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I'm working with some survival probabilities, and i've gotten kind of confused.

For simplicity, the death rate ( mu(x) ), where x is age in years, are considered constant in intervals of 1 year. In my assignment, the probability of surviving 1 year at age x is given as:

p(x,1) = e^(-integral[0,1] mu(x,1) ds). Since mu is constant in intervals of a year, then just e^( -mu(x,1) ).

But obviously, since mu(x,1) is constant in this one year, then the probability of surviving should also be 1 - mu(x,1)? But 1-x <= e^(-x).

What is the flaw in my calculation here?

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You say the death rate is $\mu(x)$ where $x$ is age in years and that $\mu(x)$ is constant intervals of years. If we modify this statement and say that $\mu(x)$ is the probability density of death, then you are correct that $\mu(x)$ is also the 1-year death probability for a subject with age $x$. It would not be true that a survival probability is $\exp(-x)$. Note: a probability density is akin to the height of a normal "bell curve", and the area under the curve is considered a probability.

Another type of density that characterizes survival is the exponential with rate parameter $\lambda$. Exponential RVs are unique because they have a constant rate unlike Weibull. For exponentially distributed failure times, and if $\lambda=1$ then the survival probability beyond time $x$ is $\exp(-x)$.

One scenario that might characterize your confusion is if a survival regression model is used to model the $\lambda$ as a function of one or more regressors. In this case, you might use $\mu(x)$ to represent the model for the rate or intensity of the survival process, and age is just such an $x$ in the model, the 1 year survival probability for a subject with age $x$ would be $\exp(-\mu(x))$.

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  • $\begingroup$ Thank you for the answer. I get that the probability of death is the area under the density. But in the case of mu being constant over a 1 year period, then the intergral over that 1 year, must simply be mu itself. So the probability of survival is 1-mu. I am a bit confused about the answer, though i appreciate it! Are you telling me that it is because the assignment assumes the probability of survival follows a different distribution? Thanks in advance! $\endgroup$ Commented Jun 16, 2023 at 12:52

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