www.lesswrong.com/posts/RrL7xqdPycGNHQkXR/the-lethal-reality-hypothesis
2 corrections found
You would be in the first 0.001% or less of all humans who will ever live.
The percentage is off by orders of magnitude. If humanity eventually numbers in the trillions, being roughly the 100–117 billionth human means being in the first ~10% of a 1-trillion total or ~1.2% of a 10-trillion total, not 0.001% or less.
Full reasoning
The article itself frames the comparison as "trillions of people". But the cited birth rank is about the 100-billionth human ever born, and current demographic estimates are closer to 117 billion total births so far.
Using that scale:
- If total humans ever = 1 trillion, then 117 billion is 11.7% of all humans.
- If total humans ever = 10 trillion, then 117 billion is 1.17% of all humans.
- To get down to 0.001%, total humans would need to be on the order of 10 quadrillion, not merely "trillions."
So the sentence's numerical conclusion does not follow from its own premise. It understates the percentile by roughly three to four orders of magnitude.
1 source
- How Many People Have Ever Lived on Earth? | PRB
This semi-scientific approach yields an estimate of about 117 billion births since the dawn of modern humankind. ... Given a current global population of about 8 billion, the estimated 117 billion total births means that those alive in 2022 represent nearly 7% of the total number of people who have ever lived.
For the simplest case, geometric Brownian motion (a multiplicative process with Gaussian log-returns of mean μ and variance σ²), there is an exact and famous result: the time-average growth rate of a single path is not μ but μ − ½σ².
This mixes up two different meanings of μ in geometric Brownian motion. In standard GBM, log returns have mean (μ − ½σ²)Δt, not μΔt; if log returns really had mean μ, then the time-average log-growth rate would be μ, not μ − ½σ².
Full reasoning
In standard geometric Brownian motion,
[
\frac{dP_t}{P_t}=\mu,dt+\sigma,dW_t,
]
Itô's lemma gives
[
d\ln P_t = \left(\mu - \tfrac{1}{2}\sigma^2\right)dt + \sigma dW_t.
]
So over an interval (\Delta t), the log return is normally distributed with mean ((\mu - \tfrac{1}{2}\sigma^2)\Delta t) and variance (\sigma^2 \Delta t).
That means the sentence in the post conflates two different quantities:
- (\mu) as the drift parameter of the price process (dP_t/P_t), and
- the mean of log returns, which in GBM is (\mu - \tfrac{1}{2}\sigma^2), not (\mu).
So the statement is mathematically inconsistent as written. Either:
- if the process is standard GBM with drift (\mu), then log returns do not have mean (\mu); or
- if log returns really do have mean (\mu), then the time-average log-growth rate is (\mu), not (\mu - \tfrac{1}{2}\sigma^2).
The familiar (-\tfrac{1}{2}\sigma^2) correction comes from converting from arithmetic-return drift to log-return drift via Itô's lemma; it is not a subtraction from an already-defined mean of log returns.
2 sources
- Stochastic Calculus | MIT OpenCourseWare
Geometric Brownian Motion (continued) ... Log return on (t, t + ∆t]: ln(Pt+∆t/Pt) ∼ N([µ − σ²/2](∆t), σ²∆t).
- Essentials of Diffusion Processes | University of Illinois
Thus, F = ln x follows arithmetic Brownian motion. Since we know that F(T) − F(0) ∼ N((µ − 1/2σ²)T, σ²T) ... then x(t) = e^{F(t)} has a lognormal distribution over any discrete interval.