/ Probability

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spearing05 - on 03 Feb 2013
Another question for the font of all knowledge. I'm looking at calculating risk and having a little problem.

If I say that 1000 people are exposed to a risk every year for 10 years and 1 dies during that time what is the risk per person per year?

I'm ok with the overall risk - 1000 people x 10 years/1 death ie 1:10000 or 1 x 10^-4

How do you then work out the risk per year? Am I right in thinking it is this number times 1/10?

Thanks in advance
as646 on 03 Feb 2013
In reply to spearing05: I personally can't remember any particular formulae for risk from when I was in uni, but then I did mostly mathematical physics modules.

It seems intuitive that the odds would be the total number of fatalities over the given timeframe, divided by the amount of time elapsed (giving the total fatalities per year), divided by the sample size.

So 1/10 * 1/1000 = 1x10^-4

This is essentially what you've done, except your unit of time was 10 years as opposed to my 1, and adjusted the sample size accordingly. Makes no difference.
as646 on 03 Feb 2013
In reply to as646: Actually no, that's incorrect!
SARS on 03 Feb 2013
In reply to spearing05:

Assume a Poisson distribution for number of deaths in one year. Then this has been 1/10 and therefore assuming independence amongst years average 1 over 10 years. From that you can calculate whatever risk figure you're trying to estimate.

Why use a Poisson? Well it's an extension of Bernoulli trial - ie live or die.
DancingOnRock - on 03 Feb 2013
In reply to spearing05: Isn't it simply that 0.1 person dies per year out of 1000 people exposed so 0.1/1000 or 0.00001 or 1:10,0000
SARS on 03 Feb 2013
In reply to DancingOnRock:

Quite. I hadn't read the OP properly and didn't realise he was after per person per year probability of death. As you say, simply 1/10 / 10000
Robert Durran - on 03 Feb 2013
In reply to spearing05:
> If I say that 1000 people are exposed to a risk every year for 10 years and 1 dies during that time what is the risk per person per year?


It is approximately 1/10000=0.0001 (In 10x1000=10000 years of life one death occurs, so probability of death in one year of life is 1/10000)

However, this does not take into account that a given person cannot die of the risk more than once! (very unlikely even if it were possible, hence above approximation being good). Taking this into account gives 1- 0.999^0.1 which is about 4.5x10^(-8) bigger.
Robert Durran - on 03 Feb 2013
In reply to DancingOnRock:
> ......so 0.1/1000 or 0.00001 or 1:10,0000

Careless error: 0.1/1000 = 0.0001 or 1:10000

spearing05 - on 03 Feb 2013
In reply to spearing05: Thanks all, looks like I wad going the right way then. ;-)
tom_in_edinburgh - on 03 Feb 2013
In reply to Robert Durran:
> (In reply to spearing05)
> [...]
> However, this does not take into account that a given person cannot die of the risk more than once! (very unlikely even if it were possible, hence above approximation being good). Taking this into account gives 1- 0.999^0.1 which is about 4.5x10^(-8) bigger.

Don't you need to know when the one person died to compensate.

For example, if they died in year 1 then in 9 out of the 10 experimental years there were only 999 people exposed to the risk. But if they died on the last second of the last year of the experiment for substantially all the experimental time there were 1000 people exposed to the risk.

There's also an assumption that the risk of death is constant and doesn't depend on length of exposure.

Robert Durran - on 04 Feb 2013
In reply to tom_in_edinburgh:
> (In reply to Robert Durran)

> Don't you need to know when the one person died to compensate.
> There's also an assumption that the risk of death is constant and doesn't depend on length of exposure.

I am taking the OP to mean that the expected number of deaths among 1000 people over the ten years is 1 (I'm not sure that there is any other sensible interpretation!)

I am then assuming that any given person in any given year (at the start of which they are alive) has the same risk of dying during that year. This is what is being calculated.

Suppose required risk is p.
Then probability that a person does not die in the ten years is (1-p)^10
So probability they do die is 1-(1-p)^10
So expected number of deaths is 1000(1-(1-p)^10)=1
So 1-(1-p)^10=0.001
(1-p)^10=0.999
1-p=0.999^0.1
p=1-0.999^0.1




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