similar lines when planning my budget for 2020. Eventually, I decided yes, and expanded the budget at the end of the planning session (and this is already after a 50% increase over 2018). My reasoning was not too different from the author’s (and the logic behind having a need for such an argument was similar as well – saving, which is a form of delaying gratification, seems to come naturally as a default to me).
In theory at least, I can work out how much £1 spent today would cost me in retirement if I was to do that at 65 if I make some assumptions. There are 37 years to go there, and if we use a 4.5% real (that is, accounting for inflation) return that becomes about £5.10. In other words, I could buy roughly five times the number of goods I could today, if I was willing to wait 37 years.
However, a large problem with excessively delaying gratification (both for money and in general) is that the expected benefit may not materialise – this is known as an interruption risk. These risks can typically take two forms:
- Collection risk, where the projected reward doesn’t actually become available. In context this means that the £1 I invested doesn’t become the inflation-adjusted equivalent of £5.10 37 years down the road.
- Termination risk, where the person delaying gratification does not get to benefit from the projected reward, even though it did become available. In context this means that although the £1 I invested does become £5.10 (or more), I don’t survive 37 more years or I do but my ability to use it is limited or hampered for some reason.
The former could arise for various reasons, as my projection was dependent on a 4.5% real return assumption. This is in line with the past 119 years – in fact that was slightly higher at 5%, though I would use 4.5% because of platform and fund fees (see Credit Suisse’s 2019 Global Investment Returns Yearbook). This pattern may not continue going forward if there are large, value-destroying events (e.g. faster than expected climate change effects, nuclear war). Even if the pattern continues, I might not be able to get this return for more local concerns (e.g. aggressive investment taxes as part of left-wing policy, if my fund of choice turns out to be poorly managed).
The latter is also very relevant. The author mentions that he’s “within striking distance of 47” and is concerned about planning 40 years ahead when using a compound interest calculator; I’m probably between fifteen or twenty years younger, and the pessimist (or realist?) in me also gets uneasy about putting 40 in that box, let alone the analogous 55-60 (e.g. see James 4:14 – whether you’re a Christian or not I think the sentiment is relatable). Even if I do make it there, it’s possible that I’d be in less of a position to enjoy the money, for time or health reasons. If I had my current financial state at the end of National Service in Singapore, for example, that 9-month period before starting university would have looked quite different and I would have done more things, some of which I’m not in a position to do right now because of work (e.g. I’d probably have done a two-month round-the-world trip).
I am still intending to be mostly frugal and saving a fair amount. Travel will likely be the largest source of expenses for me – flights and hotels are not cheap. I also spent quite a lot of money on clothing last year, though I’m trying to avoid that as I’m running up against (reasonable) storage limitations and there are quite a number of clothes that I haven’t worn for a while. Thus my plans are nowhere near as far as the author’s claim that he’ll make little effort to grow his wealth. I don’t think I’m in a position where that would be responsible, in that my current savings/portfolio are nowhere near enough to be sufficient for retirement.
as workplace pension contributions, 
as individual pension contributions (e.g. SIPPs), 
as other savings (ISAs, taxable accounts), 
as consumption and 
as tax changes derived from taxable benefits. Furthermore, define 
as the tax rate expected when one withdraws from one’s pensions. Then
, the second most recent a weight of 
, the third 
and so on. As 
approaches 0, we approach a simple historical average; as 
.
? One could use the first observation, or perhaps an average of the first few observations if short-term divergence from reality is unacceptable.
for the first part, but in the second part things might become less obvious.
.
to be the total space on the continuum that the ducks occupy. For base cases, we define 
, and 
, we can attempt to find the range of allowable third-duck positionings such that 
. Without loss of generality, suppose the two ducks are sitting at points 0 and 
. Then, the lowest possible point the third duck can sit at would be 
, and the highest possible point 
(assuming of course 
; if this is not the case then there are clearly no possible positions). The range of this interval is 
, and the probability that a duck lands in that range is 
. This of course makes the assumption that 
and 
we don’t have to worry about this.
