How old is the SECOND oldest person in the world compared to the oldest? Same for the united states?
I bogged down long before I got the answer. Below is the gibberish I generated towards bogging down.
So OK, I don’t even know offhand how old is the oldest, but I would bet it is in the 114 years old (yo) to 120 yo range.
Then figure in some hand-wavey way that people die at ages normally distributed with a mean of 75 yo. We can estimate how many sigma (standard deviations) away from that is the oldest person.
Figure there are 6 billion people now, but I know this number has grown a lot in my lifetime, it was less than 4 billion when I was born 55.95 years ago. So say the 75 yo’s come from a population consistent with 3 billion people. 1⁄2 die younger than 75, 1⁄2 die older, so the oldest person in the world is 1 in 1.5 billion on the distribution.
OK what do I know about normal distributions? Normal distribution goes as exp ( -((mean-x)/(2sigma))^2 ). So at what x is exp( -(x/2sigma)^2 ) = 1e-9? (x / 2sigma) ^ 2 = -ln ( 1e-9). How to estimate natural log of a billionth? e = 2.7 is close enough for government work to the sqrt(10). So ln(z) = 2log_10(z). Then -ln(1e-9) = −2log_10(1e-9) = 29 = 18. So (x/2sigma)^2 = 18, sqrt(18) = 4 so
So I got 1 in a billion is 4 sigma. I didn’t trust that so I looked that up, Maybe I should have trusted it, in fact 1 in a billion is (slightly more than ) 6 sigma.
mean of 75 yo, x=115 yo, x-mean = 40 years. 6 sigma is 40 years. 1 sigma=6 years.
So do I have ANYTHING yet? I am looking for dx where
exp(-((x+dx)/(2sigma))^2) - exp( -(x/2sigma)^2)
So, this isn’t quite appropriate for Fermi calculations, because the math involved is a bit intense to do in your head. But here’s how you’d actually do it:
Age-related mortality follows a Gompertz curve, which has much, much shorter tails than a normal distribution.
I’d start with order statistics. If you have a population of 5 billion people, then the expected percentile of the top person is 1-(1/10e9), and the expected percentile of the second best person is 1-(3/10e9). (Why is it a 3, instead of a 2? Because each of these expectations is in the middle of a range that’s 1/5e9, or 2/10e9, wide.)
So, the expected age* of death for the oldest person is 114.46, using the numbers from that post (and committing the sin of reporting several more significant figures), and the expected age of death for the second oldest person is 113.97. That suggests a gap of about six months between the oldest and second oldest.
* I should be clear that this is the age corresponding to the expected percentile, not the expected age, which is a more involved calculation. They should be pretty close, especially given our huge population size.
But terminal age and current age are different- it could actually be that the person with the higher terminal age is currently younger! So we would need to look at permutations and a bunch of other stuff. Let’s ignore this and assume they’ll die on the same day.
So what does it look like in reality?
The longest lived well-recorded human was 122, but note that she died less than 20 years ago. The total population whose births were well-recorded is significantly smaller than the current population, and the numbers are even more pessimistic than the 3 billion figure you get at; instead of looking at people alive in the 1870s, we need to look at the number born in the 1870s. Our model estimates she’s a 1 in 2*10^22 occurrence, which suggests our model isn’t tuned correctly. (If we replace the 10 with a 10.84, a relatively small change, her age is now the expectation for the oldest terminal age in 5 billion- but, again, she’s not out of a sample of 5 billion.)
The real gaps are here; about a year, another year, then months. (A decrease in gap size is to be expected, but it’s clear that our model is a bit off, which isn’t surprising, given that all of the coefficients were reported at 1 significant figure.)
Upvoted (among other things) for a way of determining the distribution of order statistics from an arbitrary distribution knowing those of a uniform distribution which sounds obvious in retrospect but to be honest I would never have come up with on my own.
I am looking for dx where exp(-((x+dx)/(2sigma))^2) - exp( -(x/2sigma)^2)
Assuming dx << x, this is approximated by a differential, (-xdx/sigma^2) * exp( -(x/2sigma)^2, or the relative drop of dx/sigma^2. You want it to be 1⁄2 (lost one person out of two), your x = 4 sigma, so dx=1/8 sigma, which is under a year. Of course, it’s rather optimistic to apply the normal distribution to this problem, to begin with.
