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135 points andsoitis | 1 comments | | HN request time: 0s | source
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Mistletoe ◴[] No.41848746[source]
I wonder how Ozempic will change this? I really do expect we will all be on this soon and maybe we can resume the increase in lifespan that has been stalled by obesity, lack of exercise, and processed food.
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zoklet-enjoyer ◴[] No.41848795[source]
I'm not fat or diabetic. Why would I take ozempic?
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nostrademons ◴[] No.41848912[source]
Assuming you're not on drugs or thinking of killing yourself, you're probably not in the cohort that's dragging the life expectancy stats down.

Life expectancy is a weighted average (no pun intended), and so it's unusually sensitive to outliers. People who die early drag the average down much more than people who live close to the mean life expectancy. The biggest premature killers of Americans are obesity, drugs, car accidents, and suicide. Anything that addresses one of those causes of death has an outsize effect on life expectancy. There are 100M+ obese Americans. There are about 100,000 overdose deaths per year. Obesity, while not as lethal as drugs or suicide, afflicts 1000x as many patients, and so a treatment for it can have a large effect on the numbers.

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drawkward ◴[] No.41849027[source]
>Life expectancy is a weighted average

Sure, if all the weights are 1. Where i come from, we just call that an average.

>People who die early drag the average down much more than people who live close to the mean life expectancy.

This is true of all averages where all weights are the same.

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nostrademons ◴[] No.41849150[source]
I should probably have said the change in life expectancy is a weighted average, weighted by how far you are from the average. If average life expectancy is 80, removing a data point where somebody died at 40 has 8x the effect of removing a data point where somebody died at 75.
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1. cscheid ◴[] No.41849248[source]
In case anyone else is curious about the specific term for the concept you are describing, it's https://en.wikipedia.org/wiki/Leverage_(statistics)

(To reproduce exactly the scenario being discussed, you fit a constant-only model to the data using least squares: that gives the average as the best fit. Then, you measure the leverage of each point of interest.)