then that will be equal to![\displaystyle \int_{0}^{n} (2n-x) (2) \text{d}x = \left[ 4nx - x^2 \right]_{0}^{n} = 3n^2](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cint_%7B0%7D%5E%7Bn%7D+%282n-x%29+%282%29+%5Ctext%7Bd%7Dx+%3D+%5Cleft%5B+4nx+-+x%5E2+%5Cright%5D_%7B0%7D%5E%7Bn%7D+%3D+3n%5E2&bg=ffffff&fg=000&s=0&c=20201002 "\displaystyle \int_{0}^{n} (2n-x) (2) \text{d}x = \left[ 4nx - x^2 \right]_{0}^{n} = 3n^2")
of cases with three ducks. If 
, then we need to make several refinements – most obviously, the upper bound of the integral stops at 0.5, as there’s no span with two ducks larger than that. Furthermore, there may be values of 
in which case we need to clamp it down to 1. For simplicity, we focus on the case where 
, we can similarly formulate
may not seem immediately obvious, but we can rely on the CDF of 
, and the conditional probability can be handled with the same logic that we used to go from ![\displaystyle P(S_4 \leq n) = \int_{0}^{n} (2n-x) (6x) \text{d}x = 6 \int_{0}^{n} 2nx - x^2 \text{d}x = 6 \left[ nx^2 - \frac{x^3}{3} \right]_{0}^{n} = 4n^3](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+P%28S_4+%5Cleq+n%29+%3D+%5Cint_%7B0%7D%5E%7Bn%7D+%282n-x%29+%286x%29+%5Ctext%7Bd%7Dx+%3D+6+%5Cint_%7B0%7D%5E%7Bn%7D+2nx+-+x%5E2+%5Ctext%7Bd%7Dx+%3D+6+%5Cleft%5B+nx%5E2+-+%5Cfrac%7Bx%5E3%7D%7B3%7D+%5Cright%5D_%7B0%7D%5E%7Bn%7D+%3D+4n%5E3&bg=ffffff&fg=000&s=0&c=20201002 "\displaystyle P(S_4 \leq n) = \int_{0}^{n} (2n-x) (6x) \text{d}x = 6 \int_{0}^{n} 2nx - x^2 \text{d}x = 6 \left[ nx^2 - \frac{x^3}{3} \right]_{0}^{n} = 4n^3")
seems to take on the form 
. We can prove this result by induction on 
. This clearly holds for 
, based on our work so far – though note that 
is enough for our proof. So we now take some arbitrary integer 
. We need to show that 
. The way we can do this is very similar to what we did for the concrete steps.
. We can simplify out the conditional probability term, based on a similar argument on the conditional probability as before. Thus
![\displaystyle P(S_{k+1} \leq n) = k(k-1) \int_{0}^{n} 2n x^{k-2}- x^{k-1} \text{d}x = k(k-1) \left[ \frac{2n x^{k-1}}{k-1} - \frac{x^k}{k} \right]_{0}^{n}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+P%28S_%7Bk%2B1%7D+%5Cleq+n%29+%3D+k%28k-1%29+%5Cint_%7B0%7D%5E%7Bn%7D+2n+x%5E%7Bk-2%7D-+x%5E%7Bk-1%7D+%5Ctext%7Bd%7Dx+%3D+k%28k-1%29+%5Cleft%5B+%5Cfrac%7B2n+x%5E%7Bk-1%7D%7D%7Bk-1%7D+-+%5Cfrac%7Bx%5Ek%7D%7Bk%7D+%5Cright%5D_%7B0%7D%5E%7Bn%7D&bg=ffffff&fg=000&s=0&c=20201002 "\displaystyle P(S_{k+1} \leq n) = k(k-1) \int_{0}^{n} 2n x^{k-2}- x^{k-1} \text{d}x = k(k-1) \left[ \frac{2n x^{k-1}}{k-1} - \frac{x^k}{k} \right]_{0}^{n}")
![\displaystyle P(S_{k+1} \leq n) = \left[ 2nk x^{k-1} - (k-1)x^k \right]_{0}^{n} = 2kn^k - (k-1)n^k = (k+1)n^k](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+P%28S_%7Bk%2B1%7D+%5Cleq+n%29+%3D+%5Cleft%5B+2nk+x%5E%7Bk-1%7D+-+%28k-1%29x%5Ek+%5Cright%5D_%7B0%7D%5E%7Bn%7D+%3D+2kn%5Ek+-+%28k-1%29n%5Ek+%3D+%28k%2B1%29n%5Ek&bg=ffffff&fg=000&s=0&c=20201002 "\displaystyle P(S_{k+1} \leq n) = \left[ 2nk x^{k-1} - (k-1)x^k \right]_{0}^{n} = 2kn^k - (k-1)n^k = (k+1)n^k")
for 
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