How old is the SECOND oldest person in the world compared to the oldest? Same for the united states?
I bogged down long before I got the answer. Below is the gibberish I generated towards bogging down.
So OK, I don’t even know offhand how old is the oldest, but I would bet it is in the 114 years old (yo) to 120 yo range.
Then figure in some hand-wavey way that people die at ages normally distributed with a mean of 75 yo. We can estimate how many sigma (standard deviations) away from that is the oldest person.
Figure there are 6 billion people now, but I know this number has grown a lot in my lifetime, it was less than 4 billion when I was born 55.95 years ago. So say the 75 yo’s come from a population consistent with 3 billion people. 1⁄2 die younger than 75, 1⁄2 die older, so the oldest person in the world is 1 in 1.5 billion on the distribution.
OK what do I know about normal distributions? Normal distribution goes as exp ( -((mean-x)/(2sigma))^2 ). So at what x is exp( -(x/2sigma)^2 ) = 1e-9? (x / 2sigma) ^ 2 = -ln ( 1e-9). How to estimate natural log of a billionth? e = 2.7 is close enough for government work to the sqrt(10). So ln(z) = 2log_10(z). Then -ln(1e-9) = −2log_10(1e-9) = 29 = 18. So (x/2sigma)^2 = 18, sqrt(18) = 4 so
So I got 1 in a billion is 4 sigma. I didn’t trust that so I looked that up, Maybe I should have trusted it, in fact 1 in a billion is (slightly more than ) 6 sigma.
mean of 75 yo, x=115 yo, x-mean = 40 years. 6 sigma is 40 years. 1 sigma=6 years.
So do I have ANYTHING yet? I am looking for dx where exp(-((x+dx)/(2sigma))^2) - exp( -(x/2sigma)^2)
So, this isn’t quite appropriate for Fermi calculations, because the math involved is a bit intense to do in your head. But here’s how you’d actually do it:
Age-related mortality follows a Gompertz curve, which has much, much shorter tails than a normal distribution.
I’d start with order statistics. If you have a population of 5 billion people, then the expected percentile of the top person is 1-(1/10e9), and the expected percentile of the second best person is 1-(3/10e9). (Why is it a 3, instead of a 2? Because each of these expectations is in the middle of a range that’s 1/5e9, or 2/10e9, wide.)
So, the expected age* of death for the oldest person is 114.46, using the numbers from that post (and committing the sin of reporting several more significant figures), and the expected age of death for the second oldest person is 113.97. That suggests a gap of about six months between the oldest and second oldest.
* I should be clear that this is the age corresponding to the expected percentile, not the expected age, which is a more involved calculation. They should be pretty close, especially given our huge population size.
But terminal age and current age are different- it could actually be that the person with the higher terminal age is currently younger! So we would need to look at permutations and a bunch of other stuff. Let’s ignore this and assume they’ll die on the same day.
So what does it look like in reality?
The longest lived well-recorded human was 122, but note that she died less than 20 years ago. The total population whose births were well-recorded is significantly smaller than the current population, and the numbers are even more pessimistic than the 3 billion figure you get at; instead of looking at people alive in the 1870s, we need to look at the number born in the 1870s. Our model estimates she’s a 1 in 2*10^22 occurrence, which suggests our model isn’t tuned correctly. (If we replace the 10 with a 10.84, a relatively small change, her age is now the expectation for the oldest terminal age in 5 billion- but, again, she’s not out of a sample of 5 billion.)
The real gaps are here; about a year, another year, then months. (A decrease in gap size is to be expected, but it’s clear that our model is a bit off, which isn’t surprising, given that all of the coefficients were reported at 1 significant figure.)
Upvoted (among other things) for a way of determining the distribution of order statistics from an arbitrary distribution knowing those of a uniform distribution which sounds obvious in retrospect but to be honest I would never have come up with on my own.
Assuming dx << x, this is approximated by a differential, (-xdx/sigma^2) * exp( -(x/2sigma)^2, or the relative drop of dx/sigma^2. You want it to be 1⁄2 (lost one person out of two), your x = 4 sigma, so dx=1/8 sigma, which is under a year. Of course, it’s rather optimistic to apply the normal distribution to this problem, to begin with.